Newton's Waste Book (Part 2)

Author: Isaac Newton

Source: MS Add. 4004, ff. 15v-50r, Cambridge University Library, Cambridge, UK

Published online: June 2013

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Related Texts
- This document is part of MS Add. 4004
  The previous part of this document is Newton's Waste Book (Part 1) [MS Add. 4004, ff. {cover}-15r]
  The next part of this document is Newton's Waste Book (Part 3) [MS Add. 4004, ff. 50v-198v]

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How to find y^e axes vertices Diamiters, \Centers,/ or Asymptotes of any Crooked Line supposeing it have them.

^[1] Definitions. If kgach is a crooked line, & ef touch it at y^e point a . & if y^e line lines cg hk & all other lines w^ch being par{illeg}|a|lle{illeg}|l|l to one another & extended from one si{illeg}de of y^e \{illeg}/ croked line to y^e other and also bisected by y^e streight line ai . Then is ai a Diameter, & cbg one of those lines w^ch are ordinately applyed to it, & if y^e angles {illeg}, cbd , &c are right ones Then is abd t{illeg}|he| axis, & cbg one of those lines w^ch are ordinately applyed to y^e axis.

^[2] 3 The Vertex of a crooked line is y^t point where y^e crooked line intersect the diameter or axis as at $(a)$

^[3] 4 An {illeg} The \The/ Asymptote{illeg} of {illeg} crooked lines are such lines w^ch being produced both ways infinitely have noe least distance twixt y^m & y^e crooked line & yet {doe} noe where intersect it. \or touch it/ as dδ , dθ .

^[4] 5 Those lines w^ch are limited on all sides \& have axes but/ as acxkλ are Ellipses if y^e of y^e first, 2^d, 3^d, 4^th kind &c

6 Those w^ch are not ellipses & have \axes but/ noe Asyptotes are {illeg} Parabolas of y^e first, 2^d, 3^d, 4^th kind {illeg}|&|c. as zkah .

7 Those w^ch have {illeg} Asymptotes, are Hyperbolas of y^e 1^st, 2^d, 3^d 4^th kind, &c as (upon) whose asymptotes are βd , γd .

8 There are some lines of a middle nature twixt {illeg}|a| Parab: & hyperb: have an asymto{te} haveing an Asymptote for one of its s{illeg}|i|d{illeg}|e|s w^ch are but none for y^e other as βαγe , one side αγ haveing y^e asymptoe {sic} δe , y^e other side αβ haveing none.

10 If an Ellipsis have 2 axes (a{illeg}|s| am & xλ ) y^e longer is y^e transverse axis {illeg}|(|as am ) y^e shorter is y^e right axis (as xλ ).

If all those lines w^ch are parallell to one of y^e diameters of y^e crooked line & are terminated by y^e crooked line be bisected

9 If two diameters of y^e same Ellipsis be ordinately applyed {illeg}|y^e| one to y^e other y^e shortest of them is called y^e right diameter, y^e longest y^e transverse one. ({illeg} {illeg}) (as am & xλ ).

If {illeg} all y^e parallells w^ch are terminated by \y^e same or by/ 2 divers figures, are bisected by y^e same streight line, yⁿ is y^t line \called/ a \right/ diameter. But if That y^t diameter w^ch is ordin{illeg}|a|tely applied to it (i.e. w^ch is par{illeg}|a|llell to these parallells) is a transverse diameter.

1 If all y^e parallell lines w^ch are terminated by {illeg}|t|he same or by 2 divers figures, bee bisected by a streight line; y^t bisecting line is a diameter, & those paralle{l} lines, are lines ordinately a{illeg}|p|plied to y^t diameter.

2 If those parallell lines intersect y^e diameter at right angles y^t|e| diameter is {illeg} an axis

A line is said to bee of y^e first kind wh

In an Ellipsis y^e point where 2 diameters intersect, is {illeg} \y^e/ center.

The center of an{illeg} figure \Ellipsis/ is y^t point where {illeg} two \of {illeg} its/ diameters intersect.

{illeg} The center of two di{illeg}ers figure{illeg}|s| of t \opposite/ /Hyperbol{illeg}|a|s\ is y^t point where two of their diameters intersect one another or else where there {sic} Asymptotes intersect.

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Propositions. The lines ordinately ap{illeg}|p|lie{illeg}|d| to y^e axis of a crooked line are parallell to y^e tangent of y^e crooked line at its vertex.

^[5] Demonstr. Suppose \ chad a Parab &/ dc (being ordinately applied to y^e axis ab ) not parallell to y^e tan{illeg}|g|ent an but to some oth{illeg}|e|r line like ah. If dc bee understood to move {illeg}|to|wards a db continually decreas|i||ng| \{illeg}/ un{illeg}|t|ill it vanish into nothing at y^e conjucti{illeg}|o|n of y^e points a & d . but cb sinc & since cb mus{illeg}|t| always be equall to ah at y^e conju{illeg}|n|tion of y^e point {sic} a & d . it followeth y^t cb cannot decrease so as to vanish into nothing at y^e same time w^ch bd doth & therefore cannot allways be = to bd .

^[6] Otherwise. if dc is not parallell to y^e tangent an but to some other line as ah . Then ab doth not bisec{illeg}|t| all y^e parallell li{illeg}|n|es (as oe ) w^ch are terminated by y^e crooked line cad . & therefore cannot bee its diam:

^[7] |2^dly.| If ad is y^e axis of a crooked line & $cb = y$ , is ordinately a{illeg}|p|plied to ad . y^t is if $bc = ce = y$ . Then y {illeg} must be found noe where of odd dimensions in y^e Equation \expressing y^e nature of y^e line cod/. For (suppose{in}g $y = bc = ce$ to be y^e unknowne quantity) y hath 2 valors bc & cd equall {illeg}|to| one anot{illeg}|h|er excepting y^t y^e one bc is affirmative, y^e other ce is negative. w^ch two valors canot {sic} bee exprest by an equation in w^ch y is of od {sic} dimensions for suppose $y y = a a$ . yⁿ i{illeg}|s| is {sic} $y = + a$ , or $y = - a$ $y =$ $\sqrt{a a} = a$ , since $a \times a = a a$ . & $\sqrt{a a} = - a$ , since $- a \times - a = a a$ . & $y = \sqrt{a a}$ therefore is $y = + a$ , or $y = - a$ . soe if $y^{4} = a^{4}$ . yⁿ is $y y = a a$ , & $y = a$ or $y = - a$ but if $y^{3} = a^{3}$ . yⁿ $y = \sqrt{c: a^{3}} = a$ . but not $y = \sqrt{c: - a^{3}} = - a$ . soe i{illeg}|f| $y^{5} = a^{5}$ . yⁿ $y = a = \sqrt{qc: a^{5}}$ but not $y = - 5 = \sqrt{qc: - a^{5}}$ . The same reason is cogent in compound equations. as if $y y - 2 x y + x x = a x$ . Then, $y = x \overset{\cup}{O} \sqrt{a x}$ . where though y^e root $a + \sqrt{a x}$ is affirmative & y^e roote $a - \sqrt{a x}$ may bee negative yet they can never be equall in length, & though y^e 2 roots of an equation w^ch differ in signes should bee equally long y{illeg}|e|t y^t is w{illeg}|h|en there is but one quantity considere y^e Equation is fully determined.

^[8] Prop: 4^th. If x is of more dimensions in a quantity not multiplied by x yⁿ in one multiplied by it (as in $y^{2} =$ {illeg} $x y + a a$ ) yⁿ y is not parallel to one of y^e lines Asymptotes. & e contra. Otherwise x & y are parallel to y^e Asymptotes of y^e line. et e contra.

Prop 3^d. If ag is y^e Asymptote of y^e crooked line dcf , & $ab = x$ is coincident w^th it, then & $bc = y$ . then in y^e Equation (expressing y^e relation twixt x & y ,) x must be of more dimensions in some \one/ quantity \onely/ where it is multiplyed by y or $y y$ : &c in any other quan{illeg}|t|ity in y^e Equation \& y of as many/. & if ae is an Asymptote to y^e same line dcf , & $bc = y$ bee parallell to it then y must noe where be of soe many dimensions as in if \one/ quantity \onely/ in w^ch {illeg}|t|is drawne into x , $x x$ , \or/ $x^{3}$ &c. Demonstraco. If this be false yⁿ suppose y^e equation expressing y^e nature relation twixt x & y to be $x y y + a y y$ {illeg} $- b x y - d^{3} = 0$ . first if x is coincident w^th y^e Asymptote ag yⁿ, if y | x | $= \frac{e e}{o}$ (y^t is if x being infinite in length) i{illeg}|t| is $y = 0$ , if $x = \frac{e e}{o}$ (y^t {illeg}|i|s y vanisheth into nothing if x be infinite. {illeg}|)| therefore tansposeing {sic} $\frac{e e}{o}$ into y^e place of x , I find $\frac{e e y y}{o} + \binom{a y y}{} - \frac{b e e y}{o} - \binom{d^{3}}{} = 0$ . Or $e e y y + o \times a y y - b e e y - o \times d^{3} = 0$ that is $e e y y = b e e y$ /+|−|{ $e e$ }\, or $y = b$ . Now since y vanishe{illeg}|th| not it followeth y^{illeg}|t| x is not coin
cident w^th y^e Asymptote. But if I blot out {one} of y^e terme{illeg} {illeg} as $- b x y$ So: if y^e equation be $x x + x y = a a$ . yⁿ by substituteing $\frac{e e}{o}$ into y^e place of x I have $\frac{e^{4}}{o o} + \frac{e e y}{o} = a a$ , or $y = \frac{o^{2} a a - e^{4}}{o e e} = \frac{e e}{o}$ soe y^t y not vanishing but being infinite x cannot be asymptote. But if y^e equation {illeg} was $a y y - b x y = d^{3}$ . yⁿ by writeing $\frac{e e}{o}$ for x , there results, $\frac{e e y y}{0} + a y y = d^{3}$ . or $y y = \frac{o d^{3}}{o a + e e} = 0$ . & therefore in this case x is coincident y^e asymptote ag . Soe if $a y y - b x y - d^{3} = 0$ , by making $\frac{e e}{o} = x$ . it is $a y y - \frac{b e e y}{o}$ {illeg} $- d^{3} = 0$ . Or by extracting y^e roote $\frac{b e e}{2 a o} \overset{\cup}{O} \sqrt{\frac{e^{4} b b - 4 d^{3} o o a}{4 a a o o}}$ . Soe y^t y hath 2 roots y^e one infinitly \grate {sic}/ w^ch is $\frac{b e e}{2 a o} + \frac{\sqrt{e^{4} b b - 4 d^{3} a o o}}{2 a o} = \frac{b e e}{a o} = y$ . y^e other is infinitely little w^ch is $\frac{b e e}{2 a o} - \sqrt{\frac{e^{4} b b - o}{4 a a o o}} = \frac{b e e - b e e}{2 a o} = o = y$ . bee multiplied by y wherever i{illeg}|t| is of its greatest dimensions. & if ae is an asymptote to y^e line dcf , & $bc = y$ be parallel to it, y^e & $ab = x$ terminated by it at y^{illeg}|e| point a , {illeg}|th|en must y be multiplied by x wherever it is of its greatest dimensions. Example: Suppose $a x x + y x x = b^{3}$ . because in these 2 termes $a x x + y x x$ x is of it {sic} greatest dimensions; but in one of y^m (viz: $a x x$ ) it is not multiplyed by y therefore x is not coincident w^th ag y^e asymptote. If $y y x x + a y x x$ + / − \ $a^{3} x - a^{4} = 0$ . then since x is of its greate{illeg}|s|t dimens: in $y y x x$ & $a y x x$ onely & is drawne into y in both of y^m therefore x is coincident w^th y^e Asymptote{;} I Also since y is of its greatest dimension{illeg}|s| in $x x y y$ onely, {illeg}d (w^ch {illeg} {b}{l} multiplied {illeg} by x ) therefore y is parallel to {illeg} x {illeg} {terminated} by an Asymptote. &c.

Demonstr: If x is coincident w^th y^e Asymptote yⁿ { $\frac{e e}{o} = x$ } when $o = y$ . i:e: x is infinite when {illeg} {wisheth}. Now suppose $y y x x + a y x x -$ $a^{3} x$ $= a^{4}$ . yⁿ i{illeg}|f| $y = o$ is $x = \frac{a a}{\sqrt{o o + a o}}$ . i:e: x is {illeg} but if $y y x x + a a x x = a^{4}$ . yⁿ if $y = o$ it is { $x = \frac{a a}{\sqrt{o o + a a}} = \overset{\cup}{O} a$ }. soe {y^t} x is finite & therefor{e}{illeg} coincident w^th y^e {illeg} de{illeg}{illeg} of {illeg} is like{illeg}

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^[9] Haveing {an} the nature of a crooked line expresed in al{illeg}|g|ebraicall termes to find its axi|e|s if {i {sic}}{i} {shave} { anq }

^[10] Draw a line infinitely both ways fix upon some point (as b ) for y^e begining of one of y^e unknowne quantitys (w^ch I call x . Then reduce y^e Equation to such an order (if it bee not already so) y^t x may be always found in y^e line bc . w^th one end fixed at b , & having y moving making right angles w^th it at y^e other end: y^t end of y w^ch is remote from x , describing y^e crooked line. w^ch may bee {illeg}|al|ways done w^thout any great difficulty. As may be percei{illeg}|v|ed by these examples.

^[11] Suppose y^e given Equation was $y^{3} = a x x$ . x & y {illeg} soe y^t $bg = x$ . $gd = y$ . dc being perpendicular to bc & y^e angle dgc being given, y^e proportion twixt dg & dc is given, w^ch I supose as d to e . yⁿ is $d ∶ e ∷ y ∶ \frac{e y}{d}$ {illeg}c {illeg} $\begin{matrix} {gd}^{2} - {gc}^{2} = {dc}^{2} \\ y y - \frac{e e y y}{d d} = w^{2} \end{matrix}$ . $\frac{d w}{\sqrt{d d - e e}} = y$ . $\begin{matrix} bg + gc \\ x + \frac{e y}{d} \end{matrix}$ $d ∶ e ∷ y ∶ dc = \frac{e y}{d} = w$ . & $\frac{d w}{e} = y$ . ${gc}^{2} \begin{array}{l} = {gd}^{2} - {dc}^{2} \\ = y y - w w \end{array} = \frac{d d w w - e e w w}{e e}$ . or $\frac{g}{c} \sqrt{d d - e e}$ . & $bc = v = \frac{w \sqrt{d d - e e}}{e}$ . Or $\frac{e v - w \sqrt{d d - e e}}{e} = x$ .^[12] Therefore I write $\frac{d^{3} w^{3}}{e^{3}}$ for $y^{3}$ , {illeg}|&| $\frac{e e v v - 2 e v w \sqrt{d d - e e} + d d w w - e e w w}{e e}$ for $x x$ in y^e equation $y^{3} = a x x$ , & soe I have this equation $\frac{d^{3} w^{3}}{e^{3}} = \frac{a e e v v - 2 a e v w \sqrt{d d - e e} + a d d w^{2} - a e e w^{2}}{e e}$ . w^ch expresseth y^e relation twixt w & v , y^t is twixt dc & bc , {illeg}|w|riteing therefore y for dc , & x for bc : I have this equation $d^{3} y^{3} - a e e e x x =$ \{o}{ 0 }/ $d^{3} y^{3} + a e^{3} y y - a e d d y y + 2 a e e x y \sqrt{d d - e e} - a e^{3} x x$

^[13] Soe if x \ $= bd$ / turned about y^e pole b & y \ $= dg$ / about y^e pole g j{illeg} describing y^e croo{illeg}|k|ed line ad by y^e conjunction at y^e extremitys. & y^e equation expresing y^e relation w^ch they bea{illeg}|r|e to one another is $x x$ {illeg} $x y$ $x x = a y$ . y^e distance of y^e poles is given w^ch I call $b = bg$ .^[14] perpendic: to bg I draw $dc = w$ & make $bc = v$ . Then, is $\begin{matrix} {dc}^{2} + {bc}^{2} = {bd}^{2} . \\ w^{2} + v^{2} = x x . \end{matrix}$ $\begin{array}{r} {dc}^{2} + {cg}^{2} = {dg}^{2} . \\ w^{2} + b b - 2 b v + v v = y y . \end{array}$ soe y^t for $x x = a y$ {illeg} I write $w^{2} + v^{2} = a \sqrt{w^{2} + b b - 2 b v + v v}$ . Or $w^{4} + 2 w^{2} v^{2} + w^{4} = a a w^{2} + a a b b - 2 a a b v + a a v v$ . w^ch expresseth y^e relation w^ch bc beareth to dc , & by makeing $bc = x$ , $dc = y$ , it is, $\begin{array}{l} y^{4} & + & 2 x x y y & + & x^{4} & = & 0 \\ - & a a y y & + & 2 a a b x \\ - & a a x x \\ - & a a b b \end{array}$ .

Examp 3^d If $bg = x$ {illeg}|b|e always in y^e line bc . & fd turning about y^e pole f {illeg}|&| passing by y^e end of $bg = x$ w^th its other end d describes y^e crooked line bdh , soe y^t calling gd y $x = y$ . yⁿ drawing ef & dc perpendicular to bh . $be = a$ , & $ef = b$ are given. & I make $bc = v$ {.} $dc = w$ . yⁿ is $eg = x - a$ . $gc = v - x$ . $\begin{matrix} ef ∶ eg ∷ gc ∶ cd . \\ b ∶ x - a ∷ v - x ∶ w . \end{matrix}$ $bw = v x - a v + a x - x x$ . or by extracting y^e roote $x = \frac{+ a + v}{2} \overset{\cup}{O} \sqrt{\frac{a a + 2 a v + v v - 4 b w}{4}}$ . againe $\begin{matrix} {gc}^{2} + {dc}^{2} = {gd}^{2} \\ v v - 2 v x + x x + w^{2} = y y = x x . \end{matrix}$ . Or $\frac{v v + w^{2}}{2 w} = x = \frac{a + v}{2} \overset{\cup}{O} \sqrt{\frac{a a + 2 a v + v v - 4 b w}{4}}$ . & by transposeing $\frac{a + v}{2}$ {illeg}|to| y^e other side & so squareing both pts $w^{4} - 2 a v w^{2} = 2 a w^{3} + v^{4} - 4 b w v^{2}$ . w^ch equation expresseth y^e relation twixt $w = dc$ , & v {illeg} $= bc$ . & so by {illeg} calling dc y , & bc x , it is, $y^{4} - 2 a x y y + 4 b x x y - x^{4} - 2 a x^{3} = 0$ .

Example 4^th. if $bd = x$ turnes about y^e pole b , & gd (a given line $= a$ ) slides upon bg w^th one end & intersecting bd at right angles at y^e other end describes y^e crooked line \ bde / by its intersection w^th bd . then makeing $bc = v$ , $dc = w$ . $\begin{matrix} {bc}^{2} + {dc}^{2} = {bd}^{2} . \\ v^{2} + w^{2} = x^{2} . \end{matrix}$ $\begin{matrix} bc ∶ bd ∷ dc ∶ dg . \\ v ∶ x ∷ w ∶ a . \end{matrix}$ & $\frac{a v}{w} = x$ therefore $v v + w w = \frac{a a v v}{w w}$ or $w^{4} + v v w w = a a v v$ . & so by writing x for v & y for w , I have y^e relation twixt $x = bc$ & $y = dc$ ex{illeg}|p|rest in this equation $y^{4} + x x y y - a a x x = 0$ .

Or if y^e relation twixt bd & dg {illeg} was exprest in this equation ({illeg}|m|a{illeg}|k|ing $dg = y$ . $bd = x$ ) $\begin{matrix} x x y \\ + a y y \end{matrix} = a^{3}$ . then {illeg} as before $\begin{matrix} {bc}^{2} + {dc}^{2} = {bd}^{2} . \\ v^{2} + w^{2} = x^{2} . \end{matrix}$ $\begin{matrix} {bc}^{} ∶ {bd}^{} ∷ {dc}^{} ∶ {dg}^{} . \\ v^{} ∶ \sqrt{v v + w^{2}} ∷ w^{} ∶ y . \end{matrix}$ therefore $y = \frac{w \sqrt{v v + w^{2}}}{v}$ . first therefore I take away $x x$ by making $\frac{a^{3} - a y y}{y} = x x = v v + w^{2}$ . or by ordering it $a^{3} - a a y - v v y - w w y = 0$ . Then I take away y by substituteing it {sic} valor $\frac{w \sqrt{v v + w^{2}}}{v}$ into its r{illeg}|o|ome & it will be $a^{3} - \frac{a w^{2} v^{2} - a w^{4}}{v v} = \frac{v v w \sqrt{v v + w^{2}} + w^{3} \sqrt{v v + w^{2}}}{v}$ . & by □ing both pts. $\begin{matrix} a^{6} v^{4} & - & 2 a^{4} v^{4} w^{2} & - & 2 a^{4} v v w^{4} & + & a a v^{4} w^{4} & - & 2 a a v v w^{6} & + & a a w^{8} & - & v^{8} w w & - & 3 v^{6} w^{4} & = & 0 \\ - & 3 v^{4} w^{6} & - & v v w^{8} \end{matrix}$ . & by writeing x for v & y for w y^e equation will be $\begin{array}{l} a a y^{8} & + & 2 a a x x y^{6} & + & a a x^{4} y^{4} & - & x^{8} y y & + & a^{6} x^{4} & = & 0 \\ - & x x y^{8} & - & 3 x^{4} y^{6} & - & 3 x^{6} y^{4} & - & 2 a^{4} x^{4} y y \\ - & 2 a^{4} x x y^{4} \end{array}$ .

The like may as easily be performed in any other case.

After y^e equation is brought to this order observe y^t if y is noe where of odd dimensions yⁿ y^e line bc in (w^ch {illeg} is coincident w^th x ) is ꝑte of an axis of y^{{illeg} e} crooked line, as in y^e 2^d Example. And if {illeg} x is noe \where/ of odd dimensions (as in this, $a^{4} + y y a a = a a x^{2}$ {illeg}|)| Then I draw bk from y^e point b at y^e begine|i|ng of x . I draw {illeg} \ bk / perpendicular to {illeg} \to {sic} bc / w^ch is coincident w^th y^e axis of y^e crooked line. And if neither x nor y bee of unequall dimensions in any terme of y^e equation then both bk & bc may bee taken for axes of y^e crooked line or lin{e}s \whose nature are/ expressed by y^e equation. As in y^e 4^th Example.

But if y is of odd dimension{s} in y^e Equation then ordering y^e Equation according to y see if y is of eaven dimensions in y^e first te{r}me {illeg} { x } not found in y^e 2^nd. if so take away y^e 2^nd terme of y^e equation & if there result an Equation in w^ch y is noe where of {odd} dim{en}sions. Then I draw ce perpendicular to bc & equall to y^t quantity w^ch added or substracted from y y^t m{illeg}|i|ght take away y^e {illeg}|2|^d terme; through y^e point {illeg} P{illeg} f{illeg}{illeg}{illeg}bee {illeg}.

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^[15] As in this Example, $y y +$ \ $2$ / $a y$ $- \frac{x^{3}}{a} = 0$ . Then to take away y^e {illeg}|2|^d terme I make $z - a = y$ . & soe I have, $z z * - a a - \frac{x^{3}}{a} = 0$ . in w^ch z is not of odd dimensions. Then drawing bc for x , dc for y , {illeg} & de for z : or w^ch is y^e same (since $z - a = y$ ) I make {illeg} $ce = - a$ that is I draw dc & ce on 2 contrary sides of y^e line bc . & through y^e point e I draw ae parallell to bc & make it y^e axis of y^e line dag

^[16] Example y^e 2^d. $y^{4} - 8 a y y y + 24 a a y^{2} - 3 a x y^{2} - 32 a^{3} y$ \ $+ 16 a^{4}$ / $+ 12 a a x y = 0$ . y^t y^e first terme may bee of eaven dimensions I multiply y^e {illeg} Then by making $z + 2 a = y$ I take away the 2^d terme. & y^e Equation $z^{4} * - 3 a x z z * + 12 a^{3} x = 0$ . in w^ch z is onely of eaven dimensions. Then I draw bc for x . dc for y de for, or w^ch is y^e same (sinc {sic} $z +$ \ $2$ / $a = y$ ) I make $ec = 2 a$ , y^t is I draw ec & dc on y^e same side of bc then through y^e point e para{illeg}|l|lell to bc I draw ea for y^e axis of y^e lines dac d khg .

^[17] In like manner, if x is of odd dimensions in some terme of y^e Equation, y^e Axis bk perpendicular to $x = bc$ may bee found. As for Example. $x x +$ {{illeg}}{{illeg}| $2$ |} $a x - \frac{a^{3}}{a} = 0$ . by makei{ng} $z - \frac{a}{2} = x$ , I take away y^e 2^d terme and soe have {illeg}|this| equation $z z * - \frac{y^{3}}{a} - \frac{a a}{4} = 0$ . therefore I draw $bc = x$ from y^e fixed point b , & $ce = z$ , or w^ch is y^e same (since $z - \frac{a}{2} = x$ ) I draw $eb = - \frac{1}{2} a$ , y^t is I draw eb & bc on two contrary sides of y^e line k b yⁿ throug{h} y^e point e , parallell to bk I draw ea y^e axis of y^e line

^[18] Example 2^d. $x^{4} - 4 a x^{3} + 4 a a x x$ $- 4 a^{3} x$ $- a a y y - a a b y = 0$ . by makeing $x = z + a$ I have this Equat $x^{4} - 2 a a z z - a a y y - a a b y + a^{4} = 0$ . In w^ch z is noe where of odd dimensions. therefore assumeing b for y^e begining of x & making $b
c = x$ , & $ce =$ {illeg} z , or w^ch is y^e same if {illeg} $bc = + x$ I I make $be = + a$ , since $x = z + a$ ; that is if bc is affirmative I take be & bc on y^e same side of y^e line kb . otherwise I describe y^m on contrary sides of it. then through y^e point e parallell to bk I draw eg {{illeg}|a|n} axis of y^e lines dbmd & nhr . Againe I order y^e Equation according to y & it is $\begin{array}{l} a^{2} y y & + & a a b y & - & z^{4} & = & 0 \\ + & 2 a a z z \\ - & a^{4} \end{array}$ . {illeg}|&| soe \since x is not in y^e 2^d terme/ makeing $v - \frac{b}{2} = y$ . I take away y^e 2^d terme, & it is $\begin{array}{l} a a v v & * & - & \frac{a a b b}{4} & - & z^{4} & = & 0 \\ + & 2 a a z z \\ - & a^{4} \end{array}$ . Therefore I draw $ec = z$ , $dc = y$ , & $df = v$ . or w^ch differs not {illeg} (since $v - \frac{b}{2} = y$ ) I make $cf = - \frac{b}{2}$ & through y^e point f parallell to ce I draw { fg } for another axis of y^e lines dbmd , & ndhdr .

^[19] But if {illeg} but if {sic} {illeg} {illeg} y^e unknowne quantity ( x or y ) is of odd dimensions in y^e first terme or if {illeg}|b|oth y^e unknowne quantitys are in y^e 2^d terme, or {illeg}|if| by this {sic} meanes y^e equation is irreducible to such a forme y^t x , or, y , or both of y^m bee of odd dimensions noewhere in y^e Equ{illeg}|a|t: Then try to find y^e axes by y^e following method. Observing by y^e way y^t

If $+ x$ begins at y^e {illeg} point b {illeg}|&| extends to{illeg}|w|ards c in y^e line sr yⁿ $- x$ is taken y^e contrary way towards s , & all y^e \affirmative/ lines parallell to sr are drawne y^e same way w^ch $+ x$ is but y^e negative lines parallell t{illeg}|o| sr are drawn y^e same way w^th $- x$ as if from y^e point m I must draw a line $= a$ , I draw it towards n but if from y^e s^me point m I must draw a line $= - a$ I draw it towards l . soe if from y^e point {illeg}| d | I must draw $d λ = + b$ , yⁿ I draw it towards θ , but if $d λ = - b$ then I draw it towards γ . A{illeg}|g|aine if $+ y$ is drawne toward p from y^e line sr , yⁿ $- y$ is drawne from y^e same line sr y^e contrary way towards o , & those lines w^ch are affected w^th an affirmative signe {illeg}|&| are paralell {sic} to y they are {illeg}|d|rawne y^e same way w^ch y is but those lines w^ch are negative are drawn y^e contrary way. as if $a = aπ$ then I draw a{illeg} aπ towards e but if $- a = aπ$ yⁿ I draw it towards t . soe if $vμ = + d$ yⁿ I draw it from v towards δ, if $vμ = - d$ , I draw it towards w.

A generall rule to find y^e axes of any line.

^[20] ^[21] Suppose $bc = x$ . $cd = y$ . & kg to be y^e axis. yⁿ parallel to y from y^e point b to y^e axis kg draw $bf = c$ . from d y^e end of y , d{illeg}d perpendicular to kg draw $dh = ϩ$ . & make $fh = ϱ$ & suppose $fe ∶ fg ∷ d ∶ e$ . yⁿ is $fg = \frac{e x}{d}$ . & $d ∶ e ∷ dh = ϩ ∶ \frac{e ϩ}{d} = dg$ . $\begin{matrix} {dg}^{2} - {dh}^{2} = {gh}^{2} \\ \frac{e e ϩ ϩ}{d d} - ϩ ϩ = {gh}^{2} \end{matrix}$ . $gh = \frac{ϩ \sqrt{e e - d d}}{d}$ $gh = \begin{matrix} - fh & + fg \\ - ϱ & + \frac{e x}{d} \end{matrix} = \frac{ϩ \sqrt{e e - d d}}{d}$ . therefore $\frac{d ϱ + ϩ \sqrt{e e - d d}}{e} = x$ . Againe $ge + ec - dg = dc$ , that is $\sqrt{\frac{e e x x}{d} - x x} + c - \frac{e ϩ}{d} = y$ . or for x writeing its valor, $y = \frac{c e - d ϩ + ϱ \sqrt{e e - d d}}{e}$ . Now assumeing any quantity for e , y^t I may find y^e valors of c & d . I substitute these valors of x & y into theire roome in y^e Equation. as if y^e equation be $x^{2} - 2 x y + a y + y y = 0$ . by making $e = a$ . y^e valor of x is $\frac{d ϱ + ϩ \sqrt{a a - d d}}{a}$ & y^e valor of y is $\frac{a c - d ϩ + ϱ \sqrt{a a - d d}}{a}$ . w^ch 2 valors substituting into their roome in y^e equation, {illeg} their {sic} results { $\begin{matrix} a a ϩ ϩ & + & 2 d ϩ ϩ \sqrt{a^{2} - d^{2}} & + & 4 d d ϱ ϩ & - & 2 d ϱ ϱ \sqrt{a a - d d} \\ - & 2 a a ϱ ϩ & + & 2 a c ϱ \sqrt{a a - d d} \\ - & a^{2} d ϩ & + & a a ϱ \sqrt{a a - d d} \\ - & 2 a c d ϩ & + & a a ϱ ϱ \\ - & 2 a c ϩ \sqrt{a^{2} - d^{2}} & - & 2 d a c ϱ \\ + & a^{3} c \\ + & a a c c \end{matrix}} = 0$ } Now y^t ϩ I may have an equation in w^ch {illeg}| ϩ | is of \{illeg} {2}/ eaven dimensions o{ne}ly I suppose y^e 2^d terme $= 0$ & soe have this equation $4 d d ϱ ϩ - 2 a a ϩ ϱ$ {illeg} $- a a d ϩ - 2 a c d ϩ - 2 a c ϩ \sqrt{a a - d d} = 0$ & y^t y^e teres in this {feigned} equation may destiny one another I order it according to { ϱ } & soe suppose each terme $= 0$ . & so I have these equations $4 d d ϱ ϩ - 2 a a ϱ ϩ = 0$ & $- a a d ϩ - 2 a c d ϩ - 2 a c ϩ \sqrt{a a - d d} = 0$ . by y^e first I find $2 d d = a a$ , or $d = \frac{a a}{\sqrt{2}}$ . by y^e 2^d $- a a d - 2 a c d - 2 a c \sqrt{a a - d d} = 0$ . & by substituteing y^e valor of d into its {roome} I find $\frac{a^{3} - 4 a a c}{\sqrt{2}} = 0$ . or $c = \frac{- a}{4}$ therefore from { b } perpendic: to bc I draw $bf = \frac{- a}{4}$ . through y^e point f paralell {sic} to bc I draw $fe = \frac{a}{\sqrt{2}}$ & since $f$ {illeg} $= a$ therefore I draw $ge = \sqrt{a a - \frac{a a}{2}} = \frac{a}{\sqrt{2}}$ . & lastly {illeg}{illeg}y^e points f & g I draw fg y^e axis of y^e crooked line bah .

<17v>

But since there is noe use of those termes in w^ch ϩ is of eaven dimensions y^e Calculation will bee much abre{illeg}|v|iated by this following table.

$x = \frac{ϩ \sqrt{e e - d d}}{e}$ . $y = - \frac{d ϩ}{e}$ . $y y = - \frac{2 c d ϩ}{e}$ . $y^{3} = - \frac{3 c c d ϩ}{e}$ . $y^{4} = - \frac{4 c^{3} d ϩ}{e}$ . $y^{5} = - \frac{5 c^{4} d ϩ}{e}$ . &c $x y = \frac{c ϩ \sqrt{e e - d d}}{e}$ . $x y y = \frac{c c ϩ \sqrt{e e - d d}}{e}$ . $x y^{3} = \frac{c^{3} ϩ \sqrt{e e - d d}}{e}$ . $x y^{4} = \frac{c^{4} ϩ \sqrt{e e - d d}}{e}$ . $x y^{5} = \frac{c^{5} ϩ \sqrt{e e - d d}}{e}$ . &c.

^[22] {illeg} $x = \sqrt{e e - d d}$ . $y = - d$ . $y y = - 2 c d$ . $y^{3} = - 3 c c d$ . $y^{4} = - 4 c^{3} d$ . $y^{5} = - 5 c^{4} d$ . $y^{6} = - 6 c^{5} d$ &c $x y = c \sqrt{e e - d d}$ . $x y y = c c \sqrt{e e - d d}$ . $x y^{3} = c^{3} \sqrt{e e - d d}$ . $x y^{4} = c^{4} \sqrt{e e - d d}$ . $x y^{5} = c^{5} \sqrt{e e - d d}$ . &c.

^[23] $x x = 2 d \sqrt{e e - d d}$ . $y y = \begin{matrix} - 2 \\ - 1 \times 2 \end{matrix}} d \sqrt{e e - d d}$ . $y^{3} = \begin{matrix} - 6 \\ - 2 \times 3 \end{matrix}} c d \sqrt{e e - d d}$ . $y^{4} = \begin{matrix} - 12 \\ - 3 \times 4 \end{matrix}} c c d \sqrt{e e - d d}$ . $y^{5} = \begin{matrix} - 20 \\ - 4 \times 5 \end{matrix}} c^{3} d \sqrt{e e - d d}$ . $y^{6} = \begin{matrix} - 30 \\ - 5 \times 6 \end{matrix}} c^{4} d \sqrt{e e - d d}$ . $x y = - 2 d d + e e$ . $x y y = - 4 c d d + 2 c e e$ . $x y^{3} = - 6 c c d d + 3 c c e e$ . $x y^{4} = - 8 c^{3} d d + 4 c^{3} e e$ . $x y^{5} = 5 c^{4} e e - 104 d d$ . &c. $x x y = 2 c d \sqrt{e e - d d}$ . $x x y y = 2 c c d \sqrt{e e - d d}$ . $x x y^{3} = 2 c^{3} d \sqrt{e e - d d}$ . $x x y^{4} = 2 c^{4} d \sqrt{e e - d d}$ . $x x y^{5} = 2 c^{5} d \sqrt{e e - d d}$ . &c.

^[24] $x^{3} = 3 d d \sqrt{e e - d d}$ . $y^{3} = - 3 d e e + 3 d^{3}$ . $y^{4} = 2 \times 6 c d^{3} - 2 \times 6 c d e e$ . $y^{5} = 3 \times 10 c c d^{3} - 3 \times 10 c c d e e$ . $y^{6} = 4 \times 15 c^{3} d^{3} - 4 \times 15 c^{3} d e e$ . $x y y = - 3 d d \sqrt{e e - d d} + e e \sqrt{e e - d d}$ . $x y^{3} = - 3 c e e \sqrt{e e - d d} - 9 c d d \sqrt{e e - d d}$ . $x y^{4} = 6 c c e e \sqrt{e e - d d} - 18 c c d d \sqrt{e e - d d}$ . $x y^{5} = 10 c^{3} e e \sqrt{e e - d d} - 30 c^{3} d d \sqrt{e e - d d}$ . &c $x x y = 2 d e e - 3 d^{3}$ . $x x y y = 4 c d e e - 6 c d^{3}$ . $x x y^{3} = 6 c c d e e - 9 c c d^{3}$ . $x x y^{4} = 8 c^{3} d e e - 12 c^{3} d^{3}$ . $x^{3} y = 3 c d d \sqrt{e e - d d}$ . $x^{3} y y = 3 c c d d \sqrt{e e - d d}$ . $x^{3} y^{3} = 3 c^{3} d^{2} \sqrt{e e - d d}$ . $x^{3} y^{4} = 3 c^{4} d d \sqrt{e e - d d}$ .

^[25] $x^{4} = 4 d^{3} \sqrt{e e - d d}$ . $y^{4} = - 1 \times 4 d \sqrt{- e^{6} - 3 e^{4} d d + 3 e e d^{4} - d^{6}}$ . $y^{5} = - 2 \times 10 c d \sqrt{- e^{6} - 3 e^{4} d d + 3 e e d^{4} - d^{6}}$ . $y^{6} = - 3 \times 20 c c d \sqrt{- e^{6} &c}$ : $y^{7} = - 4 \times 35 c^{3} d \sqrt{- e^{6} &c}$ . $x y^{3} = e^{4} \begin{matrix} - 2 d d e e + 1 d^{4} \\ - 3 d d e e + 3 d^{4} \end{matrix}$ . $x y^{4} = 4 c e^{4} \begin{array}{l} - 8 \\ - 12 \end{array} c d d e e \begin{array}{l} + 4 \\ + 12 \end{array} c d^{4}$ . $x y^{5} = 10 c c e^{4} \begin{array}{l} - 20 \\ - 30 \end{array} c c d d e e \begin{array}{l} + 10 \\ + 30 \end{array} c c d^{4}$ . $x y^{6} = 20 c^{3} e^{4} \begin{array}{l} - 40 \\ - 60 \end{array} c^{3} d d e e \begin{array}{l} + 20 \\ + 60 \end{array} c^{3} d^{4}$ &c. $x x y y = \begin{matrix} 1 \times 2 d e e \\ - 4 d^{3} \end{matrix} \sqrt{e e - d d}$ . $x x y^{3} = \begin{matrix} 2 \times 3 c d e e \\ - 12 c d^{3} \end{matrix} \sqrt{e e - d d}$ . $x x y^{4} = \begin{matrix} + 12 c c d e e \\ - 24 c c d^{3} \end{matrix} \sqrt{e e - d d}$ . $x x y^{5} = \begin{matrix} + 20 c^{3} d e e \\ - 40 c^{3} d^{3} \end{matrix} \sqrt{e e - d d}$ . $x^{3} y = 3 d d e e - 4 d^{4}$ . $x^{3} y y = 6 c d d e e - 8 c d^{4}$ . $x^{3} y^{3} = 9 c c d d e e - 12 c c d^{4}$ . $x^{3} y^{4} = 12 c^{3} d d e e - 16 c^{3} d^{4}$ . $x^{4} y = 4 c d^{3} \sqrt{e e - d d}$ . $x^{4} y y = 4 c c d^{3} \sqrt{e e - d d}$ . $x^{4} y^{3} = 4 c^{3} d^{3} \sqrt{e e - d d}$ . $x^{4} y^{4} = 4 c^{4} d^{3} \sqrt{e e - d d}$ .

^[26] $x^{5} = 5 d^{4} \sqrt{e e - d d}$ . $y^{5} = - 1 \times 5 d e^{4} + 10 d^{3} e e - 5 d^{5}$ . $y^{6} = - 2 \times 15 c d e^{4} + 60 c d^{3} e e - 30 c d^{5}$ . $y^{7} = - 3 \times 35 c c d e^{4} + 210 c c d^{3} e e - 105 c c d^{5}$ . $y^{8} = - 4 \times 70 c^{3} d e^{4}$ \ $+ 560 c^{3} d^{3} e e - 280 c^{3} d^{5}$ / &c $x y^{4} = e^{4} \begin{matrix} - 2 d d e e + 1 d^{4} \\ - 4 d d e e + 4 d^{4} \end{matrix} in \sqrt{e e - d d}$ . $x y^{5} = 5 e c e^{4} \begin{matrix} - 10 c d d e e + 5 c d^{4} \\ - 20 c d d e e + 20 c d^{4} \end{matrix} in \sqrt{e e - d d}$ . $x y^{6} = 15 c c e^{4} \begin{matrix} - 30 c c d d e e + 15 c c d^{4} \\ - 60 c c d d e e + 60 c c d^{4} \end{matrix} in \sqrt{e e - d d}$ . $x y^{7} = 35 c^{3} e^{4} \begin{matrix} - 70 c^{3} d d e e + 35 c^{3} d^{4} \\ - 140 c^{3} d d e e + 140 c^{3} d^{4} \end{matrix} in \sqrt{e e - d d}$ . $x x y^{3} = 2 d e^{4} \begin{matrix} - 4 d^{3} e e + 2 d^{5} \\ - 3 d^{3} e e + 3 d^{5} \end{matrix}$ . $x x y^{4} = - 2 \times 4 c d e^{4} \begin{matrix} - 2 \times 8 c d^{3} e e + 8 c d^{5} \\ - 12 c d^{3} e e + 12 c d^{5} \end{matrix}$ . $x x y^{5} = 20 c c d e^{4} \begin{matrix} - 40 c c d^{3} e e + 2 \times 10 c c d^{5} \\ - 30 c c d^{3} e e + 30 c c d^{5} \end{matrix}$ . $x x y^{6} = 40 c^{3} d e^{4} \begin{matrix} - 80 c^{3} d^{3} e e + 2 \times 20 c^{3} d^{5} \\ - 60 c^{3} d^{3} e e + 60 c^{3} d^{5} \end{matrix}$ $x^{3} y y = \begin{matrix} 1 \times 3 d d e e \\ - 5 d^{4} \end{matrix} \sqrt{e e - d d}$ . $x^{3} y^{3} = \begin{matrix} 3 \times 3 d d c e e \\ - 15 c d^{4} \end{matrix} \sqrt{e e - d d}$ . $x^{3} y^{4} = \begin{matrix} 6 \times 3 c c d d e e \\ - 6 \times 5 c c d^{4} \end{matrix} \sqrt{e e - d d}$ . $x^{3} y^{5} = \begin{matrix} 10 \times 3 c^{3} d d e e \\ - 50 c^{3} d^{4} \end{matrix} \sqrt{e e - d d}$ . $x^{3} y^{6} = \begin{matrix} 15 \times 3 c^{4} d d e e \\ - 75 c^{4} d^{4} \end{matrix} \sqrt{e e - d d}$ . $x^{4} y = 1 \times 4 d^{3} e e - 5 d^{5}$ . / $- 5 d^{5}$ \ $x^{4} y y = 2 \times 4 c d^{3} e e - 10 c d^{5}$ . $x^{4} y^{3} = 3 \times 4 c c d^{3} e e - 15 c c d^{5}$ . $x^{4} y^{4} = 4 \times 4 c^{3} d^{3} e e - 20 c^{3} d^{5}$ . $x^{4} y^{5} = 5 \times 4 c^{4} d^{3} e e - 25 c^{4} d^{5}$ . $x^{5} y = 5 c d^{4} \sqrt{e e - d d}$ . $x^{5} y y = 5 c c d^{4} \sqrt{e e - d d}$ . $x^{5} y^{3} = 5 c^{3} d^{4} \sqrt{e e - d d}$ .

^[27] $x^{6} = 6 d^{5} \sqrt{e e - d d}$ . $y^{6} = - 6 \times 1 d \sqrt{e^{10} - 5 e^{8} d d + 10 e^{6} d^{4} - 10 e^{4} d^{6} + 5 e e d^{8} - d^{10}}$ . $y^{7} = - 6 \times 7 d c \sqrt{e^{10} &c}$ : $y^{8} = - 6 \times 28 c c d \sqrt{e^{10} &c}$ : $y^{9} = - 6 \times 84 c^{3} d \sqrt{e^{10} &c}$ : $x y^{5} = 1 \times 1 e^{6} \begin{matrix} - 3 d d e^{4} + 3 d^{4} e e - d^{6} \\ - 5 d d e^{4} + 3 d^{4} e e - 5 d^{6} \end{matrix}$ . $x y^{6} = 1 \times 6 c e^{6} \begin{matrix} - 18 c d d e^{4} + 18 c d^{4} e e - 6 c d^{6} \\ - 6 \times 5 c d d e^{4} + 60 c d^{4} e e - 30 c d^{6} \end{matrix}$ . $x y^{7} = 1 \times 21 c c e^{6} \begin{matrix} - 63 c c e^{4} d d + 63 c c e e d^{4} - 21 c c d^{6} \\ - 21 \times 5 c c e^{4} d d + 210 c c e e d^{4} - 105 c c d^{6} \end{matrix}$ . $x y^{8} = 1 \times 53 c^{3} e^{6} &c$ $x x y^{4} = 2 \times 1 d e^{4} \begin{matrix} - 4 d^{3} e e + 2 d^{5} \\ - 4 d^{3} e e + 4 d^{5} \end{matrix} \sqrt{e e - d d}$ . $x x y^{5} = 2 \times 5 c d e^{4} \begin{matrix} - 20 c d^{3} e e + 10 c d^{5} \\ - 5 \times 4 d^{3} e e + 20 c d^{5} \end{matrix} in \sqrt{e e - d d}$ . $x x y^{6} = 2 \times 15 c c d e^{4} \begin{matrix} - 60 c c d^{3} e e + 30 c c d^{5} \\ - 4 \times 15 c c d^{3} e e + 60 c c d^{5} \end{matrix} \times \sqrt{e e - d d}$ . $x^{3} y^{3} = 3 \times 1 d d e^{4} \begin{matrix} - 6 d^{4} e e + 3 d^{6} \\ - 3 d^{4} e e + 3 d^{6} \end{matrix}$ . $x^{3} y^{4} = 3 \times 4 c d d e^{4} \begin{matrix} - 24 c d^{4} e e + 12 c d^{6} \\ - 3 \times 4 c d^{4} e e + 12 c d^{6} \end{matrix}$ . $x^{3} y^{5} = 3 \times 10 c c d d e^{4} \begin{matrix} - 60 c c d^{4} e e + 30 c c d^{6} \\ - 30 c c d^{4} e e + 30 c c d^{6} \end{matrix}$ . $x^{4} y y = 4 \times 1 d d d e e \begin{matrix} - 4 d^{5} \\ - 2 d^{5} \end{matrix} in \sqrt{e e - d d}$ . $x^{4} y^{3} = 4 \times 3 d^{3} c e e \begin{matrix} - 12 c d^{5} \\ - 6 c d^{5} \end{matrix} in \sqrt{e e - d d}$ . $x^{4} y^{4} = 4 \times 6 c c d^{3} e e \begin{matrix} - 24 c c d^{5} \\ - 12 c c d^{5} \end{matrix} in \sqrt{e e - d d}$ . $x^{4} y^{5} = 4 \times 10 c^{3} d^{3} e e &c$ . $x^{5} y = 1 \times 5 d^{4} e e \begin{matrix} - 5 d^{6} \\ - 1 d^{6} \end{matrix}$ . $x^{5} y y = 5 \times 2 c d^{4} e e \begin{matrix} - 10 c d^{6} \\ - 2 c d^{6} \end{matrix}$ . $x^{5} y^{3} = 5 \times 3 c c d^{4} e e \begin{matrix} - 15 c c d^{6} \\ - 3 c c d^{6} \end{matrix}$ . $x^{5} y^{4} = 5 \times 4 c^{3} d^{4} e e \begin{matrix} - 20 c c d^{6} \\ - 4 c c d^{6} \end{matrix}$ . $x^{6} = 1 \times 6 d^{5} \sqrt{e e - d d}$ . $x^{6} y = 6 c d^{5} \sqrt{e e - d d}$ . $x^{6} y y = 6 c c d^{5} \sqrt{e e - d d}$ . $x^{6} y^{3}$ − {sic} $6 c^{3} d^{5} \sqrt{e e - d d}$ . &c.

^[28] $y^{3}$ {illeg} $- d^{3}$ . $y^{4} = - 4 c d^{3}$ . $y^{5} = - 10 c c d^{3}$ . $y^{6} = - 20 c^{3} d^{3}$ . $y^{7} = - 35 c^{4} d^{3}$ . $y^{8} = - 56 c^{5} d^{3}$ . $x y y = d d \sqrt{e e - d d}$ . $x y^{3} = c d d \sqrt{e e - d d}$ . $x y^{4} = c c d d \sqrt{e e - d d}$ . $x y^{5} = c^{3} d d \sqrt{e e - d d}$ . $x x y = d^{3} - e e d$ . $x x y y = 2 c d^{3} - 2 c e e d$ . $x x y^{3} = 3 c c d^{3} - 3 c c e e d$ . $x x y^{4} = 4 c^{3} d^{3} - 4 c c e e d$ . $x^{3} = e e - d d in \sqrt{e e - d d}$ . $x^{3} y = c e e - c d d \sqrt{e e - d d}$ . $x^{3} y y = c c e e - c c d d in \sqrt{e e - d d}$ . $x^{3} y^{3} = c^{3} \sqrt{e^{6} - 3 e^{4} d d + 3 e e d^{4} - d^{6}}$ .

^[29] $y^{4} =$ {illeg} $- 1 \times 4 d^{3} \sqrt{e e - d d}$ . $y^{5} = - 5 \times 4 c d^{3} \sqrt{e e - d d}$ . $y^{6} = - 15 \times 4 c c d^{3} \sqrt{e e - d d}$ . $y^{7} = - 35 \times 4 c^{3} d^{3} \sqrt{e e - d d}$ . {illeg} $x y^{3} = 1 \times 3 d d e e \begin{matrix} - 3 d^{4} \\ - 1 d^{4} \end{matrix}$ . $x y^{4} = 4 \times 3 c d d e e \begin{matrix} - 12 c d^{4} \\ - 4 \times 1 d^{4} \end{matrix}$ . $x y^{5} = 10 \times 3 c c d d e e \begin{matrix} - 30 c c d^{4} \\ - 10 c c d^{4} \end{matrix}$ . $x y^{6} = 20 \times 3 c^{3} d d e e \begin{matrix} - 60 c^{3} d^{4} \\ - 20 c^{3} d^{4} \end{matrix}$ . $x x y y = - 1 \times 2 d e e \begin{matrix} + 2 d^{3} \\ + 2 d^{3} \end{matrix} in \sqrt{e e - d d}$ . $x x y^{3} = - 3 \times 2 c d e e \begin{matrix} + 6 c d^{3} \\ + 6 c d^{3} \end{matrix}$ . $x x y^{4} = - 6 \times 2 c c d e e \begin{matrix} + 12 c c d^{3} \\ + 12 c c d^{3} \end{matrix}$ . $x x y^{5} = - 10 \times 2 c^{3} d e e \begin{matrix} + 20 c^{3} d^{3} \\ + 20 c^{3} d^{3} \end{matrix}$ . $x^{3} y = 1 \times 1 e^{4} \begin{matrix} - 2 d d e e + d^{4} \\ - 3 d d e e + 3 d^{4} \end{matrix}$ . $x^{3} y y = 1 \times 2 c e^{4} \begin{matrix} - 4 c e e d d + 2 d^{4} c \\ - 6 c e e d d + 6 c d^{4} \end{matrix}$ . $x^{3} y^{3} = 3 c c e^{4} \begin{matrix} - 6 c c e e d d + 3 c c d^{4} \\ - 9 c c e e d d + 9 c c d^{4} \end{matrix}$ . $x^{3} y y = 4 c^{3} e^{4} &c.$ $x^{4} = 4 e e d - 4 d^{3} in \sqrt{e e - d d}$ . $x^{4} y = 4 c e e d - 4 c d^{3} in \sqrt{e e - d d}$ . $x^{4} y y = - 4 c c d \sqrt{e^{6} - 3 e^{4} d d + 3 e e d^{4} - d^{6}}$ . $x^{4} y^{3} = - 4 c^{3} d \sqrt{e^{6} &c}$ .

^[30] $y^{5} = - 1 \times 10 e e d^{3} + 10 d^{5}$ . $y^{6} = - 6 \times 10 c e e d^{3} + 60 c d^{5}$ . $y^{7} = - 21 \times 10 c c e e d^{3} + 210 c c d^{5}$ . $y^{8} = - 56 \times 10 c^{3} e e d^{3} + 560 c^{3} d^{5}$ . $x y^{4} = - 1 \times 6 e e d d \begin{matrix} - 6 d^{4} \\ - 4 d^{4} \end{matrix} in \sqrt{e e - d d}$ . $x y^{5} = - 5 \times 6 c e e d d \begin{matrix} - 30 c d^{4} \\ - 20 c d^{4} \end{matrix} in \sqrt{e e - d d}$ . $x y^{6} = - 15 \times 6 c c e e d d \begin{matrix} - 90 c c d^{4} \\ - 60 c c d^{4} \end{matrix} in \sqrt{e e - d d}$ . $\begin{matrix} x x y^{3} = - 1 \times 3 e^{4} d & + 6 e e d^{3} & - 3 d^{5} \\ + 6 e e d^{3} & - 6 d^{5} \\ - 1 d^{5} \end{matrix}$ . $\begin{matrix} x x y^{4} = - 4 \times 3 c e^{4} d & + 24 c e e d^{3} & - 12 c d^{5} \\ + 24 c e e d^{3} & - 24 c d^{5} \\ - 4 c d^{5} \end{matrix}$ . $\begin{matrix} x x y^{5} = - 10 \times 3 c c e^{4} d & + 60 c c e e d^{3} & - 30 c c d^{5} \\ + 60 c c e e d^{3} & - 60 c c d^{5} \\ - 10 c c d^{5} \end{matrix}$ . $\begin{array}{l} x^{3} y y = e^{4} & - 2 d d e e & + d^{4} \\ - 6 d d e e & + 6 d^{4} \\ + 3 d^{4} \end{array} in \sqrt{e e - d d}$ . $\begin{array}{l} x^{3} y^{3} = 3 \times 1 c e^{4} & - 6 c d d e e & + 3 c d^{4} \\ - 18 c d d e e & + 18 c d^{4} \\ + 9 c d^{4} \end{array} in \sqrt{e e - d d}$ . $\begin{array}{l} x^{3} y^{4} = 6 c c e^{4} & - 12 c c d d e e & + 6 c c d^{4} \\ - 36 c c d d e e & + 30 c c d^{4} \\ + 18 c c d^{4} \end{array} in \sqrt{e e - d d}$ . $\begin{array}{l} x^{4} y = 4 d e^{4} & - 9 d^{3} e e & + d^{5} \\ - 6 d^{3} e e & + d^{5} \end{array}$ . $\begin{array}{l} x^{4} y y = 2 \times 4 c d e^{4} & - 16 c d^{3} e e & + 8 c d^{5} \\ - 12 c d^{3} e e & + 12 c d^{5} \end{array}$ . $\begin{array}{l} x^{4} y^{3} = 3 \times 4 c c d e^{4} & - 24 c c d^{3} e e & + 12 c c d^{5} \\ - 18 c c d^{3} e e & + 18 c c d^{5} \end{array}$ . $x^{5} = 10$ {illeg} $d d - 10 d^{4}$ {illeg}{ $\sqrt{e e - d d}$ }. $10 c$ {illeg} $d d - 10 c d^{4}$ {illeg} $\sqrt{e e - d d} = x^{5} y$ . $x^{5} y y = 10$ {illeg} $c d d \sqrt{e^{6} - 3 e^{4} d d + 3 e e d^{4} - d^{6}}$

<18v>

T{illeg}|h|e use of y^e precedent table in fin{illeg}ding y^e Axes of crooked Lines, declared by Examples.

^[31] Suppose I had this Equation given, $x x - 2 x y + a y + y y$ . That I may find y^e axis of y^e line signified {illeg}|b|y it, first I observe of how many dimensions one of y^e unknowne quanties {sic} \or y^e rectang of y^m both/ is found at most in y^e Equation, (as in this Example they have noe more yⁿ 2) then I take every quantiy {sic} in w^ch one of y^e unknowne quantitys is of or y^e rectangle of y^m both is of soe many dimensions (w^ch in this case are $x x - 2 x y + y y$ .) Then lookeing in y^e Table, (either amongst y^e rules of y^e first or 2^d sort &c) for such a rule in w^ch y^e \f{illeg}|i|rst/ quantity {illeg} is {illeg} of soe many dimensions {illeg} I substitute y^e valors of y^e unknowne quantitys, found by y^t rule, into their place in y^e Equation \selected quantitys/ & supposeing y^e product $= 0$ , I find y^e proportion of d to e thereby, that is I {illeg} find y^e angle w^ch y^e axis makes w^th y^e unknowne quantity called x . As in this case I take y^e 2^d Rule of y^e first sort, & by it I find y^e valors of $x x$ {illeg} $x x = 2 d \sqrt{e e - d d}$ . $x y = e e - 2 d d$ . $y y = - 2 d \sqrt{e e - d d}$ . w^ch valors substituting into y^e roome of y^e unknown quantitys {illeg}in these selected termes $x x - 2 x y + y y$ . I have this equation. $2 d \sqrt{e e - d d} - 2 e e + 4 d d - 2 d \sqrt{e e - d d} = 0$ . or, $2 d d = e e$ . & $e = d \sqrt{2}$ so y^t by assuming any quantity for $e e$ as a I have y^e valor of d , for {illeg} $d = \frac{a}{\sqrt{2}}$ . therefore $d ∶ d \sqrt{2} ∷ d ∶ e ∷ \frac{a}{\sqrt{2}} ∶ a ∷ a ∶ a \sqrt{2}$ &c. In y^e next place y^t I may find y^e length of y^e line $bf = c$ . I take another rule whose first quantity is {illeg} not of soe many nor of fewer dimensions yⁿ one of y^e unknowne quantitys or y^e rectangle y^m both is any \some{illeg}/ where in y^e Equa{illeg}|t|ion. Then select every quantity out of y^e Equation, y^e valor of whose unknowne quantity may be found by this rule, & substituting their valors, found thereby, into y^r pl{illeg}|a|ces in this /these\ selected termes make y^e p{illeg}|r|oduct $= 0$ . & find y^e valor of c thereby. {illeg}|A|s in this example I must take y^e first rule of y^e 1^st sort. By w^ch I find $x y = c \sqrt{e e - d d}$ , $y = - d$ , $y y = - 2 d c$ : but y^e valor of $x x$ ci|a|{illeg}|n|not be found by it. therefore I onely take y^e termes $- 2 x y + a y + y y$ , & by substituting y^e valors of y^e unknown quantitys into their roomes I have $- 2 c \sqrt{e e - d d} - a d - 2 d c = 0$ . Then by {illeg}|s|ubstituting y^e \about found/ valors of $d = a$ & $e = \sqrt{2 a a}$ into y^r places, it is $\overset{\cup}{O} 2 a c + a a + 2 a c$ . Or {illeg} $+ 2 a c + a a + 2 a c$ . or $c = - \frac{a}{4}$ . Soe y^t if I make b y^e beginning of x , & $+ x$ to tend towards c \in y^e line bc /, & $+ y$ towards k perpendicularly to bc . then must I draw $bf = - \frac{a}{4}$ from y^e point b perpendicular to bc ; & $fe = a$ , & parallell to bc ; yⁿ $eg = \sqrt{{gf}^{2} - {fe}^{2}}$ $eg = a$ , & parallell to bf . Lastly through y^e points f & g draw gf y^e axis of y^e line sought. Otherwise suppose $\sqrt{e e - d d}$ : it may be done thus {illeg} $eg ∶ ef ∷ - bf ∶ bh$ . therefore I take $bh = \frac{- c d}{\sqrt{e e - d d}} = \frac{a}{4}$ , & through y^e points f & h $\sqrt{e e - d d} ∶ d ∷ - c ∶ \frac{- c d}{\sqrt{e e - d d}}$ . I draw af y^e axis sought.

^[32] Example y^e 2^d. If y^e Equation bee $x^{3} - a x y + y^{3} = 0$ . y^e Rule whose first quantity is of as many dimensions as either of y^e {illeg} unknowne quantitys {illeg}|i|n {illeg} \this/ Equation, is y^e 3^d of y^e first sort or y^e first of y^e 2^d sort. Selecting therefore \onely/ $x^{3} + y^{3}$ out of y^e Equation (since {illeg} \in/ neither of the{illeg}|se| \rules/ y^e valor{illeg} of $x y$ {illeg}|i|s found) by y^e 3^d rule of y^e first sort I find $x^{3} = 3 d d \sqrt{e e - d d}$ , $y^{3} = 3 d^{3} - 3 d e e$ . therefore y^e selected termes $x^{3} + y^{3} = 3 d d \sqrt{e e - d d} + 3 d^{3} - 3 d e e = 0$ . & $d \sqrt{e e - d d}$ {illeg} $e e - d d$ . Or, $e e = 2 d d$ . In like manner by y^e first rule of y^e 2^d sort tis found $y^{3} = - d^{3}$ . $x^{3} = e e - d d in \sqrt{e e - d d}$ . & therefore $x^{3} + y^{3} =$ $\sqrt{e^{6} - 3 e^{4} d d + 3 e e d^{4} - d^{6}} - d^{3} = 0$ . & $\sqrt{c: e^{6} - 3 e^{4} d d + 3 e e d^{4} - d^{6}} = d d$ . Or $e e = 2 d d$ as before. Soe y^t $eg ∶ fe ∷ \sqrt{d d - e e} ∶ d ∷ d ∶ d$ . therefore $eg = fe$ . Now y^t I may find $bf = c$ I take y^e 2^d Rule of y^e first sort (whose first quantity $y y$ is of fewer dimensions yⁿ $x^{3}$ or $y^{3}$ but not of fewer $x y$ ,) The quatitys {sic} in y^e Equation whose valors are expressed in this rule are $x y$ , & $y^{3}$ for $x y = e e - 2 d d$ . $y^{3} = - 6 c d \sqrt{e e - d d}$ . {illeg}|S|oe y^t I write $- 6 c d \sqrt{e e - d d} - a e e + 2 a d d$ instead of $y^{3} - a x y$ . soe y^t $c = \frac{2 a d d - a e e}{6 d d \sqrt{e e - d d}}$ . Or since $2 d d - e e = 0$ , it is, $c = \frac{0 \times a}{6 d d} = 0$ . Had I taken y^e first rule of y^e fir{illeg}st sort I had found $x y = c \sqrt{e e - d d}$ . & $y^{3} = - 3 c d d$ . therefore $y^{3} - a x y = - 3 c c d - a c \sqrt{e e - d d} = 0$ w^ch is right since $c = 0$ . but by this equation c hath other valors for $3 c d - a \sqrt{e e - d d} = 0$ or $3 c \overset{\cup}{O} a = 0$ , & $c = \frac{\overset{\cup}{O} a}{3}$ . &c. Whence obse{illeg}|r|ve y^t {illeg}|fo|r y^e most {pt} it will bee most convenient to find y^e {illeg} c by y^t rule whose 1^st quantity hath one dimension lesse yⁿ y^e first quantity of y^t rule by w^ch y^e proportion twixt d & e were found.

<20r>

Example y^{illeg}|e| 3^d. If y^e Equation be $\begin{matrix} x x y y & + & 4 b x y y & + & 4 b b y y & = & 0 \\ - & 2 a x x y & - & 8 a b x y & - & 8 a b b y \\ - & a^{4} \end{matrix}$ . $x x y y$ being of 4 dimensions I take y^e 4^th rule of y^e first sort, or y^e 2^d rule of y^e 2^d sort. By y^e 4^th rule of y^e 1^st sort I find $x x y y = 2 d e e \sqrt{e e - d d} - 4 e^{3} \sqrt{e e - d d}$ & since by that rule I can find y^e valor of noe other quantity in y^e Equation I {illeg} {illeg}|m|ake $x x y y = \begin{matrix} 2 d e e \\ - 4 d^{3} \end{matrix} \sqrt{e e - d d} = 0$ . Or {illeg} $e^{6} - 5 d d e^{4} + 8 d^{4} e e - 4 d^{6} = 0$ . Whose rootes are $e e - d d$ , $e e - 2 d d$ , & $e e - 2 d d$ . therefore either $e e = d d$ , or $e e = 2 d d$ Which is divisible by d , & $e e - 2 d d$ , & by $\sqrt{e e - d d}$ . therefore either $d = 0$ ; or, $e e - 2 d d = 0$ ; or, $e e - d d = 0$ . The operation is y^e same if I make use of y^e 2^d rule of y^e 2^d sort. Againe I take y^e 3^d rule of y^e 1^st sort & by it I find, $x y y = e e \sqrt{e e - d d} - 3 d d \sqrt{e e - d d}$ . $x x y = 2 d e e - 3 d^{3}$ . therefore $4 b x y y - 2 a x x y \begin{matrix} - 4 e e \\ - 12 d d \end{matrix} \sqrt{e e - d d} + 6 d^{3} - 4 d e e = 0$ . & if $d = 0$ . then & $x x y y = 4 c d e e - 6 c d^{3}$ . therefore $\begin{matrix} x x y y \\ - 2 a x x y \end{matrix} + 4 b x y y = \begin{matrix} 4 c d e e - 4 a d e e + 4 b e e \\ - 6 c d^{3} + 6 a d^{3} - 12 b d d \end{matrix} \sqrt{e e - d d} = 0$ . Or $c = a \frac{+ 2 b e e \sqrt{e e - d d} - b b d d \sqrt{e e - d d}}{3 d^{3} - 2 d e e}$ . & if $d = 0$ , then $c = a + \frac{2 b e e \sqrt{e e - d d}}{0}$ . or c is infinitely long. but if $e e - 2 d d = 0$ . then $c = a \overset{\cup}{O} 4 b \overset{\cup}{O} 6 b$ . & if $e e - d d = 0$ , then $c = a$ . Againe I take y^e fir{illeg}|s|t rule of y^e 2^d sort & by it I find $x x y y = 2 c d^{3} - 2 c d e e$ . $x x y = d^{3} - e e d$ . $x y y = d d \sqrt{e e - d d}$ . therefore $x x y y - 2 a x x y + 4 b x y y = 2 c d^{3} - 2 c d e e - 2 a d^{3} + 2 a e d d + 4 b d d \sqrt{e e - d d} = 0$ And $c = a + \frac{2 b d}{\sqrt{e e - d d}}$ Now if $d = 0$ . then Or $c = a d^{3} + a e d d$ Now if $d = 0$ . or if $e e - d d = 0$ , then y^e termes \of this Equat/ destroy one another soe y^t y^e valor of c may not be found thereby. but if $e e - 2 d d = 0$ , then I find $2 c d^{3} - 4 c d^{3} - 2 a d^{3} + 4 a d^{3} + 4 b d d \sqrt{d d} = 0$ . Or $2 c d^{3} = 2 a d^{3} \overset{\cup}{O} 4 b d^{3}$ . Or $c = a \overset{\cup}{O} 2 b$ . Againe I take y^e 2^d rule of y^e 1^st sort & by it I find $\begin{matrix} x x y y - 2 a x x y + 4 b x y y \\ + 4 b b y y - 8 a b x y \end{matrix} = 2 c d d \sqrt{e e - d d} - 8 b b d \sqrt{e e - d d} - 4 a c d \sqrt{e e - d d} + 16 d d a b - 8 a b e e - 16 c d d b + 8 c e e b = 0$ . & if $d = 0$ then $- 8 a b e e + 8 c e e b = 0$ , or / $c c = 2 a c \sqrt{e e} + \frac{c e e b}{0 \times \sqrt{e e}}$ \ If $d = 0$ then $c c = 2 a c \frac{- 4 c e e b}{0 \times \sqrt{e e}} + 4 b b + \frac{4 a b e e}{0 \sqrt{e e}}$ . or c {illeg} $a \overset{\cup}{O} \frac{2 e b}{0} \overset{\cup}{O} \sqrt{\frac{4 e e b b}{0 \times 0} \frac{\begin{matrix} + 4 a e b \\ \overset{\cup}{O} 4 a e b \end{matrix}}{0} \begin{matrix} + a a \\ + 4 b b \end{matrix}}$ . that is c is infinitely long as was found before. If $e e - d d = 0$ also i{illeg}|t| may bee found to bee $8 c e e b - 8 a e e b = 0$ , or $c = a$ but upon this supposition $d = 0$ it was not before found $c = a$ & therefore $c = a$ is false, when $d = 0$ . If $e e - d d = 0$ . then I find $8 a b d d - 8 c b d d = 0$ . or $c = a$ . &c. If $e e - 2 d d = 0$ . then $c c \sqrt{d d} - 4 b b \sqrt{d d} - 2 a c \sqrt{d d} = 0$ . $c = a \overset{\cup}{O} \sqrt{a a - 4 b b}$ . Which valor not being found before I conclude $e e - 2 d d = 0$ to bee false. Lastly by useing y^e first eq rule of y^e first sort I find, $4 b c c \sqrt{e e - d d} - 8 a b c \sqrt{e e - d d} - 8 c d b b + 8 a b b d = \begin{matrix} 4 b x y y + 4 b b y y \\ - 8 a b x y - 8 a b b y \end{matrix} = 0$ . & by supposeing $d = 0$ I have {{illeg} $c =$ {illeg} $2 a$ }. & $c = a$ . & if $e e = d d$ , then $c = a$ w^ch being always found upon y^e supposition $d d = e e$ . I conclude y^e valor of $d d$ to be $e e$ & of c {illeg} to be a {illeg}. & so draw y^e axis gf parallell to x & distant from it y^e length of a . But here of|b|serve y^t this might have beene better performed by taking away y^e 2^d terme of y^e Equation $\begin{matrix} x x y y & + & 4 b x y y & + & 4 b b y y & = & 0 \\ - & 2 a x x y & - & 8 a b x y & - & 8 a b b y \\ - & a^{4} \end{matrix}$ . Or $x x + 4 b x + 4 b b \frac{- a^{4}}{y y - 2 a y} = 0$ as was {illeg} observed before.

<21r>

^[33]To find y^e Diameter or axis of any crooked line which hath it

^[34] Suppose y^e \crooked/ line to bee $(lgc)$ , y^e \diameter or/ Axis $(kd)$ , y^e unde{illeg}|t|ermined quantitys describing y^e line to be $ab = x$ , $bc = y$ . \to/ from y^e begining point a (y^e begining of x), perpendicular to kb draw $ah = pb = c$ , {illeg} cutting y^e axis kd in h . paralell {sic} to kb draw $hp = x$ . & produce cb {illeg} soe y^t it intersect y^e axis in y^e point d . & suppose y^t hp is to hd . as d to e : or y^t $hd = \frac{e x}{d}$ . & therefore $dp = \frac{x \sqrt{e e - d d}}{d}$ . le{illeg}|t| ec bee one $= el = ϩ$ be one of those lines w^ch are ordinately applied to y^e diameter $he = ϩ$ . & lastly suppose that {illeg} ec is to ef , {illeg}|as| {illeg}| d | to f : yⁿ is $ef = \frac{f ϩ}{d}$ . & $fc = \frac{ϩ \sqrt{e e - f f}}{f}$ . then $hp ∶ pd ∷ fc ∶ fd$ {illeg}|.| / $x ∶ \frac{x \sqrt{e e - d d}}{d} ∷ \frac{ϩ \sqrt{d d - f f}}{f} ∶ fd$ .\ therefore $fd = \frac{ϩ \sqrt{e e d d - e e f f - d d d d + d d f f}}{d f}$ . & $hd = \frac{e x}{d} = \frac{d f ϱ + ϩ \sqrt{e^{4} - e e f f - e e d d + d d f f}}{d f}$ draw cf perpendicular to hd y^e Axis &
& {sic} by ordering y^e Equation it will bee & $hd = \frac{e x}{d} =$ {illeg} $ϱ - \frac{f ϩ}{d} + \frac{ϩ \sqrt{e e d d - e e f f - d^{4} + d d f f}}{d f}$ & by ordering y^e equation lastly suppose y^t ec is to ef as e is to f : yⁿ is $ef = f ϩ e$ ; & $fc =$ {illeg} $\frac{ϩ \sqrt{e e - f f}}{e}$ then $hp ∶ pd ∷ fc ∶ fd$ .
$x ∶ \frac{x \sqrt{e e - d d}}{d} ∷ \frac{ϩ \sqrt{e e - f f}}{f} ∶ \frac{ϩ \sqrt{e^{4} - e e f f - e e d d + d d f f}}{d e} = fd$ & $hd = ϱ - \frac{f ϩ}{e} + \frac{ϩ \sqrt{e^{4} - e e f f - e e d d + d d f f}}{d e} = \frac{e x}{d}$ . & by ordering y^e Equation it will be, $x =$ $d d$ {illeg} $\frac{e d ϱ - d f ϩ + ϩ \sqrt{e^{4} - e e f f - e e d d + d d f f}}{e e}$ . Againe, $d ∶ e ∷ fc ∶ cd = \frac{ϩ \sqrt{e e - f f}}{d}$ . & $dp = \frac{x \sqrt{e e - d d}}{d}$ , or by substituteing {illeg}|t|he valor of x into its ro $dp + pb - dc = y = \frac{x \sqrt{e e - d d} + dc - ϩ \sqrt{e e - f f}}{d}$ or by substituteing y^e valor of x into its place it is $y = \frac{e ϱ \sqrt{e e - d d} + f ϩ \sqrt{e e - d d} - d ϩ \sqrt{e e - f f} + e e c}{e e}$ . {illeg}|A|nd y^t I may abbreviate y^e termes I make $\sqrt{e e - d d} = t$ ; & $\frac{+ f \sqrt{e e - d d} + d \sqrt{e e - f f}}{e} = v$ ; & so y^e Equation is $y = \frac{t ϱ - v ϩ + e c}{e}$ . Also by supposeing $s = \frac{\sqrt{e^{4} - e e f f + d d f f - e e d d} - d f}{e}$ , I lesson {sic} y^e termes of y^e Equation $x = \frac{e d ϱ - d f ϩ + ϩ \sqrt{e^{4} - e e f f - e e d d + d d f f}}{e e}$ , by writeing instead $\frac{d ϱ + s ϩ}{e} = x$ .

Now therefore by meanes of t substituting these valors of x & y into y^r stead I take y^m out of y^e Equation expressing y^e {illeg} reltion {sic} twixt y^m soe y^t yⁿ I have an equation expressing y^e relation twixt ϱ & ϩ . And to that end it will bee convenient to have a table of y^e squares, cube, squaresquares, square=cubs, rectangles \&c/ of y^e valors of x & y , {illeg} After y^e manner of y^t w^ch follows.

$x = \frac{d ϱ + s ϩ}{e}$ . $x x = \frac{s s ϩ ϩ + 2 d s ϱ ϩ + d d ϱ ϱ}{e e}$ . $x^{3} = \frac{s^{3} ϩ^{3} + 3 d ϱ s s ϩ ϩ + 3 d d ϱ ϱ s ϩ + d^{3} ϱ^{3}}{e^{3}}$ . $x^{4} = \frac{s^{4} ϩ^{4} + 4 d ϱ s^{3} ϩ^{3} + 6 d d ϱ ϱ s s ϩ ϩ + 4 d^{3} ϱ^{3} s ϩ + d^{4} ϱ^{4}}{e^{4}}$ . $x^{5} = \frac{s^{5} ϩ^{5} + 5 d ϱ s^{4} ϩ^{4} + 10 d d ϱ ϱ s^{3} ϩ^{3} + 10 d^{3} ϱ^{3} s s ϩ ϩ + 5 d^{4} ϱ^{4} s ϩ + d^{5} ϱ^{5}}{e^{5}}$ . \&c./ $y = \frac{t ϱ - v ϩ + e c}{e}$ . {illeg} $y y = \frac{\begin{matrix} v v ϩ ϩ & - & 2 t ϱ v ϩ & + & t t ϱ ϱ \\ - & 2 e c v ϩ & + & 2 e c t ϱ \\ + & e e c c \end{matrix}}{e e}$ . $y^{3} = \frac{\begin{matrix} - & v^{3} ϩ^{3} & + & 3 t ϱ v v ϩ ϩ & - & 3 t t ϱ ϱ v ϩ & + & t^{3} ϱ^{3} \\ + & 3 e c v v & - & 6 e c t ϱ v & + & 3 t t ϱ ϱ e c \\ - & 3 e e c c v & + & 3 t ϱ e e c c \\ + & e^{3} c^{3} \end{matrix}}{e^{3}}$ . $\begin{array}{l} y^{4} & = & v^{4} ϩ^{4} & - & 4 t ϱ v^{3} ϩ^{3} & + & 6 t t ϱ ϱ v v ϩ ϩ & - & 4 t^{3} ϱ^{3} v ϩ & + & t^{4} ϱ^{4} \\ - & 4 e c & + & 12 c e t ϱ & - & 12 t t ϱ ϱ c e v & + & 4 c e t^{3} ϱ^{3} \\ + & 6 c c e e & - & 12 c c e e t ϱ v & + & 4 c^{3} e^{3} t ϱ \\ - & 4 c^{3} e^{3} v & + & c^{4} e^{4} \end{array}$ . $\begin{array}{l} y^{5} & = & - & v^{5} ϩ^{5} & + & 5 t ϱ v^{4} ϩ^{4} & - & 10 t t ϱ ϱ v^{3} ϩ^{3} & + & 10 t^{3} ϱ^{3} v v ϩ ϩ & - & 5 t^{4} ϱ^{4} v ϩ & + & t^{5} ϱ^{5} \\ + & 5 c e & - & 20 c e t ϱ & + & 30 t t ϱ ϱ c e & - & 20 c e t^{3} ϱ^{3} & + & 5 c e t^{4} ϱ^{4} \\ - & 10 c c e e & + & 30 c c e e t ϱ & - & 30 c c e e t t ϱ ϱ & + & 10 c c e e t^{3} ϱ^{3} \\ + & 10 c^{3} e^{3} & - & 20 c^{3} e^{3} t ϱ & + & 10 c^{3} e^{3} t^{2} ϱ^{2} \\ - & 5 c^{4} ϱ^{4} & + & 5 c^{4} e^{4} t ϱ \\ + & c^{5} e^{5} \end{array}$ . \&c/ $x y = \frac{\begin{matrix} - & s v ϩ ϩ & + & e ϱ s ϩ & + & d t ϱ ϱ \\ + & e c s & + & d e c ϱ \\ - & d v ϱ \end{matrix}}{e e}$ . $x x y = \frac{\begin{matrix} - & s s v ϩ^{3} & + & s s t ϱ ϩ ϩ & + & 2 e c d s ϱ & + & e c d d ϱ ϱ \\ + & s s e c & + & 2 t d s ϱ ϱ & + & t d d ϱ^{3} \\ - & 2 d s v ϱ & - & d d v \end{matrix}}{e e e}$ . $x^{3} y = \frac{\begin{matrix} - & v s^{3} ϩ^{4} & + & 2 t ϱ s^{3} ϩ^{3} & + & 3 e c d s s ϱ ϩ^{2} & + & 3 e c d d s ϱ ϱ ϩ & + & e c d^{3} ϱ^{3} \\ + & e c s^{3} & + & 3 t d s ϱ^{2} ϩ^{2} & + & 3 t d d s ϱ^{3} & + & t d^{3} ϱ^{4} \\ - & 3 d ϱ s s v & - & 3 d d s v ϱ ϱ & - & d^{3} v ϱ^{3} \end{matrix}}{e^{4}}$ . &c. $x y y = \frac{\begin{matrix} v v s ϩ^{3} & - & 2 t ϱ v s ϩ ϩ & + & s t t ϱ ϱ ϩ & + & t t d ϩ^{3} \\ - & 2 e c v s & + & 2 s t c e ϱ & + & 2 e c t d ϱ^{2} \\ + & d v v ϱ & + & s c c e e & + & e e c c d ϱ \\ - & 2 d t v ϱ ϱ \\ - & 2 d ϱ e c v \end{matrix}}{e^{3}}$ . $x x y y = \frac{\begin{matrix} s s v v ϩ^{4} & - & 2 s s t v ϱ ϩ^{3} & + & s s t t ϱ ϱ ϩ^{2} & + & 2 t t d s ϱ^{3} ϩ & + & d d t t ϱ^{4} \\ - & 2 s s e c v & + & 2 s s e c t ϱ & + & 4 e c t d s ϱ^{2} & + & 2 d d e c t ϱ^{3} \\ + & 2 d s ϱ v v & + & s s e e c c & + & 2 e e c c d s ϱ & + & d d e e c c ϱ ϱ \\ - & 4 d s ϱ ϱ t v & - & 2 d d t v ϱ^{3} \\ - & 4 d s ϱ e c v & - & 2 d d e c v ϱ ϱ \\ + & d d v v \end{matrix}}{e^{4}}$ . $x y^{3}$
$x y^{3} = \frac{\begin{array}{l} - & s v^{3} ϩ^{4} & + & 3 s t ϱ v v ϩ^{3} & - & 3 s t t ϱ ϱ v ϩ^{2} & + & s t^{3} ϱ^{3} ϩ & + & d t^{3} ϱ^{4} \\ + & 3 s e c v v & - & 6 s e c t ϱ v & + & 3 s t t ϱ ϱ e c & + & 3 d t t ϱ^{3} c e \\ - & d ϱ v^{3} & - & 3 s e e c c v & + & 3 s t ϱ e e c c & + & 3 d t ϱ ϱ c c e e \\ - & 3 d t ϱ ϱ v v & - & 3 d t t ϱ^{3} v \\ + & 3 d e c v v & - & 6 d e c t ϱ ϱ v \\ - & 3 d e e c c v \end{array}}{e^{4}}$ . &c $\begin{matrix} If there bee occasions to doe \\ these opera \overline{co} ns in Equations \\ of 5 or 6 or more dimensions \\ this table may be easily enlarged. \end{matrix}$

As for example. If y^e relation twixt x & y bee exprest in this Equation, $x x + a x - 2 x y + y y = 0$ . then into y^e place of $x x$ , x , $x y$ , $y y$ , I substitute their valors found by this table, & there results $\begin{array}{l} + & s s ϩ ϩ & + & 2 d s ϱ ϩ & + & d d ϱ ϱ & = & 0 \\ + & 2 s v & + & a e s & + & a d e ϱ \\ + & v v & - & 2 t s ϱ & - & 2 d t ϱ ϱ \\ - & 2 e d s & - & 2 d e c ϱ \\ + & 2 d v ϱ & + & t t ϱ ϱ \\ - & 2 t ϱ v & + & 2 e c t ϱ \\ - & 2 e c v & + & e e c \end{array}$ . Which Equation expresseth y^e relation twixt ϱ & ϩ . y^t is twixt ge & le or ec . Now y^t $ge = ϱ$ be y^e diameter of y^e line & {illeg} & $le = ec = \overset{\cup}{O} ϩ$ be ordinately applied to it, it is required (by Prop 2^d) y^t in this Equation ϩ be not of odd dimensions. & that may bee soe {illeg} y^e quan\ti/tys in th{illeg}|e| 2^d terme \w^ch ϩ is {illeg} of {illeg} / must {illeg} one another w^ch canot {sic} be unlesse thes{illeg}|e| {illeg} quantitys destroy one another in w^ch y^e unknowne quantitys ϱ & ϩ destroy one /{illeg}are of y^e {same}\ <21v> dimensions. Which things being considered it will appeare y^t I must divide y^e 2^d terme into two pts, makeing, $\begin{matrix} 2 d s ϱ ϩ & - & 2 t s ϱ ϩ & + & 2 d v ϱ ϩ & = & 0 \\ - & 2 t v ϱ ϩ \end{matrix}$ ; &, $a e s ϩ - 2 e c s ϩ - 2 e v c ϩ = 0$ . & divideing y^e first by $2 ϱ ϩ$ & y^e 2^d by $e ϩ$ they will be, $d s - t s - t v + d v = 0$ . {illeg} Hitherto useing useing y^e letters s , t , & v for b{illeg}|r|evitys sake, I must now write their valors in theire stead (y^t I may \find/ y^e length of c , & y^e proportion of d to e w^ch determine y^e position of y^e axis, & also y^e proportion of e to f w^ch determines y^e position of y^e lines applyed to y^e axis.) & soe {instead} of y^e Equation $d s - t s - t v + d v = 0$ ; there results, $\begin{array}{r} + d \\ - e \end{array} \sqrt{e^{4} - e e f f + d d f f - d d e e} \begin{array}{r} + d d \\ - e e \\ + d e \end{array} \sqrt{e e - f f} + 2 d f \sqrt{e e - d d} - e e f = 0$ w^ch by squareing /both pts is\ $e^{6} - 2 d e^{5} \begin{array}{r} + d d \\ - f f \\ - d d \end{array} e^{4} \begin{array}{r} + 2 d f f \\ + 2 d^{3} \end{array} e^{3} \begin{array}{r} - d d f f \\ - d^{4} \end{array} e^{2}$ Or $\sqrt{e e - f f} = \frac{e e f - d d f - d f \sqrt{e e - d d}}{d d - e e + d e \begin{array}{r} + d \\ - e \end{array} \sqrt{e e - d d}}$ . Or $\begin{array}{r} + d \\ - e \end{array} \sqrt{e^{4} - e e f f + d d f f - d d e e} \begin{array}{r} + d d \\ - e e \\ + d e \end{array} \sqrt{e e - f f} = e e f + 2 d f \sqrt{e e - d d}$ . And by squareing both pts {illeg} {illeg} it is /& ordering y^e product it is.\ { $2 e^{6} - 4 d e^{5} \begin{matrix} + 2 d d e^{4} \\ - 3 f f e^{4} \\ - 3 d e^{4} \end{matrix} \begin{matrix} + 4 d f f e^{3} \\ + 4 d^{3} \end{matrix} \begin{array}{l} - d d f e e \\ + d^{4} e e \\ + d d f f e e \end{array} - d^{3} f f e \begin{array}{l} - d^{4} f f \\ + 2 d^{3} e e \\ - 2 d^{3} f f \\ + d e^{4} \\ + d e e f f \end{array} \begin{matrix} + 2 e^{5} \\ - 2 e^{3} f f \end{matrix} \sqrt{e e - d d} = \begin{matrix} 2 d^{3} f f \\ 2 d e e f f \end{matrix} \sqrt{e e - f f}$ Or $2 e^{6} - 4 d e^{5} \begin{matrix} - 3 f f e^{4} \\ - d d e^{4} \end{matrix} \begin{matrix} + 4 d f f e^{3} \\ + 4 d^{3} \end{matrix} + 2 d d e e f f - 4 d^{3} f f = \begin{matrix} - 2 d^{3} e e \\ + 4 d^{3} f f \end{matrix} \begin{matrix} + 4 d^{5} \\ - 6 d e e f f \end{matrix} \begin{matrix} + 2 e^{3} f f \\ - 2 e f - 2 e^{5} \end{matrix} in \sqrt{e e - d d}$ .} $\begin{matrix} d d e^{4} \\ - f f e^{4} \end{matrix} \begin{array}{l} + & 3 d d f f e e \\ - & 2 d^{4} e e \end{array} - 2 d^{4} f f = \begin{array}{l} + & 2 e^{4} f \\ + & 2 e^{3} d f \\ - & 6 e e d d f \end{array} \begin{array}{l} - & 4 e d^{3} f \\ + & 4 d^{4} f \end{array} in \sqrt{e e - f f}$ . Which b{illeg}|ei|ng divided by $e e - 2 d d$ the product is $\begin{array}{l} d d e e \\ - & f f e e \end{array} + d d f f = \begin{array}{l} 2 e e f \\ + & 2 e d f \\ - & 2 d d f \end{array} in \sqrt{e e - f f}$ . Wherefore I conclude y^t $e e = 2 d d$ . or y^t $d ∶ e ∷ d ∶ \sqrt{e d d}$ . Againe by inserting \y^e valors/ of s & v into y^e Equation $a s - 2 c s - 2 c v$ there results $a \sqrt{e^{4} - e e f f + d d f f - d d e e} - a d f = 2 c \sqrt{e^{4} - e e f f + d f f - d d e e} + 2 c f \sqrt{e e - d d} + 2 c e \sqrt{e e - f f} - 2 c d f$ . or by $2 d f \sqrt{e e - d d} = e e f - e e \sqrt{e e - f f} - 2 d d \sqrt{e e - f f}$ . & by squareing both pts & ordering y^e product it is, $e^{4} - 4 d d e e + 4 d^{4} = 4 d d f \sqrt{e e - f f} - 2 e e f \sqrt{e e - f f}$ . Which is divisible by $2 d d - e e = 0$ , for y^e quote will bee $2 d d - e e = 2 f \sqrt{e e - f f}$ . & therefore $2 d d = e e$ . Or, $2 d d = e e + 2 f \sqrt{e e - f f}$ . Againe by inserting \y^e valors/ {illeg}|of| s & v into y^e Equation, $a s - 2 c s - 2 c v = 0$ , there resulteth, $\begin{matrix} + & a \\ - & 2 c \end{matrix} \sqrt{e^{4} - e e d d - e e f f + d d f f} - 2 c f \sqrt{e e - d d} - 2 c d \sqrt{e e - d d} - a d f + 2 c d f = 0$ . & by writing $2 d d$ instead of $e e$ & d{illeg}|i|videing it by d there resulteth $\begin{matrix} + & a \\ - & 4 c \end{matrix} \sqrt{2 d d - f f} - a f = 0$ Or $\frac{a}{4} \frac{- a f}{4 \sqrt{2 d d - f f}} = c$ . Thus haveing found y^e proportion of d to e , & y^e valor of c since theire {sic} remaines noe more equations by w^ch I may find y^e proportion of e to f I concluded {illeg} it to be undetermined, soe y^t I may assume any proportion betwixt y^m. As if I make $f = 0$ . Then y^e angle ceh is a right one & eh y^e axis of y^e line, & $c = \frac{a}{4} = ah$ . & $d ∶ \sqrt{2 d d} ∷ d ∶ e ∷ hp ∶ hd$ . or $d ∶ \sqrt{2 d d} ∶ hp ∶ dp$ . that is $hp = dp$ ; As in y^e 1^st figure.^[35] Or if I make $e e ∶ f f ∷ 2 ∶ 1$ . that is $e e = 2 f f$ . or $f = d$ , then I find y^t $c = 0$ . that is that y^e diameter ed intersects y^e line ap at y^e point a y^e begining of x . & y^t y^e lines ec are parallell to ma as in y^e 2^d figure^[36] &c. Soe y^t by assumeing any proportion twixt ec & ef , that is, {illeg}|s|upposeing y^e angles fec of any bigness, y^e position of y^e dia{illeg}|mi|ter fa , {illeg} may be found i|a|fter y^e same manner. As If I would have y^e angle fec to be an anangle of 60 degrees. yⁿ must $ec = ϩ$ be double to $fe = \frac{1}{2} ϩ$ , & $fc = \sqrt{\frac{3}{4} y y}$ . i.e. $e ∶ f ∷ 2 ∶ 1$ . & $2 d d = e e$ , therefore $\frac{d d}{2} = f f$ . I found before y^t $\frac{a}{4} \frac{- a f}{4 \sqrt{2 d d - f f}} = c$ . or writeing y^e valor of f in its roome, tis $\frac{a}{4} \frac{- a \sqrt{\frac{1}{2} d d}}{4 \sqrt{\frac{3 d d}{2}}} = c$ that is $\frac{a}{4} \overset{\cup}{O} \frac{a}{4 \sqrt{3}} = c$ . Or $\frac{a}{4}$ since c must be lesse yⁿ $\frac{a}{4}$ it must $\frac{a}{4} - \frac{a}{4 \sqrt{3}} = c = ah$ . & $ph = dp$ since $e e = 2 d d$ . As in y^e 3^d figure.^[37] But if I would make y^e angle ceh of 60 degr: th{illeg}|e|n as before $e = 2 f$ , & $\frac{d d}{2} = f f$ , & $\frac{a}{4} \overset{\cup}{O} \frac{a}{4 \sqrt{3}} = c$ , or since c must be greater yⁿ $\frac{a}{4}$ tis $\frac{a}{4} + \frac{a}{4 \sqrt{3}} = c$ , as in y^e 4^th fig. &c.^[38]

Example 2^d. If y^e Equation expressing y^e nature of y^e line be $x^{3} - 3 x y y + 3$ $\begin{array}{l} x^{3} & - & 3 x x y & + & 3 x y y & - & y^{3} & = & 0 \\ - & a y y \end{array}$ .

<23v>

^[39] To find y^e Axis or Diame{illeg}|t|er of any crooked Line supposeing it have y^m.

^[40] ^[41] ^[42] Suppose $bc = x$ ; $cd = y$ ; {illeg} nad y^e line whose axis \or Diameter/ is sought; pk its axis or Diameter; a its vertex; $hd = hn = ϩ =$ lines ordinately applied to its Diameter; bm a perpendicular to pc drawne from y^e point b , i.e. from y^e begining of x ; $bf = c =$ pte of y^e line bm intercepted twixt y^e diameter & pc ; $mh = ϱ =$ a line parallell to bc & drawne from y^e bm to y^e intersection of fk & nd ; & $he = mf$ & parallell to mf ; & dgh a right angled triangle. $dh ∶ hg ∷ d ∶ f$ &, $mh = fe ∶ he = mf ∷ d ∶ e$ .

Then since {illeg} $d ∶ e ∷ ϱ ∶ he$ therefore $\frac{e ϱ}{d} = he$ {illeg}

Then $d ∶ e ∷ ϩ ∶ \frac{f ϩ}{d} = hg = eq$ . &, $eq + fe = \frac{d ϱ + f ϩ}{d} = x$ . Againe $gd = \sqrt{\frac{d d ϩ^{2} - f f ϩ^{2}}{d d}}$ $d ∶ e ∷ ϱ ∶ he = \frac{e ϱ}{d} = gq$ . &, $gq + qc + dg = cd$ ; or, $\frac{e ϱ + d c + ϩ \sqrt{d d - f f}}{d} = y$ .

Now therefore {illeg} by substituteing $\frac{d ϱ + f ϩ}{d}$ into y^e place of x , & $\frac{e ϱ + d c - ϩ \sqrt{d d - f f}}{d}$ into y^e place of y , & theire □s & cubes &c: into y^e place of $x^{2}$ , $x^{3}$ , $y^{2} - y^{3}$ &c. I take x & { y } out of y^e Equation expressing y^e relation twixt y^m & {illeg}|S|oe have an Equation expressing y^e relation twixt ϱ & ϩ . And to y^t end it will be convenient to have a table of y^e squares, cubes, & rectangles &c: of y^e valors of x & y , like y^t w^ch follows.

$\begin{array}{l} \begin{array}{l} \begin{array}{l} d x & = & d ϱ + f ϩ \\ d d x x & = & d d ρ ϱ + 2 d ϱ f ϩ + f f ϩ^{2} \\ d^{3} x^{3} & = & f^{3} ϩ^{3} + 3 f f ϩ ϩ d ϱ + 3 d d ϱ ϱ f ϩ + d^{3} ϱ^{3} \\ d^{4} x^{4} & = & f^{4} ϩ^{4} + 4 d ϱ f^{3} ϩ^{3} + 6 d d ϱ^{2} f^{2} ϩ^{2} + 4 d^{3} ϱ^{3} f ϩ + d^{4} ϱ^{4} \\ d y & = & \begin{matrix} + & ϩ \sqrt{d d - f f} & + & e ϱ \\ + & d c \end{matrix} \\ d d y y & = & \begin{array}{l} + & d d ϩ ϩ & + & 2 e ϱ ϩ \sqrt{d d - f f} & + & e e ϱ ϱ \\ + & 2 d c & + & 2 e d c ϱ \\ + & d d c c \end{array} \end{array} & \begin{matrix} d d x y & = & \begin{array}{l} + & f ϩ^{2} \sqrt{d d - f f} & + & f e ϱ ϩ & + & d e ϱ ϱ \\ + & f d c ϩ & + & d d c ϱ \\ + & d ϱ ϩ \sqrt{d d - f f} \end{array} \\ d^{3} x^{2} y & = & \begin{array}{l} + & f f ϩ^{3} \sqrt{d d - f f} & + & f f e ϱ ϩ^{2} & + & 2 e f ϱ^{2} ϩ & + & d d e ϱ^{3} \\ + & f f d c & + & 2 d d c f ϱ & + & d^{3} c ϱ ϱ \\ + & 2 d f ϱ \sqrt{d d - f f} & + & d d ϱ ϱ \sqrt{d d - f f} \end{array} \\ d^{4} x^{3} y & = & \begin{array}{l} + & f^{3} ϩ^{4} \sqrt{d d - f f} & + & e f^{3} ϱ ϩ^{3} & + & 3 d e f f ϱ ϱ ϩ ϩ & + & 3 d d e f ϱ ϱ ϩ ϩ & + & d^{3} e ϱ^{4} \\ + & d c f^{3} & + & 3 d d c f f ϱ & + & 3 c d^{3} f ϱ ϱ & + & d^{4} c ϱ^{3} \\ + & 3 f f d ϱ \sqrt{d d - f f} & + & 3 d d f ϱ ϱ \sqrt{d d - f f} & + & d^{3} ϱ^{3} \sqrt{d d - f f} \end{array} \end{matrix} \end{array} \\ \begin{array}{l} d^{3} y^{3} & = & \begin{array}{l} \begin{array}{r} d d \\ - f f \end{array} ϩ^{3} \sqrt{d d - f f} & + & 3 d d e ϱ ϩ ϩ & + & 3 e e ϱ ϱ ϩ \sqrt{d d - f f} & + & e^{3} ϱ^{3} \\ + & 3 d d d c & + & 6 e d c ϱ & + & 3 e e d c ϱ^{2} \\ - & 3 f f e ϱ & + & 3 d d c c & + & 3 d d c c e e ϱ \\ - & 3 f f d c & + & d^{3} c^{3} \end{array} \\ d^{4} y^{4} & = & \begin{array}{l} d^{4} ϩ^{4} & + & 4 e ϱ d d ϩ^{3} \sqrt{d d - f f} & + & 6 d d e e ϱ ϱ ϩ ϩ & + & 4 e^{3} ϱ^{3} ϩ \sqrt{d d - f f} & + & e^{4} ϱ^{4} \\ - & 2 d d f f & + & 4 d^{3} c & - & 6 f f e e ϱ ϱ & + & 12 d c e e ϱ ϱ & + & 4 d c e^{3} ϱ^{3} \\ + & f^{4} & - & 4 e ϱ f f & + & 12 d^{3} e c ϱ & + & 12 d d e c ϱ & + & 6 d d c c e e ϱ ϱ \\ - & 4 d c f f & - & 12 d f f e e ϱ & + & 4 d^{3} c^{3} & + & 4 d^{3} c^{3} e ϱ \\ + & 6 d^{4} c c & + & d^{4} c^{4} \\ - & 6 d d f f c c \end{array} \\ d^{3} x y y & = & \begin{array}{l} f d d ϩ^{3} & + & d^{3} ϱ ϩ ϩ & + & e e ϱ ϱ ϩ & + & d e e ϱ^{3} \\ - & f^{3} & - & d f f ϱ & + & 2 c d e f ϱ & + & 2 c d d e ϱ ϱ \\ + & 2 e f ϱ \sqrt{d d - f f} & + & c c d d f & + & c c d^{3} ϱ \\ + & 2 c d f \sqrt{d d - f f} & + & 2 c ϱ^{2} \sqrt{d d - f f} \\ + & 2 c d d ϱ \sqrt{d d - f f} \end{array} \\ d^{4} x x y y & = \\ d^{4} x y^{3} & = \end{array} \end{array}$

As for example if y^e relation twixt x & y bee exprest by, $x x + a x - 2 x y + y y = 0$ then in stead of $x x$ , { x }, $x y$ , $y y$ , writeing their valors found by this table there resulteth { $\begin{array}{l} d d ϩ ϩ & + & 2 d f ϱ ϩ & + & d d ϱ ϱ & = & 0 \\ + & 2 f ϩ ϩ \sqrt{d d - f f} & + & a d f & + & a d d ϱ \\ - & 2 e f ϱ & - & 2 d e ϱ ϱ \\ - & 2 d c f & - & 2 d d c ϱ \\ - & 2 d ϱ \sqrt{d d - f f} & + & e e ϱ ϱ \\ + & 2 ϱ \sqrt{d d} & + & 2 e d c ϱ \\ - & 2 d c & + & d d \end{array}$ .} Which equation espresseth y^e relation twixt ϱ & ϩ when any valors are assumed for c , d , e , & f . And if y^e valors of c , d , e , & f bee such that y^e 2^d terme in ϩ is not of odd dimensions {illeg} in any terme y^e Equation (that is y^t y^e 2^d terme of this Equation be wa{illeg}) then (by Prop: y^e 2^d) $\overset{\cup}{O} y = hn = hd$ is ord{illeg} <24r> is ordinately applyed to y^e Diameter pk .Now y^t y^e 2^d terme of this Equation vanish it is necessary y^t those termes destroy one another in w^ch y^e unknowne quan{illeg}|t|ys {sic} are ϱ & ϩ are not diverse nor differ in dimensions. Whence it appeares y^t I must divide {illeg}|t|he 2^d terme into 2 pts making $2 d f ϱ ϩ - 2 e f ϱ ϩ - 2 d ϱ ϩ \sqrt{d d - f f} + 2 e ϱ ϩ \sqrt{d d - f f} = 0$ . & $a d f ϩ - 2 d c f ϩ + 2 d c ϩ \sqrt{d d - f f} = 0$ . Or by divideing the first of these by $2 ϱ ϩ$ , & y^e 2^d by $d ϩ$ . they are, $d f - e f \begin{array}{r} - d \\ + e \end{array} \sqrt{d d - f f} = 0$ , & $a f - 2 c f + 2 c \sqrt{d d - f f} = 0$ . The first being divided by $d - e = 0$ . there results, $f + \sqrt{d d - f f} = 0$ . Therefore one or both these propositions $d = e$ ; $d d = 2 f f$ , is {illeg} trew. by y^e 2^d tis found y^t $\frac{a f}{2 f - 2 \sqrt{d d - f f}} = c$ . Now since by assumeing some quantitys for y^e valors of d , c , or f I cannot find y^e valor of e unless by y^e Equation $d = e$ . therefore I conclude $d = e$ . whence it is not necessary y^t $d d = 2 f f$ , \or y^e proportion of d to f bee limited/ soe y^t by assuming y^e angle ahd of any bigness I may find y^e position of y^e axis ahd . As if I suppose y^e angle fhd to be a right one (i.e. y^t ah is y^e axis of y^e line) then are y^e △s feh & hgd alike, & therefore $fh ∶ fe ∷ dh ∶ gh ∷ \sqrt{d d + e e} ∶ e ∷$ & $gh = \frac{e ϩ}{\sqrt{d d + e e}}$ $d ∶ f$ . & $\frac{e d}{\sqrt{d d + e e}} = f$ . Or because $d = e$ . therefore $+ f = \frac{d}{\sqrt{2}}$ . Soe y^t I draw fq parallell to pb & qh equall & $c = \frac{a}{4}$ . There Soe y^t I draw $bf = \frac{a}{4}$ . & fq parallell to pb $qk = fq$ & parallell bf & through y^e points f & k I draw kh y^e axis of y^e line nad , w^ch is a Parab. /as in figure 1^st\^[43] So if I would have hd {illeg} paralell {sic} to qk i.e. y^e an{illeg}|g|leg dhf of 45 degre{illeg}|e|s. then this evident y^t $hg = 0 = f$ . & $c = \frac{a f}{2 f \overset{\cup}{O} 2 \sqrt{d d - f f}}$ . Threfore through y^e point b I draw y^e axis kh , so y^t $bq = kq$ , as before. {illeg} /&c.\ {illeg}|&| note y^t since kh t{illeg}|h|e axis is always paralell {sic} to it selfe y^e line dbn is a parabola.^[44]

^[45] Example y^e 2^d, $x^{3} + y^{3} = a^{3}$ . B{illeg}|e|ing first to write y^e valors of $x^{3}$ & $y^{3}$ (found by y^e precedent table) in{illeg}|to| their roome, since I have {illeg} noe neede of those termes in w^ch ϩ is of eaven dimensions I leave y^m out, & soe for $x^{3} + y^{3} - a^{3} = 0$ I write onely $\begin{array}{l} f^{3} ϩ^{3} & + & d d \sqrt{d d - f f} & + & 3 d d f ϱ ϱ ϩ & + & 6 e d c ϱ ϩ \sqrt{d d - f f} & + & 3 d d c c ϩ \sqrt{d d - f f} & = & 0 \\ - & f f & + & 3 e e ϱ ϱ ϩ \sqrt{d d - f f} \end{array}$ . Then sorting these quantitys together in w^ch y^e unknowne quantitys are y^e sam{illeg}|e| there these 4 Equations (y^e {illeg} 1^st being divided by $ϩ^{3}$ , y^{illeg}|e| 2^d by $3 ϱ ϩ^{2}$ , y^{illeg}|e| 3^d by $6 ϩ ϱ$ , y^e 4^th by $3 ϩ$ ) viz: $f^{3} \begin{array}{r} + d d \\ - f f \end{array} \sqrt{d d - f f} = 0$ ; $d d f + e e \sqrt{d d - f f} = 0$ ; $+ c d e \sqrt{d d - f f} = 0$ ; $+ d d c c \sqrt{d d - f f} = 0$ . In y^e first Equa
tion $f^{3} = \begin{array}{r} - d d \\ + f f \end{array} \sqrt{d d - f f}$ , I extract y^e cube roote & tis $f = - \sqrt{d d - f f}$ . or $d d = 2 f f$ . In y^e 2^d $d d f = e e \sqrt{d d - f f}$ , $d d f = e e f$ , or $d = e$ . By the 3^d, $+ c d e \sqrt{d d - f f} = 0$ , or $c = \frac{+ 0}{d e f} = 0$ . & so by y^e fourth.^[46] Now therefore since \ $c = 0$ / $d =$ {illeg} f $d =$ {illeg} e . In y^e line bq from some point as q perpendicular to bq I \draw/ $kq = + e$ , & $ql = - e$ , both of y^m $= bq = d$ . then from y^e points k & l through b I draw y^e two lines ak & gl both w^ch (since they it cuts one another \y^e lines hnd applied to them/ at right angles) is are axe|i|s of y^e lines ndr & {illeg} w^ch appeares al{illeg} in {illeg} y^t $d d = 2 f f$ , for therefore ${nt}^{2} = 2 {st}^{2} = {st}^{2} + {ns}^{2}$ , soe y^t $ns = st$ & nt perpen
dicular to bk .

^[47] Example 3^d If y^e nature of y^e given line bee e{illeg}|x|pressed in these termes $x^{3} - 3 x x y + 2 x y y - 2 a y y = 0$ . Then by supplanting y^e {illeg}|v|alors of x & y into theire roome & working as before, there will bee, $- f^{3} + 2 f d d - 3 f f \sqrt{d d - f f} = 0$ . & &|2|^dly $3 d d f - 6 d e f \begin{array}{r} - 3 d d \\ + 4 e d \end{array} \sqrt{d d - f f} = 0$ . & 3^dly $- 6 c d d f + 4 c d e f + 4 c d d \sqrt{d d - f f} - 4 a d e \sqrt{d d - f f} = 0$ . & 4^tly, $2 c c d d f - 4 a c d d \sqrt{d d - f f} = 0$ . The first of these divided by $f = 0$ . is $- f f + 2 d d = 3 f \sqrt{d d - f f}$ . Or □ing tis {illeg} & ordering y^e product tis $10 f^{4} - 13 f f d d + 4 d^{4} = 0$ . Which being $2 f f - d d = 0$ . there results $5 f f - 4 d d = 0$ . Wherefore I conclude one of these 3 to be y^e valors of f viz: $f = 0 = \sqrt{\frac{d d}{2}} = \sqrt{\frac{4 d d}{5}}$ . Now y^t I may know w^ch of those is y^e right valor of f I try y singly, {illeg}|&| fi\r/s{illeg}|t| suppose $f = 0$ ; If so yⁿ by y^e 4^th Equation $- 4 a c d d \sqrt{d d - f f} = 0$ , therefore $c = 0$ . If {illeg} $c = 0 = f$ , yⁿ in y^e 3^d Eq{illeg}/u\ation all y^e termes vanish except $- 4 a d e \sqrt{d d} = 0$ : therefore $e = \frac{0}{4 a d \sqrt{d d}} = 0$ . & since $0 = e = f = 0$ , all y^e termes in y^e 2^d Equat vanish except except $- 3 d d \sqrt{d d}$ , therefore also $d = 0$ , w^ch since it ought not to bee I conclude y^t $f = 0$ is false. Therefore I passe y^e 2^d valor of $f f = \frac{d d}{2}$ , or, $\overset{\cup}{O} f = \sqrt{d d - f f}$ . & soe divideing y^e 4^th Equat by $2 d d f$ {i{t}} results $c c = \overset{\cup}{O} 2 a c$ . w^ch is divisible by $c = 0$ & by $c \overset{\cup}{O} 2 a = 0$ , Now y^t I may know w^ch is y^e right valor of c first I suppose $c = 0$ : & soe all y^e termes in y^e Equation vanish except, $4 a d e \sqrt{d d - f f} = 0$ . {illeg} or, { $e = \frac{0}{4 a d f} = 0$ . } & since $e = 0$ , by y^e 2^d Equation tis $3 d d f - 3 d d \sqrt{d d - f f} = 0$ or $f = \overset{\cup}{O} \sqrt{d d - f f} = \sqrt{f f} = + \sqrt{d d - f f}$ . {Which things since} they agree I conclude y^t {illeg} $f = \sqrt{d d - f f}$ , or $d d = 2 f f$ ; $c = 0$ ; $e = 0$ . Since $e = 0$ {illeg} {illeg} must be parallell to x & {illeg} $c = 0$ {illeg} must bee {coincident} w^th it. then {illeg} y^e axis { bc I take some {illeg} {illeg} } & fro{illeg} perpendicular {illeg} {illeg} {illeg}y^e {illeg}

<24v>

^[48] Example y^e 4^th. If y^e Equation bee $b x^{3} + a y x x = a^{4}$ . by takeing \onely/ those termes (of y^e valors of $x^{3}$ & $y x x$ found by y^e precedent table) in w^ch ϩ is of odd dimensions, & sorting those together in {illeg}|w|^ch are multiplied by y^e same unkowne {sic} quantit{illeg}|y|s are y^e same & of y^e same dimensions \as before/. there will result these Equations. first $b f^{3} + a f f \sqrt{d d - f f} = 0$ . 2^dly $3 d d b f + 2 d d b f + a d d \sqrt{d d - f f} = 0$ . & 3^dly, $2 a d d f c = 0$ . y^e 1^st is divisible by $f = 0$ , $f b + a \sqrt{d d - f f} = 0$ . To know w^ch of these 2 are y^e valors of f first I suppose $f = 0$ to be trew, & yⁿ all y^e termes in y^e 2^d Equation vanish except $a d d \sqrt{d d}$ , or $d d = \frac{0}{a \sqrt{d d}} = 0$ . by y^e 3^d Equation $2 a d d f c = 0$ vanisheth since $f = 0$ & there|fore| y^e valor of c cannot bee found soe y^t if I assume some valor for it as $c = 0$ now since both d & f should never bee $= 0$ therefore I conclude y^t $f = 0$ is false & so pass to its other valor $f = \frac{- a \sqrt{d d - f f}}{b}$ . or $\frac{- b f}{a} = \sqrt{d d - f f}$ . & soe by y^e 2^d Equation tis {illeg} $d d b = - a d e$ . w^ch is divisible by $d = 0$ , $d b + a e = 0$ . If $d b = - a e$ tis $b ∶ - a ∷ e ∶ d ∷ \sqrt{d d - f f} ∶ f$ . & soe y^e lines ordinat diameter will bee parallel to y^e lines ordinately applied to it w^ch cannot bee therefore I try y^e other valor of $d = 0$ . And if $d = 0$ , yⁿ y^e 3^d Equation $2 a d d f c = 0$ vanisheth & soe c cannot bee found & is therefore unlimited. Now since I find noe repugnancys in these Equations $f = \frac{- a \sqrt{d d - f f}}{b}$ , & $d = 0$ , to be trew I conclude y^m trew. & since $d = 0$ . I draw bh perpendicular wc from b y^e begining of x , w^ch shall bee y^e Diameters of y^e lines enm & dpl . then in y^t diameter I tak{illeg}|e| some point as b or h & from y^t point dra{illeg}|w| {illeg} \ $gh = + f$ or/ $th = - f$ , i.e. of any length, & paralell {sic} to bc . then from y^e pointe t or g \perpendicular to tg / I draw $ts = \frac{a \sqrt{d d - f f}}{b}$ , or $gr = - \frac{a \sqrt{d d - f f}}{b}$ . that{illeg} is, $- a ∶ b ∷ th ∶ ts ∷ gh ∶ gr$ . & so through y^e points s & h or h & r I draw sr w^ch shall be ordinate parallel to y^e lines ordinately applied to y^e Diameter bh .

Example y^e {illeg} 5^t. Suppose $x^{3} = a a y$ . Then by selecting those termes out y^e valors of $x^{3}$ & y in w^ch ϩ is of {illeg} o{illeg}|d| dimensions , & sorting them together in w^ch y^e unknowne quantitys differ not, I have, $f^{3} ϩ^{3} = 0$ ; $3 d d f ϱ ϱ ϩ = 0$ ; & 3^dly $a a d d ϩ \sqrt{d d - f f} = 0$ . by y^e first $f = 0$ , & therefore y^e 2^d vanisheth{illeg}; & y^e 3^d divided by $a a ϩ$ is, $d d \sqrt{d d - f f} = 0$ ; or {illeg} $0 = d d d$ . w^ch may not bee since $f = 0$ . Now since $d = f = 0$ , & y^e proprortion {sic} of d to e {illeg} {illeg}|&| y^e length of c cannot bee found tis e{illeg}|v|ident y^e line hath n{illeg}|o|e axis or diameter.

<25r>

November 1664

^[49] ^[50] Observe y^e Axes, Diameters & position of y^e lines ordinately applied to y^e|m| may bee for y^e most pte easlier obteined {illeg} by making $bc = x$ . $cd = y$ . $bf = c = cq$ . $mh = ϱ = fe$ . $hd = ϩ = - hn$ . $hd ∶ dg ∷ d ∶ f ∷ ϩ ∶ \frac{f ϩ}{d} = dg$ . y^e angles bcd , mgd , fqd , {illeg} mbc , feh , right ones. $fe ∶ he ∷ d ∶ e ∷ ϱ ∶ \frac{e ϱ}{d} = he = gq$ . $qc + gq + gd = cd = \frac{f ϩ + e ϱ + c d}{d} = y$ . $hg = \sqrt{ϩ ϩ - \frac{f f ϩ ϩ}{d d}} = eq$ . $fe + eq = bc = \frac{d ϱ + ϩ \sqrt{d d - f f}}{d} = x$ . Then for readiness in these operations make a table of y^e □s, cubes, rectangles, &c of these valors of x & y . As was done before $\begin{matrix} \begin{array}{l} \begin{matrix} d x & = & ϩ \sqrt{d d - f f} & + & d ϱ . \end{matrix} \\ \begin{matrix} d d x x & = & d d ϩ ϩ & + & 2 d ϱ ϩ \sqrt{d d - f f} & + & d d ϱ ϱ . \\ - & f f \end{matrix} \\ \begin{array}{l} d^{3} x^{3} & = & d d ϩ^{3} \sqrt{d d - f f} & + & 3 d^{3} ϱ ϩ ϩ & + & 3 d d ϱ ϱ ϩ \sqrt{d d - f f} & + & d^{3} ϱ^{3} . \\ - & f f & - & 3 d f f ϱ ϩ ϩ \end{array} \\ \begin{matrix} d y & = & f ϩ & + & e ϱ & + & c d . \end{matrix} \\ \begin{array}{l} d d y y & = & f f ϩ ϩ & + & 2 e f ϱ ϩ & + & e e ϱ ϱ . \\ + & 2 c d f & + & 2 c d e ϱ \\ + & c c d d \end{array} \\ \begin{array}{l} d^{3} y^{3} & = & f^{3} ϩ^{3} & + & 3 e f f ϱ ϩ ϩ & + & 3 e e f ϱ ϱ ϩ & + & e^{3} ϱ^{3} . \\ + & 3 c d f f & + & 6 c d e f ϱ & + & 3 e e c d ϱ ϱ \\ + & 3 c c d d f & + & 3 c c d d e ϱ \\ + & c^{3} d^{3} \end{array} \end{array} & \begin{array}{l} \begin{array}{l} d d x y & = & f ϩ ϩ \sqrt{d d - f f} & + & d f ϱ ϩ & + & d e ϱ ϱ . \\ + & e ϱ ϩ \sqrt{d d - f f} & + & c d d ϱ \\ + & c d ϩ \sqrt{d d - f f} \end{array} \\ \begin{array}{l} d^{3} x y y & = & f f ϩ^{3} \sqrt{d d - f f} & + & d f f ϱ ϩ ϩ & + & 2 d e f ϱ ϱ ϩ & + & d e e ϱ^{3} . \\ + & 2 c d d f ϱ & + & 2 c d d e ϱ ϱ \\ + & 2 e f ϱ \sqrt{d d - f f} & + & e e ϱ ϱ \sqrt{e e - f f} & + & c c d^{3} ϱ \\ + & 2 c d f \sqrt{d d - f f} & + & 2 c d e ϱ \sqrt{e e - f f} \\ + & c c d d \sqrt{e e - f f} \end{array} \\ \begin{array}{l} d^{3} x x y & = & d d f ϩ^{3} & + & d d e ϱ ϩ ϩ & + & d d e ϱ^{3} . \\ - & f^{3} & - & f f e ϱ & + & c d^{3} ϱ ϱ \\ + & 2 d f ϱ \sqrt{d d - f f} & + & d d f ϱ ϱ ϩ \\ + & c d^{3} & + & 2 d e ϱ ϱ ϩ \sqrt{d d - f f} & &c. \\ - & d c f f & + & 2 c d d ϱ ϩ \sqrt{d d - f f} \end{array} \end{array} \end{matrix}$

^[51] Example If y^e relation twixt bc & cd be expressed by $\begin{matrix} a y y & - & b x y & + & b b x & = & 0 \\ - & a b y \end{matrix}$ . then by inserting those quant{illeg}|y|s {sic} (of y^e valors of x & y found by this table) in w^ch ϩ is of odd dimensions, into place of $y y$ , $x y$ , y , x in this Equation, & supposeing those to destroy one another w^ch are multiplied by y^e same unknowne quantitys there will bee these 2 Equations $2 a e f - b d f - b e \sqrt{d d - f f} = 0$ , & ∼ $2 a c d f + b c d \sqrt{d d - f f} + b b d \sqrt{d d - f f} - a b d f = 0$ . The 2^d is divisible by $d = 0$ & there ∼ results $2 a c f + b c \sqrt{d d - f f} + b b \sqrt{d d - f f} - a b f = 0$ . Now to try w^ch of these {illeg}|t|wo are true fir{illeg}|s|t I suppose $d = 0$ ^[52], & soe y^e first Equation will bee $2 a e f - b e \sqrt{- f f} = 0$ . w^ch is imposible {sic} unlesse $f = 0$ , & yⁿ y^e valors of e & c canot {sic} bee found, Therefore $d = 0$ is false. And therefore by y^e 2^d Equat $c = \frac{- b b \sqrt{d d - f f} + a b f}{2 a f + b \sqrt{d d - f f}}$ . & by y^e first $e = \frac{b d f}{2 a f - b \sqrt{d d - f f}}$ . $2 a f \underset{\cap}{O} b \sqrt{d d - f f} ∶ b f d ∷ e ∶$ . & $c = \frac{\underset{\cap}{O} b b \sqrt{d d - f f} + a b f}{2 a f \overset{\cup}{O} b \sqrt{d d - f f}}$ . Whence y^e proportion twixt d & f {illeg} y^t is y^e angle fhn is undetermined, & e & c have double valors viz: when $e = \frac{b d f}{2 a f - b \sqrt{d d - f f}}$ , then $c = \frac{a b f - b b \sqrt{d d - f f}}{2 a f + b \sqrt{d d - f f}}$ . And when $e = \frac{b d f}{2 a f + b \sqrt{d d - f f}}$ , then $c = \frac{a f b + b b \sqrt{d d - f f}}{2 a f - b \sqrt{d d - f f}}$ . wherefore y^e lin{illeg}|e| hath 2 axes.

^[53] For avoyding mistakes \ (w^ch might have happened in y^e 4^th Example where I found $d = 0$ . & $f = \frac{- a \sqrt{d d - f f}}{b}$ )/ it will no{illeg}|t| be amisse to make $hd ∶ dg ∷ f ∶ g ∷ ϩ ∶ \frac{g ϩ}{f} = dg$ . & $fe ∶ he ∷ d ∶ e ∷ ϱ ∶ \frac{e ϱ}{d} = he$ . & soe it will be $\frac{g ϩ}{f} + \frac{e ϱ}{d} + c = y$ . & $\frac{ϱ + ϩ \sqrt{f f - g g}}{f} = x$ . Or, $\frac{d g ϩ + f e ϱ + d f c}{d f} = y$ . & $\frac{f ϱ + ϩ \sqrt{f f - g g}}{f} = x$ . And then observe y^t it can never happen y^t $f = 0$ . or $d + e = 0$ . observe alsoe y^t if $- d ∶ e ∷ g ∶ + \sqrt{f f - g g}$ . yⁿ y^e line fh is y^e axis, otherwise y^e diameter of y^e crooked line. when $d = 0$ \{illeg}/ y^e axis is perpendicular to x from y^e point b as also if $c = \frac{a}{0}$ : And yⁿ it will be convenient to doe y^e worke over againe changing {illeg}{illeg}|y|^e names of x & y {y^t} is writeing y instead of x & x instead {illeg}|o|f y .

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^[54] Haveing y^e Diameter to find y^e Vertex of y^e line.

^[55] Suppose $bc = x$ , or $tr = x$ . cd or $ra = y$ . $bf = c$ . $fq ∶ kq ∷ d ∶ e ∷ fs = br = x ∷ as = y - c$ . y^t is $e x = d y - d c$ . soe y^t into y^e given equation I insert this valor of $x = \frac{d y - d c}{e}$ or of $y = \frac{e x + d c}{d}$ into y^e place of x or (w^ch may more readily bee done). {illeg}

^[56] As in y^e first example I found d{illeg} $= e$ & y^e proportion twixt d & f to bee unlimited so y^t if I would fk to bee y^e diameter Axis I make $- d ∶ e ∷ f ∶ \sqrt{d d - f f}$ . (vide {illeg} C) or {illeg} $f = \sqrt{d d - f f}$ . & there I found $c = \frac{a f}{2 f - 2 \sqrt{d d - f f}}$ . or sinc {sic} $f = - \sqrt{d d - f f}$ it is $c = \frac{a}{4}$ . As may {illeg}|b|ee seene in y^t example. Now y^t I may find y^e vertex of y^e line was there exprest in these termes. $x x + a x - 2 x y + y y = 0$ . I suppose $d ∶ e ∷ x ∶ y - c$ . {illeg}|tha|t is $x = y - c$ . or $x = y - \frac{a}{4}$ . or $y = x + \frac{a}{4}$ . & writeing this valor of y into its r{illeg}|o|ome in y^e Equation $x x + a x - 2 x y + y y = 0$ ; there results $a x + \frac{a a}{16} = 0$ . or $x = - \frac{a}{16}$ . Therefore from y^e point b I draw $br = - \frac{a}{16}$ . & from y^e point r I draw y^e perpendicular rd until it {cu{illeg}t}{cu{illeg}|t|t} y^e axis hd , y^t is, soe y^t $rd = hd$ . & y^e point d shall bee y^e vertex of y^e Parab: {sic} mdo .

^[57] Soe in y^e 2^d Example of y^e line $x^{3} + y^{3} = a^{3}$ , it was found $d = e$ . & $c = 0$ . & therefore $y = \frac{e x + d c}{d}$ . or $y = x$ . therefore I write $x^{3}$ for $y^{3}$ in y^e Equation $x^{3} + y^{3} = a^{3}$ . & it is $2 x^{3} = a^{3}$ . or $x = \frac{a}{\sqrt{c: 2}}$ . therefore I take $br = \frac{a}{\sqrt{c: 2}}$ & soe draw y^e perpendicular ar , which shall intersect y^e axis ab at y^e vertex of y^e crooked line. ^[58] & yⁿ (calling $br = h$ ) {illeg} it shall be $ar = c + \frac{e h}{d}$ . Soe y^t in this case $ar = \frac{a}{\sqrt{2}}$ .

^[59] In y^e 3^d Example y^e Equation being $x^{3} - 3 x x y + 2 x y y - 2 a y y = 0$ , It was found, $c = 0$ . $d ∶ e ∷ q ∶ 0$ . y^t is $e = 0$ . therefore $y = \frac{e x + d c}{d} = 0$ . Therefore by {illeg} writeing 0 instead of y in y^e Equation all y^e termes vanish except ∼ $x^{3} = 0$ , or $x = 0 = br$ . & $ar = c + \frac{e h}{d} = 0$ . soe y^t y^e vertex of y^e line $(bdn)$ must bee at y^e point b .

^[60] But in y^e 4^th Example, $b x^{3} + a x x y = a^{4}$ . It was found $d = 0$ $d ∶ e ∷ 0 ∶$ {1 }{i}. \or $d = 0$ ./ & c {illeg} was unlimited, I make therefore $c = 0$ . & s{illeg}|in|ce for y I make $y = \frac{e x + d c}{d} = \frac{x}{0}$ . Or $b x^{3} + \frac{a x^{3}}{0} = a^{4}$ . $0 b x^{3} + a x^{3} = 0 a^{4}$ , Or $a x^{3} = 0$ , & $x = 0$ y^e axis is perpendicular to x therefore I insert y^e valor of x into y^e equ{illeg}|a|tion $(x = \frac{d y - d c}{e} = 0)$ & there results $b 0 0 0 + a 0 0 y = a^{4}$ . or $y = \frac{a^{4} - b 0 0 0}{0 0 a} = \frac{a^{3}}{0 0}$ . Wherefore I conclude y^e vertex of y^e line to be infinitely distant from b towards m.

^[61] ☞ If y^e position of any line (as ts ) be given y^e point where it intersects y^e given crooked line dsa may be found by y^e same manner; for suppose ∼ ar {illeg}|o|r $ac = x$ . cd or $rs = y$ . & {illeg} $r x = y y$ . $ta = a$ . $tq = b$ . $pq = d$ . angles tqp , srt , dct right ones; yⁿ, $tq ∶ qp ∷ tr ∶ rs$ to find y^e point s where y^e crooked line dsa is intersected by y^e line fp , I suppose $tq ∶ pq ∷ tr ∶ rs$ . y^t is, {illeg} $y = \frac{d a + d x}{b}$ . or $\frac{b y - d a}{d} = x$ . & since by y^e nature of y^e line $r x = y y$ . it follows y^t {illeg} $y y = \frac{b r y - d a r}{d}$ . {illeg}|&| $\frac{d d x x + 2 d d a x + d d a a}{b b} = r x$ . & by extracting y^e rootes of y^m. both, $y = \frac{b r}{2 d} \overset{\cup}{O} \sqrt{\frac{+ b b r r - 4 d d a r}{4 d d}}$ . & $x = a - \frac{b b r}{d d} \overset{\cup}{O} \sqrt{}$ $x = \frac{b b r}{2 d d} - a \overset{\cup}{O} \sqrt{\frac{b^{4} r r}{4 d^{4}} - \frac{a b b r}{d d}}$ . therefore I take $ar = \frac{b b r}{2 d d} - a \overset{\cup}{O} \frac{b}{d} \sqrt{\frac{b b r r}{4 d d} - a r}$ . & $rs = \frac{b r}{2 d} \overset{\cup}{O} \sqrt{\frac{b b r r}{4 d d} - a a}$ .

By y^e same manner y^e intersec{illeg}|t|ion by 2 crooked lines may be found.

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^[62] Having y^e nature of any lines expressed in Algebraicall termes, to find its Asymptotes if have any

^[63] Suppose \rh , & rs y^e asymptotes of y^e line dtn . & hd parallel to rs drawne from y^e Asymptote to y^e line \ dtn // $bc = x$ . $cd = y$ . $bf = c$ . $fe ∶ he ∷ d ∶ e$ . $hd ∶ dg ∷ f ∶ g$ . $fe = mh = ϱ$ . $hd = ϩ$ . y^e angles dcb , {illeg} cbm , dgm , bfq , hef , {illeg} c to bee right ones. yⁿ is, $eh = \frac{e ϱ}{d}$ . $dg = \frac{g ϩ}{f}$ . $dg + gq$ $hg = \frac{ϩ \sqrt{f f - g g}}{f}$ . $dg + gq + qc = \frac{d g ϩ + e f ϱ + d f c}{d f} = y$ . & $fe + eq = \frac{f ϱ + ϩ \sqrt{f f - g g}}{f} = x$ . Now for readiness in operation it bee convenient to have in readinesse a table of the{illeg}|se|{illeg} valors of x & y w^ch will bee y^e same w^th y^t by w^ch y^e diameters of crooked lines are determined. viz.
$\begin{matrix} \begin{array}{l} f x = f ϱ + ϩ \sqrt{f f - g g} . \\ \begin{array}{l} f f x x & = & f f ϩ ϩ & + & 2 f ϱ ϩ \sqrt{f f - g g} & + & f f ϱ ϱ . \\ - & g g \end{array} \\ \begin{array}{l} f^{3} x^{3} & = & f f \sqrt{f f - g g} ϩ^{3} & + & 3 f f f ϱ ϩ^{2} & + & 3 f f ϱ^{2} ϩ \sqrt{f f - g g} & + & f^{3} ϱ^{3} . \\ - & g g & - & 3 g g f ϱ \end{array} \\ d f y = d g ϩ + e f ϱ + d f c . \\ \begin{array}{l} d d f f y y & = & d d g g ϩ ϩ & + & 2 d e f g ϱ ϩ & + & e e f f ϱ ϱ \\ + & 2 d d c f g & + & 2 c d e f f ϱ \\ + & c c d d f f \end{array} \\ \begin{array}{l} d^{3} f^{3} y^{3} & = & d^{3} g^{3} ϩ^{3} & + & 3 d d e f g g ϱ ϩ ϩ & + & 3 d e e f f g ϱ ϱ ϩ & + & e^{3} f^{3} ϱ^{3} . \\ + & 3 c d^{3} f g g & + & 6 c d d e f f g ϱ & + & 3 c d e e f f ϱ ϱ \\ + & 3 c c d^{3} f f g & + & 3 c c d d e f ϱ \\ + & c^{3} d^{3} f^{3} \end{array} \end{array} & \begin{array}{l} \begin{array}{l} d f f x y & = & d g \sqrt{f f - g g} ϩ ϩ & + & d f g ϱ ϩ & + & e f f ϱ ϱ . \\ + & e f ϱ \sqrt{f f - g g} & + & d c f f ϱ \\ + & c d f \sqrt{f f - g g} \end{array} \\ \begin{array}{l} d f^{3} x x y & = & d f f g ϩ^{3} & + & e f^{3} ϱ ϩ ϩ & + & d g f f ϱ ϱ ϩ & + & e f^{3} ϱ^{3} . \\ - & g g g d & - & e f g g ϱ & + & 2 e f f ϱ ϱ \sqrt{f f - g g} & + & c d f^{3} ϱ ϱ \\ + & c d f^{3} & + & 2 c d f f ϱ \sqrt{f f - g g} \\ - & c d f g g \\ + & 2 d f g ϱ \sqrt{f f - g g} \end{array} \\ \begin{array}{l} d d f^{3} x y y & = & d d g g ϩ^{3} \sqrt{f f - g g} & + & d d f g g ϱ ϩ ϩ & + & 2 d e f f g ϱ ϱ ϩ & + & e e f^{3} ϱ^{3} . \\ + & 2 d e f g ϱ \sqrt{f f - g g} & + & 2 c d d f f g ϱ & + & 2 c d e f^{3} ϱ ϱ \\ + & 2 c d d f g \sqrt{f f - g g} & + & e e f f ϱ ϱ \sqrt{f f - g g} & + & c c d d f^{3} ϱ \\ + & 2 c d e f f ϱ \sqrt{f f - g g} \\ + & c c d d f f \sqrt{f f - g g} \end{array} \end{array} \end{matrix}$ This table may be continued {illeg}|w|hen y^e nature of y^e lines are expre{illeg}|s|sed by Equations of 4 or more dimensions. This like y^e former rules will be {illeg} be{illeg}|s|te{illeg} perceived by Examples yⁿ precepts. As

Example y^e 1^st. To find y^e asymptotes of y^e line \whose nature is/ exprest by $r x + \frac{r x x}{q} = y y$ . first I write y^e valors of x , $x x$ & $y y$ (found by this table) into theire places in y^e Equation $r x + \frac{r x x}{q} - y y = 0$ . & there results $\begin{matrix} \frac{r f f ϩ ϩ - r g g ϩ ϩ + 2 r f ϱ ϩ \sqrt{f f - g g} + r f f ϱ ϱ}{q f f} \\ \begin{matrix} \frac{\begin{matrix} - & d d g g ϩ ϩ & - & 2 d e f g ϱ ϩ & - & e e f f ϱ ϱ \\ - & c c d d f f & - & 2 d d c f g ϩ & - & 2 c d e f f ϱ \end{matrix}}{d d f f} & \frac{+ r f ϱ + r ϩ \sqrt{f f - g g}}{f} \end{matrix} \end{matrix} = 0$ . \or {illeg} by ordering it,/ $\begin{matrix} + \frac{r}{q} \\ - \frac{r g g}{q f f} \\ - \frac{g g}{f f} \end{matrix} ϩ ϩ \begin{matrix} + \frac{2 r ϱ \sqrt{f f - g g}}{q f} \\ - \frac{2 e g ϱ}{d f} \\ - \frac{2 c g f}{f} \\ + \frac{r \sqrt{f f - g g}}{f} \end{matrix} ϩ \begin{matrix} + \frac{r ϱ ϱ}{q} \\ - \frac{e e ϱ ϱ}{d d} \\ - \frac{2 c e ϱ}{d} \\ + r ϱ - c c \end{matrix} = 0$ . Or, $\begin{array}{l} + & d d f f r & ϩ ϩ & + & 2 d d f r ϱ \sqrt{f f - g g} & ϩ & + & d d f f r ϱ ϱ \\ - & d d g g r & - & 2 d e f g q ϱ & - & e e f f q ϱ ϱ \\ - & d d g g q & - & 2 c d d f g q & + & d d f f q r ϱ \\ + & d d f q r \sqrt{f f - g g} & - & 2 c d q e f f ϱ \\ - & c c d d f f q \end{array} = 0$ . Now by assumeing any valors for c ; d , e ; f , g . I have, by this equation, y^e relation w^ch ϱ beares to ϩ , y^t is w^ch mh {illeg}|b|eares to hd . But y^t y^e valors of c , d , e , f & g , may be such {illeg} y^t hr (to w^ch hd is applyed) may be one asymptote of y^e {illeg} line \&/ hd paralell {sic} to y^e other, it is necessary (by Prop: 3^d) y^t neither ϱ nor ϩ bee any where of soe many or {illeg}|o|f mor{illeg}|e| dimensi{illeg}ons, yⁿ in those {illeg} termes in w^ch {illeg}|t|hey multiply one another. Therefore I consider of how many dimensions \ ϱ is/ at y^e most in any terme multiplied by ϩ, & how {illeg}|ma|ny ϩ is in any terme multiplied by ϱ ; & find y^m but of one. & therefore conclude y^t ϱ & ϩ {illeg} \ou/ght to be found in noe terme in this equation ∼ unlesse wher{illeg}|e| they multiply one another. ^[64]^[65] Moreover tis manifest y^t y^e Equation (expresing y^e nature of y^e {illeg} given line) will ever be of one \& but of one/ dimension more yⁿ ϱ or ϩ in some termes in w^ch they multiply one another: & therefore this may bee put for a Ge{illeg}|ne|rall Rule viz. All those termes must destroy one another in w^ch there is not $ϱ ϩ$ & w^ch are of as many, {illeg} or want but one dimension of being of as many dimensions as y^e Equation is. Now that the{illeg}|se| termes destroy one another, tis necessary y^t those be $= 0$ in w^ch y^e unknowne quantitys ϱ & ϩ {illeg} are y^e same. Upon w^ch considerations it will appeare y^t in this example I must make, $d d f f r ϩ ϩ - d d g g r ϩ ϩ - d d g g q ϩ ϩ = 0$ . 2^dly, $+ d d f q r \sqrt{f f - g g} ϩ - 2 c d d f g q ϩ = 0$ . {illeg}|t|hi{rdly} { $d d f f r ϱ ϱ - e e f f q ϱ ϱ = 0$ }. 4^thly $d d f f q r ϱ - 2 c d q e f f ϱ = 0$ . Or by dividing y^m by those quantitys w^ch {illeg} /{neede}\ {illeg} $= 0$ . they are $d d r f f - d d g g r - d d g g q = 0$ . 2^dly $d d r \sqrt{f f - g g} - 2 c d d g = 0$ . thirdly, $d d r - e e q = 0$ . 4^thly, $d d q r - 2 c d q e = 0$ . by y^e 3^d, $d = e \sqrt{\frac{q}{r}}$ . & since tis not $d = 0$ , by y^e first $f f = g g + \frac{q g g}{r}$ . by y^e 2^d $c = \frac{r \sqrt{f f - g g}}{2 g}$ . or { $c = g \frac{\sqrt{r q}}{2}$ }. by y^e 4^th $\frac{d r}{2 e} = c = \frac{\sqrt{r q}}{2}$ . Therefore from y^e point b I draw bf & $bk = c = \overset{\cup}{O} \frac{\sqrt{r q}}{2}$ . from f I draw fq parallell to bc . from q I draw qv , soe y^t $fq ∶ qv ∷ d ∶ e ∷ \sqrt{q} ∶ \sqrt{r}$ . & through y^e points f & v I draw fv w^ch shall be {illeg} \{one}/ Asymptote yⁿ <27v> or which is y^e same I make (since tis not $c = 0$ ) I make. $e ∶ d ∷ c ∶ bl = \frac{d c}{e}$ . Or, $bl = \frac{c \sqrt{q}}{\sqrt{r}}$ . & soe draw y^e asyptote {sic} passing through y^e line points l & f . Againe if from some point in y^e Asymptote lh , as h I draw {illeg} hg & from g I draw $gd =$ Then from y^e point k I draw pk paralell {sic} to {illeg} bl & pv paralell {sic} to bk soe y^t (assumeing some other proportion twixt d & e \yⁿ before/ if there be any other) $d ∶ e ∷ pk ∶ pv ∷ - \sqrt{q} ∶ \sqrt{r}$ . & soe through y^e points k & v I draw y^e other Asymptote. Or since {illeg} it is not $c = 0$ ; I make $e ∶ d ∷ bk ∶ \frac{d c}{e} = bl = \frac{+ c \sqrt{q}}{r}$ . & soe t{illeg}|h|rough y^e points l & k I draw y^e other asymptote, w^ch shall be paralle{illeg}|l|l to hd.

Example y^e 2^d. Supose {sic} y^e Asymptotes of $x x - y x + a y = 0$ were to bee determined, Since I have noe use of y^e termes in w^ch is $ϱ ϩ$ I onely select those termes out of y^e valors of $x x$ , | y | & $y x$ in w^ch $ϱ ϩ$ is not & sorting them as was before {illeg} taught I have these equations, 1^st $f f ϩ ϩ - g g ϩ ϩ - - g ϩ ϩ \sqrt{f f - g g} = 0$ 2^dly $a g ϩ - c ϩ \sqrt{f f - g g} = 0$ . 3^dly, $d ϱ ϱ - e ϱ ϱ = 0$ . 4^thly $\frac{a e ϱ}{d} - c ϱ = 0$ . by y^e {illeg}|th|ir{illeg}|d| $d = e$ . by y^e {{illeg}|4|^{illeg}|th|} $a = c$ . by the 1^st $g = + \sqrt{f f - g g}$ . by the 2^d $c = a$

Example y^e 3^d, $y y = \frac{b x^{2}}{c} + y b$ . by working as before I have these Equations {illeg} $g g - g \sqrt{f f - g g} = 0$ . $2 c g - c \sqrt{f f - g g} - b \sqrt{f f - g g} = 0$ . $+ e ϱ ϱ - d ϱ ϱ = 0$ . { $+ x a e -$ }
$g g$ {illeg} $\frac{- b f f + b g g}{c} = 0$ . $2 c g - b g = 0$ . $e e - d d = 0$ . Suppose $x x - a x + b y - y y = 0$ . then by workeing as before I have these Equations, $f f - 2 g g = 0$ . $- a \sqrt{f f - g g} + b g - 2 c g = 0$ . $d d - e e = 0$ . { $b e - a d - 2 c e = 0$ .} by y^e 2^d $c = \frac{b \overset{\cup}{O} a}{2}$ . & soe by y^e 4^th by y^e 4^th $d = \overset{\cup}{O} e$ . by y^e 1^st $f = \overset{\cup}{O} g \sqrt{2}$ . or { $g = \overset{\cup}{O} \sqrt{f f - g g}$ .} by y^e 2^d (by suposeing {sic} $g = + \sqrt{f f - g g}$ ) tis $\frac{b - a}{2} = c$ : & by y^e 4^th (by supposeing $d = + e$ ) tis $\frac{b - a}{2} = c$ . {illeg}|B|ut by y^e 2^d (by supposing $g = - \sqrt{f f - g g}$ ) tis $\frac{a + b}{2} = c$ . & by y^e 4^th (by supposeing $d = - e$ ) tis $\frac{a + b}{2} = c$ . Whence I conclude y^t when $c = \frac{b - a}{2}$ yⁿ is $d = e$ , & $g = \sqrt{f f - g g}$ ; & when $c = \frac{b + a}{2}$ yⁿ $d = - e$ & $g = - \sqrt{f f - g g}$ .

<30v>

^[66] To find y^e Quantity of crookednesse in lines.

^[67] Suppose $ab = x$ . $be = y$ . $bc = o = gh$ . $bg = c$ . \ ed & df tangents \secants/ to y^e crooked line intersecting at d ./ y^e angles abe , acf , egd , right ones. & let $r x = y y$ , be y^e relation twixt x & y . soe y^t aef is a {illeg}|P|arab. Then $be = \sqrt{r x}$ . $bn = v = \frac{r}{2}$ . {illeg} \ $eb ∶ bn ∷ eg ∶ gd$ / $\sqrt{r x} ∶ \frac{r}{2} ∷ c + \sqrt{r x} ∶ \frac{c r + r \sqrt{r x}}{2 \sqrt{r x}} = gd = \frac{r}{2} + \frac{c r}{2 \sqrt{r x}}$ . $cf = \sqrt{r x + r o}$ . {illeg} \ $cf ∶ cm ∷ fh ∶ hd$ / $\sqrt{r x + r o} ∶ \frac{r}{2} ∷ c + \sqrt{r x + r o} ∶ \frac{c r + r \sqrt{r x + r o}}{2 \sqrt{r x + r o}} = hd = \frac{r}{2} - o + \frac{c r}{2 \sqrt{r x}}$ . That is $2 c r \sqrt{r x} + 2 r \sqrt{r r x x + r r o x} = 2 r \sqrt{r r x x + r r o x} - 4 o \sqrt{r r x x + r r o x} + 2 c r \sqrt{r x + r o}$ . Or, Sqareing {sic} both sides $4 c c r^{3} x = 16 o o r r x x + 16 o^{3} r r x \begin{matrix} - 16 c r r x o \\ - 16 c r r x o \end{matrix} \sqrt{r x} + 4 c c r^{3} x + 4 c c r^{3} o$ that is ({illeg} by blotting out $4 c c r^{3} x$ on both sides, divideing y^e rest by o, & then supposeing $o = bc$ to vanish) $- 16 c r r x \sqrt{r x} + 4 c c r^{3} = 0$ . Or $c = \frac{4 x \sqrt{r x}}{r}$ therefore makeing $ab = x$ . $bg = \frac{4 x \sqrt{r x}}{r}$ . $gd = \frac{bn \times eg}{eb} = \frac{1}{2} r + 2 x$ . & describing a circle with y^e Rad $de = \sqrt{\frac{16 x^{3}}{r} + 12 x x + 3 r x + \frac{r r}{4}}$ , y^e circle shall have y^e same quantity of crookednesse w^ch y^e Parabola hath at y^e point e.

^[68] Or thus. If $ab = x$ . $cb = y$ . $bd = v$ . cd \& em/ perpendiculars to y^e crooked line cma w^ch intersect at y^e point e . $af = c$ . $fe = d$ . y^e angle{illeg}|s| abc , baf , afe , \ mna / right ones.

Supose {sic} $r x = y y$ , expresseth y^e relation twixt {illeg} ab & bc . First I find y^e length of $bd = v$ ({illeg}|se|d|e| fol: 8^th hujus, or Des=Cartes his Geom: pag 40) w^ch is $v = \frac{r}{2}$ . $cb ∶ bd ∷ gc ∶ ge$ . / $y ∶ v ∷ c + y ∶ \frac{c v + v y}{y}$ .\ {illeg} $ab + ge = x + v + \frac{c v}{y} = d$ . Or $- d y + c v + v y + x y = 0$ . Out of these termes first I {illeg}|t|ake away v by w{illeg}|r|iteing its valor in its rome {sic} w^ch in this case is $\frac{1}{2} r$ & there remaines results, $\frac{c r}{2} + \frac{y r}{2} + x y - d y = 0$ . Then I take away either x or y (w^ch may bee{illeg} e{illeg}|a|sliest {sic} done) by y^e helpe of y^e Equation expressing y^e nature of y^e li{ne w^ch} is now $r x = y y$ . or $\frac{y y}{r} = x$ . And there results $\frac{c r}{2} + \frac{y r}{2} + \frac{y^{3}}{r} - d y = 0$ . Now tis Evident y^t when y^e lines cm & ce are coincident y^t ce is {illeg}|y|^e radius of a circle w^ch hath y^e same quantity of crookednesse w^ch y^e Parabola mca hath at y^e point c . Wherefore I suppose eb & nm \2 of/ y^e {illeg} rootes of y^e Equation $2 y^{3} + r r y - 2 d r y + c r r = 0$ , to be equall to one another. & {illeg}|s|o by Huddenius his method I multiply it $\begin{matrix} 2 y^{3} & + & r r y & - & 2 d r y & + & c r r & = & 0 \\ 3 & 1 & 1 & 0 \end{matrix}$ . & there results, $\frac{6 y y - r r}{2 r} = d$ . againe otherwise $\begin{matrix} 2 y^{3} & + & r r y & - & 2 d r y & + & c r r & = & 0 \\ 2 & 0 & 0 & - 1 \end{matrix}$ , & there results $\frac{4 y^{3}}{r r} = c$ . Soe y^t if $cb = qa = y$ . yⁿ $ab = \frac{y y}{r}$ . $af = \frac{4 y^{3}}{r r}$ . $fe = \frac{6 y y - r r}{2 r}$ {illeg}. $ce = \sqrt{}$ then y^e circle described by y^{illeg}|e| radius ec shall bee as crooked as y^e Parabola \at y^e point c/

Or better thus. All th Make $ab = x$ . $bc = y$ . $cg = c$ . $fe = d$ . $bd = v$ . Then $y ∶ v ∷ c ∶ \frac{c v}{y} = ge$ . $c v + x y - d y = 0$ .

Or thus. Make $ab = x$ . $bc = y$ . $bd = v$ . $bg = e$ . $ge = f$ . & $f y = v y + e v$ . Thus in y^e former Example $r x = y y$ . $v = \frac{r}{2}$ . $f y = \frac{1}{2} r y - \frac{1}{2} r e$ Or if out of y^e Equa

^[69]^[70] The crookednesse of \equall portions of/ circles are as their diameters, reciprocally.

Demonstr. The crookednesse of {illeg}th any whole circle ( bfd , gcme ) {illeg}|a|mounts to 4 right angles, therefore there is as much crookednesse in y^e circle bfd as in cmeg . Now supposing y^e perimeter fbdf is equall to y^e arch cme , Then as y^e arch $emc = fdbf$ is to y^e circumference cmegc , soe is y^e crookednese {sic} of y^{illeg}|e| arch cme to y^e crookednesse of y^e perimeter cmegc , or of bdfb . so is ab to ac .

<31r>

To find y^e Quantity of crookednesse in lines^[71]

^[72] Suppose ndf & efm perpendiculars to y^e c{illeg}|r|ooked line adeo , w^ch intersect one another at f . $ac = x$ . $ce = y$ . $cm = v$ . $ag = ch = c$ . $gf = d$ . & y^e angles abd , ace , mag, agf right ones. Then, $ec = y ∶ cm = v ∷ eh = y - c ∶ hf = \frac{v y - v c}{y}$ . $gf = gh + hf = x + \frac{v y - v c}{y} = d$ . Or $d y - v y + v c - x y = 0$ .

Haveing therefore y^e relation twixt x & y (as if it be $r x - \frac{r}{q} x x = y y$ ) first I find y^e valor of v ({illeg} see Cartes Geom: pag 40^th. or fol: 8^th of this) {illeg} (as in this Example tis $\frac{1}{2} r - \frac{r x}{q} = v$ ) by w^ch I take v out of y^e Eq{illeg}|u|atio $d y - x y - v y + v c = 0$ , (& in this case {illeg}|t|here results $d y - x y - \frac{1}{2} r y + \frac{r x y}{q} + \frac{1}{2} r c - \frac{r x c}{q} = 0$ .) then by meanes of y^e Equation expressing y^e relation twixt x & y I take out either x or y , w^ch may easliest bee done (as in this example I take out y by writeing $\sqrt{r x - \frac{r x x}{q}}$ in its stead & there results
$\begin{array}{r} + d \\ - x \\ - \frac{r}{2} \\ + \frac{r x}{q} \end{array} \sqrt{r x - \frac{r x x}{q}} = \frac{r c x}{q} - \frac{1}{2} r c$ . Or squareing bi|o|th ptes { $\begin{array}{l} d d r x \\ - d r r \end{array} \begin{array}{l} - 2 d r x \\ + \frac{2 d r r}{q} \\ - \frac{d d r}{q} \end{array} \begin{matrix} d q \\ x q \\ - r \end{matrix}$ } Or $\begin{matrix} 2 d q \\ - & 2 x q \\ - & r q \\ + & 2 r x \end{matrix} \sqrt{r x q - r x x} = 2 r c x \sqrt{q} - r c q \sqrt{q}$ . & by squareing {illeg} both pts {illeg} $\begin{array}{l} - 1 & 0 & 1 & 2 & - 2 & - 1 & 0 \\ 4 d d q^{3} r x & - & 4 d d q q r x x & + & 8 d q q r x^{3} & - & 4 q q r x^{4} & = & q^{3} r r c c & - & 4 q q r r c c x & + & 4 r r q c c x x \\ - & 4 d q^{3} r r & - & 8 d q^{3} r & - & 8 d q r r & + & 8 q r r \\ + & q^{3} r^{3} & + & 4 d q q r r & + & 4 q^{3} r & - & 4 r^{3} \\ + & 8 d q q r r & - & 4 q q r r \\ + & 4 q^{3} r r & - & 8 q q r r \\ - & q q r^{3} & + & 4 q r^{3} \\ - & 4 q q r^{3} & + & 4 q r^{3} \end{array}$ ) Then if I assume any valors for c {illeg}|&| d y^t is {illeg} if I determine y^e point f , I have an Equation by w^ch I can find all y^e perpendiculars to y^e crooked line, drawne from y^e point f . for if I tooke x out of y^e Equation, y^e rootes of y^e Equation will bee ( $db = y$ {illeg}|&| $ec = y$ &c:) all such lines as are drawn from y^e points of interse{illeg}|ct|ion d , e , k , h , to y^e line {illeg} ao (as db , ec , &c) but if I tooke y out of y^e Equation yⁿ y^e roots of y^e Equation will bee those lines drawne from a to y^e perpendiculars (as ab , ac , &c. Now by how much y^e nigher y^e lines points d & f are to one another, soe much y^e lesse difference there will bee twixt y^e crookednesse of y^e pte of y^e line de , & a circle described by y^e radius df or ef . And should y^e line df be understood to m{illeg}|o|ve untill it bee coincident w^th ef , taking f for y^e last point where they ceased to intersect at theire coincidence, y^e circle described by y^e radius {illeg} ef , would have y^e & y^e given crooked line at y^e point e , would bee alike crooked. And when {illeg}|y|^e 2 lines df & ef are c{illeg}|o|incident |2 of| y^e rootes of y^e Equation (viz db & ec , if y^e Equation ab {illeg} & ac ) shall bee equall to one another; Wherfor {sic} to find y^e crookednesse of y^e line at y^e point e I {illeg} supose {sic} y^e e{illeg}|q|uation to have 2 equall rootes & so ordering it Ac{illeg}|c|ording D: Cartes or Huddenius his Method, y^e valor of any of these {illeg} \3/ { $\begin{matrix} x \\ y \end{matrix}}$ } c d being giv{illeg}|e|n, y^e valor of y^e other 2 may be found. ^[73] (as in this Example y^e valor of x being given I multiply y^e Equation according to Huddenius method & it is $\begin{matrix} \begin{array}{l} - & 4 d d q^{3} x & + & 8 d q q x^{3} & - & 8 q q x^{4} \\ + & 4 d r q^{3} & - & 8 d q r & + & 16 q r \\ - & r r q^{3} & + & 4 q^{3} & - & 8 r r \\ + & 8 q r r \end{array} \\ \begin{matrix} 4 q q r x & - & 2 q^{3} x \end{matrix} \end{matrix} = c c = \begin{matrix} \begin{array}{l} - & 4 q q x^{4} & + & 8 d q q x^{3} & - & 4 d d q q x x & + & 4 d d q^{3} x \\ + & 8 q r & - & 8 d q r & - & 8 d q^{3} & - & 4 d q^{3} r \\ - & 4 r r & + & 4 q^{3} & + & 12 d q q r & + & q^{3} r r \\ - & 12 q q r & + & 4 q^{3} r \\ + & 8 q r r & - & 5 q q r r \end{array} \\ \begin{matrix} 4 q r x x & - & 4 q q r x & + & q^{3} r \end{matrix} \end{matrix}$ ^[74] |Then by divideing both y^e numerators by x & y^e denominators| by $2 q r x - q q r$ , & so multiplying y^m {in crucem} & ordering y^e product it is. { $\begin{array}{l} 4 d d q^{4} & - & 4 q^{4} r d & + & q^{4} r r & = & 0 \\ - & 16 q^{4} x & + & 8 q^{4} r x \\ + & 16 q^{3} r x & - & 8 q^{3} r r x \\ + & 24 q^{3} x x & + & 12 q^{4} x x \\ - & 24 q q r x x & - & 36 q^{3} r x x \\ - & 16 q q x^{3} & + & 24 q q r r x x \\ + & 16 q r x^{3} & - & 24 q^{3} x^{3} \\ + & 56 q q r x^{3} \\ - & 32 q r r x^{3} \\ + & 16 q q x^{4} \\ - & 32 q r x^{4} \\ + & 16 r r d^{4} \end{array}$ }.

Now considering y^t if {illeg} q , r , & x bee known, y^t is, if y^e Ellipsis eak be determined, & y^e line ac given{,} there are onely two points in y^e line (viz: e & k ) to be considered. And y^e roots of this equation \valors of d / are ( gf , am , sq ) {illeg} {such} lines as are drawne from y^e line gas to y^e points where y^e perpendiculars efm kqm intersect (as m ) or to such points as of where two perpendiculars (as ef & df ) {ceased} to intersect at theire coincidence into one (as f & y ). Therefore \{illeg}of y^e first {illeg} roots/ I get y^e valor of y^e {line} $cm = x + \frac{r}{2} - \frac{r x}{q} = d$ . as {illeg} this Equation by {illeg} y^t is $d - x + \frac{r}{2} - \frac{r x}{q} = 0$ {illeg} $+ 2 d q - 2 q x + q r - r x = 0$ ; {illeg} there results <31v> $+ 2 d q^{3} - q^{3} r - 6 q^{3} x + 6 q q r x + 12 q q x x - 12 q r x x - 8 q x^{3} + 8 r x^{3} = 0$ .^[75] That is dividi{ng} it by $2 y^{3}$ ; $d = \frac{r}{2} + 3 x - \frac{3 r x - 6 x x}{q} + \frac{6 r x x + 4 x^{3}}{q q} - \frac{4 r x^{3}}{q^{3}}$ . Which Equation \expresseth/ y^e length of y^e lines ( $qs = d$ , & $gf = d$ ) w^ch are drawne from y^e line sag to y^e points q & f at w^ch y^e coincident perpendiculars last intersected \one another/ before theire coincidence. Now haveing y^e length of gf or sq it will not be difficult to find, $c = ag = ch$ , or, $as = cl = c$ ; for it was found before y^t $d y - v y - x y + v c = 0$ Or $c = \frac{v y + x y - d y}{v}$ . Likewise it will not bee difficult to find ef or kq , for (supposeing $lq = d - x$ ; $hf = d - x$ ; $lc = + c$ ; $hc = - c$ ; $ec = + y$ ; $ck = - y$ ). $ef = e$ . or $kq = e$ ) it is, $y y - 2 c y + c c + x x - 2 d x + d d = e e = {\begin{matrix} kq \times kq \\ ef \times ef \end{matrix}$ , {illeg} Lastly y^e circle described w^th y^e radius ef shall have y^e same quantity of crookedness w^ch y^e Ellipsis hath at y^e point e .

^[76] Example y^e 2^d. Were I to find y^e quantity of crookedness a{illeg}|t| some given point of y^e line exprest by $r x + \frac{r x x}{q} = y y$ ; I might consider y^t it differs from y^e former Example {illeg}|o|nely in y^t there I have $\frac{- r x x}{q}$ here or $\frac{r x x}{- q}$ , hi|e|re $\frac{r x x}{q}$ , y^t is in y^e former q was negative in this is affirmative. Soe y^t this operation will bee y^e same w^th y^e former y^e signe of q being changed soe y^t it will be found
{illeg} qf or $hq = d = \frac{r}{2} + 3 x \frac{+ 3 r x + 6 x x}{q} \frac{+ 6 r x x + 4 x^{3}}{q q} \frac{+ 4 r x^{3}}{q^{3}}$ . &c as before.

^[77] Example y^e 3^d. Had I y^e Par In y^e Parabola, $r x = y y$ . & $v = \frac{r}{2}$ . In y^e above mentioned Equation $d y - x y - v y + v c = 0$ I take out v by write{ing} $\frac{r}{2}$ in its roome & it is $d y - x y \frac{- r y + r c}{2} = 0$ . yⁿ I take out y by writeing $\sqrt{r x}$ in its stead /{illeg}\ $- 2 d \sqrt{r x} + 2 x \sqrt{r x} + r \sqrt{r x} = r c$ . & by □ing both sides, $\begin{matrix} 4 d d x & - & 8 d x x & - & 4 d r x & + & 4 x^{3} & + & 4 r x x & + & r r x & = & r c c \\ 1 & 2 & 1 & 3 & 2 & 1 & 0 \end{matrix}$ . Which is an equation |haveing 2 equal{illeg}|l| roots| & therfore multiplied accordind|g| Huddenius his method soe y^t $r c c$ be blotted out, & therere result{illeg} {illeg}|d|ivided by x it is, $\begin{matrix} 4 d d & - & 16 d x & + & 12 x x & = & 0 \\ - & 4 d r & + & 8 r x \\ + & r r \end{matrix}$ . Now tis evident |y^t {illeg} $x = ac$ | being determind {sic} {illeg} there are 2 points (viz: e & k ) from w^ch perpendiculars being drawne they intersect one another in y^e axis at {illeg}| m |, wherefore ∼ $am = x + \frac{r}{2}$ is one of y^e rootes of y^e Equation & therefore it being divided by $d - x - \frac{r}{2} = 0$ , or by $2 d - 2 x - r = 0$ there results $2 d - 6 x - r = 0$ Or $d = \frac{1}{2} r + 3 x = sq = gf$ . Then into y^e above found Equation ∼ ∼ $\frac{2 x \sqrt{r x} - 2 d \sqrt{r x} + r \sqrt{r x}}{r} = c$ , I substitud|t| th{illeg}|i|s valor of d & there results $\frac{- 4 x \sqrt{r x}}{r} = c = ag = ch = pf$ . | $ef = \frac{r + 4 x}{2 r} \sqrt{r r + 4 r x}$ .| Soe y^t I have $eh = \sqrt{r x} + \frac{4 x \sqrt{r x}}{r}$ . And $hf = \frac{1}{2} r + 2 x$ . & therefore $ef = \sqrt{\frac{1}{4} r r + 3 r x + 12 x x + \frac{16 x^{3}}{r}}$ shall be y^e Rad of a circle w^ch is as crooked as y^e Parabola at y^e point e .

^[78] Or it might have be{illeg}|e|ne done thus, haveing y^e Equation $d y - x y - \frac{r y + r c}{2} = 0$ , I might have pr{illeg}t /writ {sic}\ $\frac{y y}{r}$ in stead of x , & soe have had $\begin{matrix} d y & - & \frac{y^{3}}{r} & - & \frac{r y}{2} & + & \frac{r c}{2} & = & 0 \\ 1 & 3 & 1 & 0 \end{matrix}$ w^ch must have 2 equall \roots/ & therefore by y^e Method de max: & min: I {bott {sic}} out $\frac{r c}{2}$ & there results, $d y - \frac{3 y^{3}}{r} - \frac{r y}{2} = 0$ . Or, $d = \frac{r}{2} + \frac{r y y}{r}$ . makeing $ad = y$ , $de = x$ . $cm = v$ . $gf = d$ . $ag = fp = c$ . now if $ad = y$ bee determined it is manifest y^t there is but one point of y^e Parab: (viz: e) to bee considered from w^ch y^e perpendiculars w^ch are drawne doe noe where intersect one another & therefore this equation hath not superfluous rootes like y^e former.

^[79] Example y^e 4^th. If it bee supposed y^t $r x$ {illeg} $x x = r y$ {illeg} $r x + x x = 0$ /y^e nature of y^e line is contained in\ $r y - y y - r x = 0$ . {illeg} & if tis $ad = x$ . $y = {\begin{matrix} ed \\ ek \end{matrix}$ . $v = {\begin{matrix} dt \\ dy \end{matrix}$ . {illeg} fe & jk 2 perpendiculars to y^e crooked line, f, & q two points where y^e coincident perpendiculars last intersected $d = {\begin{cases} ap = qs \\ fg \end{cases}$ . $c = {\begin{matrix} pq \\ pf \end{matrix}$ . Then is $v = \frac{r y}{r - 2 y}$ . by w^ch I take v out of y^e above named Equation \ $d y - x y - v y + v c = 0$ /, & y^e result being divided by y , it is, $r y + 2 d y - 2 x y - d r + r x - c r$ Or ({illeg} $y = \frac{d r - r x + c r}{r + 2 d - 2 x}$ : Then I substitute this valor of y into its place in Equation $r x + y y - r y$ & there results $r x + \frac{d d r r - 2 d r r x + r r x x + 2 d c r r - 2 c r r x + c c r r}{r r - 4 r x + 4 x x + 4 d d + 4 d r - 3 d x} \frac{r r x - d r r - d r r}{r + 2 d - 2 x} = 0$ . or by ordering it {it} will bee {{illeg} $\begin{array}{l} d^{3} & + & d d x x & + & 4 d d x & - & d d r \\ - & 5 r & + & 6 d r & + & 2 c r \\ + & 2 r r & - & d r r \\ - & c r r \\ 3 & 2 & 1 & 0 \end{array} = 0$ } Which Equation |must have {illeg}|

<32r>

two equall roots & therefore by ordering it according to Huddenius Method de Maximis & Minimis, I blot out y^e last terme & y^e result is $\begin{array}{l} 6 x x & - & 8 d x & + & 2 d d \\ - & 5 r x & + & 3 d r \\ + & r r \end{array} = 0$ . Or $\begin{array}{l} 2 d d & + & 3 d r & + & 6 x x \\ - & 8 d x & - & 5 r x \\ + & r r \end{array} = 0$ . By what was said before tis evident y^t the perpendicular $(rm)$ drawne from y^e line asg to y^e point where y^e two perpendiculars intersect, is one of y^e rootes of this Equation.

^[80] And y^t I may have a general rule to find y^e line{illeg} rm (or had there beene 3 or more perpendiculars, to find all those lines w^ch are drawne from y^e line acrw to each /every\ intersection of y^e perpendiculars) I consider y^t if $[ac = λv = θb = c]$ be not drawne from y^e line $[at]$ to y^e point of intersection $[m]$ ; yⁿ d hath two valors as $[vc & bc]$ but if they bee dr{illeg}|a|wne to y^e point $[m]$ , y^t is, if they be ∼ coincident w^th nm ; yⁿ $[d]$ hath c{illeg}t yⁿ y^e two roots of d are equall to one ∼ another, being y^e same w^th y^e line rm . Likewise if $[aλ = cv = wz = d]$ be drawne from y^e line aw to y^e perpendiculars fe , qk {illeg} but not from y^e point where they ∼ intersect; then hath $[c]$ two root{illeg}s (as λv , λz ) w^ch will also be equall to one ∼ another & coincident w^th y^e line mn , when $(d)$ is y^e same w^th $(rm)$ . This being considered; if I would y^e valor of nm , I must order y^e affore found Equation (in w^ch x was supposed to have 2 equall roots) according to c & it will bee $\begin{array}{l} r c c & - & r r c & - & d r r & + & 6 d r x & - & 8 d x x \\ - & d d r & + & 4 d d x & + & 4 x^{3} \\ + & 2 r r x & - & 5 r x x \\ 2 & 1 & 0 & 0 & 0 \end{array} = 0$ . w^ch must have 2 equall roots & therefore by Hudenꝰ {sic} Meth: de Max: & Min: I take away y^e last terme & Soe I have, $2 r c c - r r c = 0$ ; or, $c = \frac{r}{2} = nm$ : {illeg}|B|ut if I would have y^e valor of rm I order y^e Equation according to y^e letter d & it is $\begin{array}{l} 4 d d r & - & d r r & + & c c r \\ - & d d r & + & 6 d r x & - & c r r \\ - & 8 d x x & + & 2 r r x \\ + & 4 x^{3} \end{array} = 0$ . w^ch Equation must likewise have two equall roots & therefore takeing away y^e last terme there re by Hud: meth: de Max: & Min: there resulteth this, $\begin{array}{l} 8 d d x & - & d r r \\ - & 2 d d r & + & 6 d r x \\ - & 8 d x x \end{array} = 0$ . Or $d = x - \frac{r}{2} = rm$ & this $(x - \frac{1}{2} r)$ is \one o of/ y^e rootes of y^e Equation $\begin{array}{l} 2 d d & + & 3 d r & + & 6 x x \\ - & 8 d x & - & 5 r x \\ + & r r \end{array} = 0$ , w^ch was required, therefore I must divide this equation by $d - x + \frac{r}{2} = 0$ . y^t is by $2 d - 2 x + r = 0$ , & there will result, $d - 3 x + r = 0$ . That is $d = 3 x - r = fg = qs$ . Whence i{illeg}|t| will not be difficult to find y^e points q & f & ∼ consequently y^e lines qk , fe w^ch shall be y^e radij of circles w^ch have y^e same quantity of crookednesse y^e line $(aek)$ hath at y^e points e & k . Makeing { $c = {\begin{matrix} as \\ ag \end{matrix}$ }.

Note y^t these Equations have not $(rm)$ or $(ar)$ as one of th{illeg}|e|ire rootes unlesse when y^e axis of y^e axis line is either paralell {sic} to x {illeg}|(| for yⁿ \onely/ a circle whose center is at y^e intersection m can touch y^e crooked line in both k & e together) & then perhaps they may easlyer bee found {y^{illeg}|n|} by y^e foregoeing rule.

May 1665

^[81] 1. Note y^t y^e crooked line φfγqπ (described by y^e points q & f ) is always touched by the (perpendicular) kq ; & that in such sort as to bee measured by it{,} they applying themselves the one to the other, point by point; soe y^t if $aγ = kξ$ the shortest of all y^e lines qk be substracted from qk the{illeg}re remaines $qξ = qγ$ . By this meanes y^e length of as many crooked lines may bee found as is desired

2. Also if y^e line qk is applyed to y^e crooked line qγ point by point, every point of y^e line qk (as k ) shall describe lines to w^ch (as akw) to w^ch qξk is perpendicular.

^[82] 3. The line {qπxfy } is y^e same (if wka be a Parab:) {Heura{illeg}} found. & {illeg} perpendicular to th{e} line efb , & abd a tangent & y^e position & {illeg} point {illeg} w^th y^e tangent (as if they were inherent in y^e same {body}) while y^e tangent gl{illeg} y^e crooked crooked {sic} line ebf, soe y^t y^e point { $\begin{matrix} a \\ f \end{matrix}$ {illeg}} describe {illeg} dag{illeg} then from {illeg} draw perpendiculars to y^e line dag (or {illeg} then shall the \{illeg}/ point c , {illeg} perpendiculars intersect {illeg} y^e po{illeg} of y^e {illeg} {illeg} {illeg} {least} {illeg}

<32v>

^[83] The Crookednesse in lines may bee otherwise found as in the following Examples

^[84] In the Parabola aeg suppose e y^e point where y^e crookednesss {sic} is sought for, & y^t f is the center & fe y^e Radius of a Circle equally crooked w^th y^e Parabola at e . Then naming y^e quantitys {illeg} $ce = y$ . $ap = d$ . $pf = - c$ . $ef = s$ . By^e {sic} nature of y^e line $ac = \frac{y y}{r}$ . $ce + pf = eh =$ {illeg} $y - c$ . { $cp = \frac{y}{r} - d = hf$ }. ${eh}^{2} + {hf}^{2} = {ef}^{2}$ , That is, $\begin{matrix} \frac{y^{4}}{r r} & * & - & \frac{2 d y y}{r} & + & y y & - & c y & \begin{matrix} + & c c \\ + & d d \end{matrix} & = & s s \\ 4 & 2 & 2 & 1 & 0 & 0 \end{matrix}$ w^ch Equation must have 2 \equall/ rootes that ef may be ⊥ to y^e Parab: & therefore multiplyed according to Hudden {sic}'s Method {illeg} it produceth $\begin{matrix} \frac{2 y^{3}}{r r} & - & \frac{2 d y}{r} & + & y & - & c & = & 0 \\ 3 & 1 & 1 & 0 \end{matrix}$ . Which Equation hath soe many rootes as there ca{illeg}|n| be drawn perpendiculars {illeg}|t|o y^e Parab: from {illeg}|t|he determined point f . And two of ther{illeg}|s|e {sic} rootes must be{illeg}|c|ome equall, y^t f may bee the center of y^e required Circle, therefore this equation \is to bee/ multiplyed againe, & it will produce $\frac{6 y^{2}}{r r} - \frac{2 d}{r} + 1 = 0$ that is $\frac{3 y y}{r} + \frac{r}{2} = d$ . Or $3 x + \frac{r}{2} = d$ : As was found in y^e 3^d precedent example.

^[85] Here observe y^t in y^e 1^st of these 3 Equations y hath 4 valors gl , ec , hs & kv . \see fig 2^d/ when d ; c , & s are det{illeg}ermined. But d , c , & $y = ec$ being determined $(s)$ hath but one valor $= ef$ . And if d , s , & $y = ec$ bee determined yⁿ c hath 2 valors pf & pm. And c, s , & $y = ec$ being determined d hath 2 valors an & ap as that {illeg} first equation denotes by^{illeg} y^e dimensions of y^e quantitys in it. By the 2^d of these Equations 2 of y^e valors of $(y)$ are united but not by y^e increasing or diminishing y^e Rad: of {illeg} y^e ci valor of $d = fe$ &c. But 2 rootes one as {illeg} {illeg} first suppose y^e circle soe little as noe where to intersect y^e Parabola, it being increased gradually will fir{illeg}|st| touch y^e Parab: at r (fig 3^d) then ceasing to touch it it intersects it in 2 points g & k (fig 2^d) w^ch two points growing more distant at last it untill it touch y^e Parab: in t (fig 3^d) w^ch being divided into two intersection points e & h (fig 2d) the points g & e draw neered|r| {sic} untill they conj{illeg}|o|yne in y^e touch point w & soe y^e circle ceaseth (by still increasing) to touch {o is} intersect y^e Parab: or intersect it unlesse in h & k . Whence from one point f may be drawne 3 perpendiculars fr , fw , ft , to y^e Parabola twar . And therefore in this 2^d Equation $(y)$ must have 3 valors wc , tl , & vr , when $ap = d$ , & $pf = - c$ , are determined then also hath $(s)$ three valors fr , fw , & ft .

By y^e 3^d Equation Two of the {illeg} valors of $(y)$ in y^e 2^d Equation are united by incresing or diminishing y^e length of $pf = - c$ . For begining at y^e point at the point p (from w^ch y^e {illeg} \3 perpendiculars/ fall upon γ , a & β ) if y^e point f doth gradually move from p , the perpendicular $\{\begin{matrix} ft \\ fw \\ fr \end{matrix}\}$ moves from $\{\begin{matrix} γ \\ a \\ β \end{matrix}\}$ towards $\{\begin{matrix} a \\ γ \\ λ \end{matrix}\}$ Soe y^t y^e two perpendiculars = wf & tf will at last conjoyne into one {illeg}f EF , Which shall be y^e Rad: of a Circle as crooked as y^e Parab: at E .

This 3^d operation might have beene done by making pf determined & by = increasing or diminishing $ap = d$ . That is by destroying y^e term $- \frac{2 d y}{r}$ in stead of $- c$ in y^e 2^d Equation. And so might y^e 2^d Operacon beene done otherwise by {illeg} determining y^e circle egh , &|O|r taking c or d out of y^e 1^st Equation instead of $s s$ .

^[86] There is another way of finding y^e crookednesse in lines & y^t is not by supposing {illeg}|t|wo perpendiculars ( wf & ft , or wf & fr ). but 3 intersections of a circle w^th y^e figure, (fig 2^d h , e , g : or e , g & h ). And then shall $\begin{matrix} x \\ y \end{matrix}}$ have 3 equall valors $\{\begin{matrix} al & , & ac & , & as & . \\ lg & , & fc & , & eh \end{matrix}\}$ Or $\{\begin{matrix} ac & , & al & , & av \\ ce & , & lg & , & vk \end{matrix}\}$ . As if (in y^e last example I had this equation $\begin{matrix} y^{4} & * & - & 2 d r y y & + & r r y y & - & 2 c y r r & + & c c r r & + & d d r r & - & s s r r & = & 0 \\ 3 & 1 & 1 & 0 & - 1 & - 1 & - 1 \end{matrix}$ . Supposing it to have 3 equall rootes by Huddenius his method tis $\begin{matrix} 3 y^{4} & * & - & 2 d r y y & + & r r y y & - & c c r r & - & d d r r & + & s s r r & = & 0 \\ 4 & 2 & 2 & 0 & 0 & 0 \end{matrix}$ . (Which equation doth not determine y^e perpendiculars to eag ) as $\frac{2 y^{5}}{r r} * - \frac{2 d y}{r} + y = c$ {illeg} doth for by { $12 y^{4} - 2 d r y y + r r y y = 0$ }. this I can find y^e valor of c ( y being determined) {illeg} but by it I {illeg} {illeg} neither find y^e valor of { c } nor {illeg}till one of y^m is {illeg}|t|aken out of y^e Equation). That Equation multiplyed {illeg} y^e dimensions of y produceth $6 y^{2} - 2 d r + r r$ Or { $\frac{3 y y}{r} + \frac{r}{2}$ {illeg} $3 x$ {illeg}{illeg} $= d$ }.

^[87] The same may be done thus. If a circle touch a crooked line at one point & intersect it {illeg}er when two points come together y^t circle {illeg} to {illeg} or {illeg}

As if $bc = y$ . { $ac = \frac{y y}{r}$ $an = mt = d$ . $pv = o$ {illeg} c{illeg}={illeg} $= \frac{r}{2}$ . {illeg} {illeg} $- \frac{y y}{r}$ . $\frac{r}{2} ∶ y ∷ \frac{d r - y y}{r} ∶ \frac{2 d r y - y^{2}}{r r} = bq$ }. { $\frac{r}{2} ∶$ {illeg} $+ y ∷$ {illeg} $= d - \frac{y y - 2 o y}{r} ∶ \frac{2 d r y + 2 r o - 2 y^{3} - o y y}{r r} = pg$ . {illeg} may ever bee {illeg}). Then {illeg}} { ${bq}^{2} + {qt}^{2}$ {illeg} $6 y^{4} - 8 d r y y o + r r y y o + 2 d d r r o - d r^{3} o = 0$ . Or $d d = \frac{8 d r y y}{2 r r}$ {illeg}.} Or $d =$ {illeg} Or, {illeg} $d =$ {illeg} w^ch cannot {illeg} T{illeg} by y^e {illeg} perpendiculars {illeg} by supposeing y^e circle described by y^e Rad: {illeg} pt & {illeg} to {illeg}

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^[88] Haveing found an Equa (by y^e former rule) {illeg}|a|n Equation expressing \by which/ y^e quantity of crookednesse in any line may bee found to find y^e greatest or least crookednes of any that line.

^[89] In y^e 3|4|^d|t|h Example I had found $gf = qs = d = 3 x - r$ . And by a rule there shewed viz { $\begin{matrix} as \\ ag \end{matrix}} = c = \frac{x y + v y - d y}{v}$ }: or writeing $3 x - r$ in stead of d It was there found $\begin{array}{l} 4 x^{3} & - & 8 d x x & + & 4 d d x & - & d d r \\ - & 5 r & + & 6 d r & + & c c r \\ + & 2 r r & - & d r r \\ - & c r r \end{array} = 0$ . {illeg}|N|ow by writeing {illeg} $3 x - r$ in stead of d & ordering y^e product according to y^e letter c it is $\begin{array}{l} c c r & - & c r r & + & 16 x^{3} \\ - & 12 r x x \\ - & 3 r r x \end{array} = 0$ . Or extracting y^e roote i{illeg}|t| is $c = \frac{r}{2} \overset{\cup}{O} \sqrt{12 x x - 3 r x + \frac{r r}{4} - 16 \frac{x^{3}}{r}} = {\begin{matrix} as & . \\ ag & . \end{matrix}$ Also by y^e nature of y^e line, $\begin{matrix} kd \\ ed \end{matrix}} = y = \frac{r}{2} \overset{\cup}{O} \sqrt{\frac{r r}{4} - r x}$ . Therefore $kh = gl = \{\begin{matrix} kd - sa \\ ga - ed \end{matrix}\} = \sqrt{\frac{r r}{4} - r x} + \sqrt{12 x x - 3 r x + \frac{r r}{4} - \frac{16 x^{3}}{r}}$ . Also $qh = le - fg = r - 2 x$ . And Since ${\begin{cases} {he}^{2} + {gl}^{2} \\ {kh}^{2} + \end{cases}$ ${he}^{2} + {gl}^{2} = {qk}^{2}$ Therefore
$\frac{3 r r}{2} - 8 r x + 16 x x - \frac{16 x^{3}}{r} + 2 \sqrt{12 x x - 3 r x + \frac{r r}{4} - 16 \frac{x^{3}}{r}} \times \sqrt{\frac{r r}{4} - r x} = qk qk = z z$ ; Supposing $qk = z$ The roote of y^e Surde quantity extracted the Equation is $- \frac{16 x^{3}}{r} + 24 x x - 12 r x + 2 r r = z z$ . Or $16 x^{3} - 24 r x x + 12 r r x - 2 r^{3} + r z z = 0$ . In w^ch Equation y^e least valor of $z = fe$ is to {illeg} bee found & y^t should happen when x hath 2 equall {illeg}|v|alors or rootes. But because $fe = z$ {illeg} being determined x can have but one valor $= ad$ y^e other 2 {illeg} rootes being imaginary tis impossible y^t it should have 2 equall rootes: Therefore I substitute {y}{ y } take away x out of y^e Equation &|b|y substituting its valor $\frac{r y - y y}{r}$ in its stead & there results $\begin{matrix} 16 y^{6} & - & 48 r y^{5} & + & 24 r r y^{4} & + & 32 r^{3} y^{3} & - & 36 r^{4} y y & + & 12 r^{5} y & - & 2 r^{6} & + & r^{4} z z & = & 0 \\ 6 & 5 & 4 & 3 & 2 & 1 & 0 & 0 \end{matrix}$ . In w^ch equation z {illeg} or $ef = qk$ being {illeg}|d|etermined y hath 2 valors de & dk y^e other foure being imaginary & when ef is {illeg} the longest or shortest that may bee then these two valors become one & then is y^e line aek {illeg} more or least crooked. If therefore (y^t y 's valors become equall) this Equation is multiplyed according to its dimensions there will result $8 y^{5} - 20 y^{4} + 8 r r y^{3} + 8 r^{3} y y - 6 r^{4} y + r^{5} = 0$ . w^ch is divisible by $y - \frac{r}{2} = 0$ , or by $2 y - r = 0$ (for there results $4 y^{4} - 8 r y^{3} + 4 r^{3} y - r^{4} = 0$ ). And if $y = \frac{r}{2}$ , yⁿ is $x = \frac{r}{4}$ . Therefore I take $ad = \frac{r}{4}$ & $de = \frac{r}{2}$ & |at| y^e point e shall bee y^e least crookednesse.

Here may bee noted Huddenius his mistake, y^t if i{illeg} some quantity in an equation designe a maximum or minimu y^t Equation hath two{illeg} equall rootes & {illeg} w^ch {illeg}|is| false in y^e equation $16 x^{3} - 24 r x x + 12 r r x - 2 r^{3} + r z z = 0$ . & in all other equation's {sic} which have but one roote.

Or because

Another way. May 1665

^[90] Or because y^e lines fn & qn described by y^e points f & q doe touch one another point's {sic} n from w^ch points onely {illeg} lines drawne perpendicular to y^e croked {sic} line kea will bee perpendicular to y^e \point of/ greatest or least point crookednesse: And also since all those are points of greatest or least crookedness f|t|o w^ch such {illeg} \perpendiculars/ are drawne: The difficulty will be to find y^e point n . Now suppose y^t $am = d$ be determined yⁿ c hath two valors for $\begin{matrix} mf \\ mq \end{matrix}} = c$ . And alsoe y hath two valors for $\begin{matrix} dk \\ de \end{matrix}} = y$ . A{illeg}|l|soe (when am is not parallell to y^e axis of y^e line) x hath {illeg} two (or more) valors $\begin{matrix} ad \\ ad \end{matrix}} = x$ . w^ch valors of c , x , or y become equall if $am = an$ : {illeg}|b|y w^ch meanes y^e point n may bee found: Excepting onely {illeg}|w|hen fm , mq , are parallel to y^e crooked line at n \({ un }{)}/ ^[91] y^t is, parallell \perpendicular/ to y^e streightest or most crooked ptes of y^e line aek . But if $as = c$ be de{illeg}|t|ermined, then $d = {\begin{matrix} st & . \\ sf & . \end{matrix}$ $x = {\begin{matrix} av & . \\ ad & . \end{matrix}$ $y = {\begin{matrix} rv & . \\ ed & . \end{matrix}$ (but if $(as)$ is parallel to y^e axis of y^e line y^e two valors of y are equall & soe not usefull). Which valors of d , x, & y become equall if $as = b un$ : excepting onely when as is perpendicular to y^e most streight or crooke{illeg}|d| pts of y^e line ake .

^[92] As for example. In y^e precedent example it was found $d + r - 3 x = 0$ . But b{illeg}|e|cause an or x is parallell to y^e axis of y^e line, in y^t {illeg}|E|quation x hath but one dimension. Therefore substitute either y^e valors of d or of x into their stead. As if I substitute y^e valor of $x = \frac{r y - y y}{r}$ into its place it will bee $d + r - 3 y + \frac{3 y y}{r} = 0$ or $\begin{matrix} 3 y y & - & 3 r y & + & r r & + & d r & = & 0 \\ 2 & 1 & 0 & 0 \end{matrix}$ . w^ch must have 2 equall roots & therefore multiplyed according to y 's dimensions tis $6 y y - 3 r y = 0$ . Or $y = \frac{r}{2}$ as before. But if I had substitu{illeg}|t|ed d 's valor into its stead it would have beene $\begin{matrix} 16 x^{3} & - & 12 r x x & + & 3 r r x & - & c r r & + & c c r & = & 0 \\ 3 & 2 & 1 & 0 & 0 \end{matrix}$ {illeg}|w|hich {illeg}ving 2 equall roots being rightly ordered is $48 x^{3} - 24 r x^{2} + 3 r r x = 0$ . Or $16 x^{2} - 8 r x + r r = 0$ . Or $4 x - r = 0$ . Or $x = \frac{r}{4} = 0$ , as before.

{In y^e first} Example of finding y^e quantity of crookednesse in lines y^e {illeg} found { $\begin{matrix} 4 r x^{3} \end{matrix}$ {illeg} $\begin{matrix} \frac{x x + 3 r x}{q} & - & 3 x & - & \frac{r}{2} & = & 0 \\ 1 & 1 & 0 \end{matrix}$ }. w^ch {must} have 2 equall rootes & therefore by Hudde{nius} method it is {{illeg} $x x - 4 q x x - 4 q r x + 4 q q x + q q r - q^{3} = 0$ }. {illeg} \w^ch being divided by { $2 x - q$ } it is/ $2 r x - 2 q x - q r$ . Or, $4 x x - 4 q x + q q = 0$ .{illeg} That is $2 x = y$ . or $x =$ {illeg} $= y$ Or {illeg} = {illeg} $= y$ {illeg} y^t if I take { $us = \frac{r}{2}$ } & {illeg} $} =$ {illeg} $= y$ {illeg} y^e points {illeg} y^e greatest or least crookednesse {illeg} y^e line {illeg} {illeg} crookednesse w^ch {illeg} found

{These Equations} {illeg} superfluous rootes {illeg} often as {illeg} y^e perpendiculars {illeg} {illeg}

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The points of greatest or least crookednesse may bee yet otherwise found by an equation of 4 equall rootes. As in y^e example of y^e 2^d way of finding y^e quantity of crookedness in lines it was found $y^{4} * - 2 d r y y + r r y y - 2 c r r y + c c r r + d d r r - s s r r = 0$ . w^ch being compared w^th an equation like it $y^{4} - 4 e y^{3} + b e e y y - 4 e^{3} y + e^{4} = 0$ . by y^e 2^d terme tis $4 e y^{3} = 0$ , or $y = 0$ . In like manner by & $\frac{y y}{r} = \frac{0 0}{r} = 0 = x$ . Soe y^t y^e Parab at y^e {illeg}|b|egining is most crooked (at a ).

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^[93]If y^e body{illeg} b {illeg} {illeg}|m|ove {fro} y^e line bd & from y^e point {illeg}| d | two lines da , dc bee drawne y^e motion of y^e body b from ad is to its m{illeg}|ot|ion from dc as $ab ∥ dc$ is to $cb ∥ ad$ .

^[94] Coroll: 1. The body b receiving two divers forces from a & c & y^e force from ba is to y^e forc{e} from { bc } as ba to bc , yⁿ draw $ad ∥ bc$ & $cd ∥ ab$ , y^e body b shall bee moved in y^e line {illeg} / bd \.

^[95] 2^d Or if y^{illeg}|e| body { b }{ d } is suspended by y^e {thred} bd & is forced from a to {illeg} & from { c } towards f , yⁿ draw $dc ∥ ab$ & $da ∥$ {illeg}, & make $da ∶ ab ∷$ force from c to f ∷ force from {illeg} y^e body {illeg} {illeg}nd in Equilibrio is b .

{illeg}Coroll: 3^d. y^e force of y^e body b from d i{illeg}|s| to its force from a as bd to ba .

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vide pag: 15. But here observe y^t unlesse y^e reflecting line adn bee drawne from y^e through y^e point $(w)$ the center of motion in y^e whole body aiwn y^e determinacon of y^e motion of adn will not be y^e same w^th y^e determinacon of y^e motion of g before reflection (as in y^e first figure^[96]) but verge from it (as in y^e 2^d fig^[97]) y^t is wl & gdi will not bee parrallell {sic}. For since y^e cheife {sic} resistan{ce} is in y^e line of y^e body adn is from its center of motion (prop 32) from d w towards d , & not from i towards d , the body g will find more opposition on y^t side towards y^e center w , yⁿ on y^e other side towards a & therefore at its reflection it{illeg} must incline toward v {illeg} from y^e \(ax 120)/ &{illeg} not returne in y^e line dg . But if y^e body awn presse g towards w yⁿ g presseth y^e body awn towards y^e contrary pte \as/ from w towards l (ax 119) & not from w towards m, if $wm ∥ dg$ . But if adn {illeg}|t|he line adn pass through y^e point w (as in fig: 1^st^[98]) yⁿ

38 If y^e superficies abr (fig 3^d^[99]) circulate all its points in y^e line cd move w^th equall velocity from c towards d. For make $∠ sfr = tsr = recto$ . & $∠ srf = crt$ & draw $ta ⊥ dc$ than is y^e motion of y^e point e from c to y^e motion of y^e point f from c as ae to sf. but $ae = sf$ (for △ rsf similis △ ret therefore $\frac{er \times sf}{fr} = et$ . also aet similis △ erf therefore $er ∶ fr ∷ et ∶ ae$ . or, $\frac{er \times ae}{fr} = et = \frac{er \times sf}{fr}$ & $ae = sf$ ) therefore y^e motion of e from c is equall to y^e motion of f from c.

39 If y^e body g move reflect on y^e immoveable surface \ $(dv)$ / a{illeg}|t| its corner {illeg}| o | (fig 4^th^[100]) its parallell motion (viz from d to v ) shall not bee hindered {illeg} by y^e surface dv , (viz: if y^e center of g 's motion were distant from y^e perpendicular dm an inch at one minute before reflection it shall bee s{illeg}|o| farr {sic} distant from it one minute after reflection). For dv i{illeg}|s| noe ways opposed to motion parallell to it, & a body might $\{\begin{matrix} slide \\ move \end{matrix}\}$ upon it w^thout l{illeg}|o|oseing any motion, & if at y^e first moment of contact y^e body g should loose its ⊥ & onely keepe its ∥ motion it would (perhaps) continue to slide upon it & not reflect.

40 The body g reflecting on y^e plaine vd at its cor{illeg}ner {illeg} o all its points in y^e ⊥ line $op ⊥ vd$ shall move from y^e plaine vd w^th y^e same velocity w^ch before reflection {illeg} \they/ moved fro to it. For y^e point o (prop 9) moves w^th y^t velocity bacwards {sic} w^ch it before did forwards (vi{illeg}|z| to vd ) & all y^e other points (prop 38) move w^th y^e same velocity from it.

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^[101] A Method for finding theorems concerning Quæstions de Maximis e{illeg}|t| minimis. And 1^st Concerning y^e invention of Tangents to crooked lines.

^[102] Suppose $ab = x$ . $eb = y$ . $bd = v$ . $bc = o$ . $cf = z$ . & $ed = df$ . y^e nature of y^e line $a x + x x = y y$ . Then is $ac = x + o$ . $a x + a o + x x + 2 o x + o o = z z$ . $v v + y y = {ed}^{2} = {fd}^{2} = z z + v v - 2 o v + o o$ . Or $y y = o o - 2 o v + z z$ Or $y y = o o - 2 o v + a x + a o + x x + 2 o x + o o$ . & since $a x = y y - x x$ . Therefore $0 = 2 o o - 2 v o + a o + 2 x o$ . Or $2 \times o - 2 v + a + 2 x = 0$ . Now y^t ed may bee perpendicular to y^e line tis required y^t y^e points e , & f conjoyne, w^ch will hapen {sic} when $bc = o$ {illeg}|v|ani{illeg}sheth into nothing. Therefore in the equation $2 \times o - 2 v + a + 2 x = 0$ . Or \{illeg}/ $v = o + \frac{a}{2} + x$ , those termes in w^ch $(o)$ is must be blotted out, & there remaines $v = x + \frac{a}{2} = bd$ . w^ch determines y^e perpendicular ed .

^[103] Hence it appeares y^t in such like operations those termes may be ever blotted out in w^ch $(o = b)$ is of more yⁿ one dimension.

As if y^e nature of y^e line was $x^{3} + x x y + x y y = a y y$ . Then is {illeg} since $ac = o + x$ it is $x^{3} + 3 x^{2} o + 3 x o o + o^{3} + x x z + 2 x o z + o o z + x z z + o z z = a z z$ . That is $x^{3} + 3 x^{2} o + x x z + 2 x o z + o o z + x z^{2} + o z^{2} = a z^{2}$ . Also $v v + y y = v v - 2 v o + o o + z z$ . or $y y + 2 v o = z z$ . Therefore {illeg}
$a y y - x y y - x x y (= x^{3}) + 3 x x o + x x z + 2 x o z + o z z (+ x z z - a z z =) + x y y + 2 v o x - a y y - 2 v o a = 0$ . That is $x x y + 3 x x o$ $x x y - 3 x x o - 2 x o z - o z z + 2 v o a - 2 v o x = x x \sqrt{y y + 2 v o}$ . That is (both pts □^ed {illeg}|&| those terms left out in w^ch o is of more yⁿ one dimension) $x + y y + \overline{2 x x y in 2 v o a - 2 v o x - 3 x x o - 2 x o z - o z z} = x^{4} y y + 2 v o x^{4}$ Or $y in 2 v o a - 2 v o x - 3 x x o - 2 x o z - o z z = v o x x$ . That is $- 3 x x y - 2 x z y - z z y = v x x + 2 v x y - 2 v a y$ . Now if $bc = o$ vanisheth yⁿ is $z = y$ . And consequently $\frac{- 3 x x y - 2 x y y - y^{3}}{x x + 2 x y - 2 a y} = v = \frac{3 x x y + 2 x y y + y^{3}}{2 a y - 2 x y - x x}$ .

^[104] Hence I observe y^t if {illeg} in y^e valor of y there be divers termes in w^ch x is then in y^e valor of z there are those same termes & also those termes \each of y^m/ multiplyed according to y^e by so many units as x hath dimensions in y^t terme & againe multiplyed by o & divided by x . As if ∼ ∼ ∼ $x^{3} + x x y + x y y - a y y = 0$ . Then, $x^{3} + x x z + x z z - a z z + \frac{3 x^{3} o + 2 x x o z + x o z z}{x} = 0$ . Which operacon may bee conveniently symbolized by (ordering y^e equation according to y^e dimensions of y ) &) making some letter \(as a . e . m . n . p )/ to signifie a terme, & y^e same letter w^th some marke (as $\ddot{a}$ , $\ddot{c}$ , $\ddot{e}$ , $\ddot{g}$ , $\ddot{m}$ , $\ddot{n}$ , $\ddot{p}$ &c), to signifie y^e same terme multiplyed according to y^e dimensions of x in it as in y^e former example (suposing {sic} $x - a = m$ . $x x = n$ . $x^{3} = p$ .) The nature of y^e line is ${\begin{matrix} in letters & x y y - a y y + x x y + x^{3} = 0 & x z z - a z^{2} + x^{2} z + x^{3} + \frac{x o z z + 2 x x o z + 3 x^{3} o}{x} = 0 . \\ & \\ in their symbols & m y y + n y + p = 0 & m z z + n z + p + \frac{\ddot{m} z z o + \ddot{n} z o + \ddot{p} o}{x} = 0 . \end{matrix}$ Soe if $\begin{matrix} \underset{︸}{a^{4} + a x^{3} + b b x^{2} - a b b x} & = & y^{4} . \\ m = y^{4} . \end{matrix}$ Then $\begin{matrix} \underset{︸}{3 a o x^{2} + 2 b b o x - a b b o} & + & \underset{︸}{a x^{3} + b b x^{2} - a b b x + a^{4}} & = & z^{4} . \\ \frac{\ddot{m} o}{x} \sim \sim & + & \sim \sim m \sim \sim \sim \sim & = & z^{4} . \end{matrix}$ $\frac{\ddot{m} o}{x} + m = z^{4}$ .

A{illeg} And as any particular Equation may be thus symbolized so divers equations may bee represented by y^e same caracters as $0 = a + c y + y y e$ may represent all equations in w^ch y is of one & two dimensions

Now if a generall Theoreme be required fo{illeg}|r| drawing tangents to such lines it may bee thus found. $eb = y$ , $bd = v$ , $ab = x$ , $bc = o$ , $fc = z$ , by supposition, $a + c y + e y y = 0$ . Then {illeg}/x\ \by observation y^e 2^d/, $a + c z + e z z + \frac{\ddot{a} o + \ddot{c} o z + \ddot{e} o z z}{x} = 0$ . Or, $- c y x - e y y x (= x a) + c z x + e z z x + \ddot{a} o + \ddot{c} o z + \ddot{e} o z z = 0$ . Againe ${eb}^{2} + {bd}^{2} = {cf}^{2} + {cd}^{2}$ that is. $y y + 2 o v = z z$ . Which valor {illeg} of $z z$ put into its stead in y^e termes $e z z x$ \& $c z x$ / in y^e former Equation the result is $+ c y x - \ddot{a} o - \ddot{c} o z - \ddot{e} o z z - 2 e o v x = c x \sqrt{y y + 2 o v}$ . And both pts squared it is (by y^e first Observacon) $c c y y x^{2} - 2 c y x \ddot{a} o - 2 c x y \ddot{c} o z - 2 c x y \ddot{e} o z z - 4 c x y e o v x = c^{2} x x y y + 2 c c x x o v$ . W^ch rightly ordered is $- \ddot{a} y - \ddot{c} y z - \ddot{e} y z z = 2 e x y v + c x v$ . And since y^e points e & f conjoin{illeg}|e| to make ed a perpendicular therefore is $z = y$ & consequently $\frac{- \ddot{a} y - \ddot{c} y y - \ddot{e} y^{3}}{c x + 2 e x y} = v$ . W^ch is y^e Theorem sought for. As for example were it required to draw a perpendicular to y^e line whose nature is $\begin{matrix} x^{3} & + & x x y & \begin{array}{l} + & x \\ - & a \end{array} y y & = & 0 . \\ a & + & c y & \begin{array}{l} + & e \end{array} y y & = & 0 . \end{matrix}$ Then is $\frac{- \ddot{a} y - \ddot{c} y y - \ddot{e} y^{3}}{c x + 2 e x y} = \frac{- 3 x^{3} y - 2 x x y y - x y^{3}}{x^{3} + 2 x x y - 2 a x y} = v$ or $v = \frac{3 x x y + 2 x y y + y^{3}}{2 a y - 2 x y - x x}$ .

In like manner to draw tangents to those lines in w^ch y is of 1, 2 & 3 dimensions suppose $a + c y + e y y + g y^{3} = 0$ . Then is {illeg} $y y + g y^{3} (= a x$ \by 2^d observacon/ $- c y x - e y y x - g y^{3} x + \ddot{a} o + c z x +$ {illeg}/{illeg}\ $= 0$ & by writeing y^e valor of $z = \sqrt{y y + 2 v o}$ ) in its stead in th{illeg}|o|se te{illeg}rmes in w^ch {illeg} not (viz {illeg} $+ g z^{3} x$ there results { $c y x + g y^{3} x - \ddot{a} o - \ddot{c} z o - e z^{2} o - g z^{3} o - 2 e x v o = c x + g y y x - 2 v o g x - \sqrt{y y + 2 v o}$ }. {illeg} {illeg} by {illeg} it is
{illeg}{ $2 g o z^{3} = 4 e o v x = c x + g y y x in c x + g y y$ }{illeg} $in y y + 2 v o$ . Or {illeg}{ $+ g y y o v x + 4 v o g x y y$ }. That is { $\frac{- \ddot{a} y - \ddot{c} y y - \ddot{e} y^{3} - \ddot{g} y^{4}}{c x + 2 e x y + 2 g x y y} = v$ }.

By y^e s{illeg}|a|me proceeding {illeg} of 1, 2, 3 {illeg} dimensions in { $a + c y +$ {illeg} $g y^{3}$ {illeg}} it would be found {illeg} $= v$ . &c {illeg} {illeg}

<47v>

^[105] Having y^e nature of a crooked line expressed in Algebraicall termes w^ch are not put one pte equall to another but all of y^m equall to nothing, if each of y^e {illeg} termes be multiplyed by soe many units as x hath dimensions in them. & then multiplyed by y & divided by x they shall be a numerator: Also if the \signes be changed &/ each terme be multiplyed by soe many units as y hath dimensions in y^t terme{illeg} & yⁿ divided by y they shall bee a denominator in y^e valor of v .

Example 1^st. If $r x + \frac{r x x}{q} - y y = 0$ . Then $\frac{\begin{matrix} 1 & 2 & 0 \\ r x & + & \frac{r}{q} x x & - & y y & in & \frac{y}{x} \end{matrix}}{\begin{matrix} - & r x & - & \frac{r}{q} x x & + & y y & in & \frac{1}{y} \\ 0 & 0 & 2 \end{matrix}} = \frac{r y + 2 \frac{r}{q} x y}{2 y} = \frac{r}{2} + \frac{r}{q} x = v$ . Example 2^d. If $x^{3} + x x y + x y y - a y y = 0$ . Then $\frac{3 x^{3} + 2 x x y + x y y in \frac{y}{x}}{- x x y - 2 x y y + 2 a y^{2} in \frac{1}{y}} = v = \frac{3 x x y + 2 x y y + y^{3}}{- 2 a y - 2 x y - x x}$ . Exam: 3^d. If $x^{3} + b x x - c d x + b c d + d x y = 0$ . Then $\frac{3 x x y - 2 b x y - c d y + d y y}{- d x} = \frac{2 b y}{d} - \frac{3 x y}{d} + \frac{c y}{x} - \frac{y y}{x} = v$ . And by taking y out of y^e valor of v yⁿ, $v = \frac{2 x^{3}}{d d} - \frac{3 b x x}{d d} + \frac{b b x}{d d} - \frac{2 c x}{d} + \frac{2 b c}{d} + \frac{b c c}{x x} - \frac{b b c c}{x^{3}}$ .^[106]

Note. That haveing $\{\begin{matrix} x \\ y \end{matrix}\}$ given, it will be often {illeg} \more/ convenient to find $\{\begin{matrix} y \\ x \end{matrix}\}$ by y^e equation expressing y^e nature of y^e line & yⁿ having x & y to find v by them both, Then to take $\{\begin{matrix} y \\ x \end{matrix}\}$ out of v 's valor & soe to find it by $\{\begin{matrix} x \\ y \end{matrix}\}$ alone.

The Perpendiculars {illeg}|t|o crooked lines & also y^e Theorems ∼ for finding them may otherwis{illeg} more conveniently \be/ found thus

^[107] Supposing $ab = x$ ; $cb = o$ , $db = v$ , $eb = y$ , $cf = z$ . {illeg} And if y^e fc bee supp distance twixt fc & fb , bee imagined to bee infinitely little, y^t is if y^e triangle efr is supposed to bee infinitely little then $be ∶ bd ∷ bg ∶ be ∷ re ∶ fr ∷ y ∶ v ∷ o ∶ z - y$ . That is $y z - y y = v o$ . Or $z = y + \frac{v o}{y}$ .

Now suppose y^e nature of y^e line bee $r x - \frac{r x x}{q} - y y = 0$ . Then is $r x + r o - r x x - 2 r o x - r r o - z^{2}$ { $= 0$ }

In w^ch equation instead of $r x (= \frac{r x x}{q} + y y)$ & $z z (= y y + 2 v o + \frac{v v o o}{y y})$ write theire valos {sic} & y^e result is $r o - \frac{\begin{matrix} 2 r o x & - & 2 v o q \\ - & r o o \end{matrix}}{q} - \frac{v v o o}{y y} = 0$ . Or $r - \frac{2 r x - r o}{q} - 2 v - \frac{v v o}{y y} = 0$ . but these two termes $r o$ , $\frac{v v o}{y y}$ are infinitely little, y^t is if compared to finite termes they vanish therefore I blot y^m out & there rests $\frac{r}{2} - \frac{r x}{q} = v = db$ .

Suppose y^e nature of y^e line be $p + q y + r y y = 0$ Then (by observation y^e 2^d) it is {illeg} $- q y x - r y y x (= p x) + \ddot{p} o + q z x + \ddot{q} z o + r z z x + \ddot{r} z z o = 0$ . Then writeing y^e valor of $z (= y + \frac{v o}{y})$ in its stead in these termes $q z x + r z z x$ , There results $\ddot{p} o + \frac{q x v o}{y} + \ddot{q} z o + 2 r x v o + \ddot{r} z z o = 0$ . Or because y^e difference twixt z & y is infinitely little it is $\frac{\ddot{p} y + \ddot{q} y y + \ddot{r} y^{3}}{- q x - 2 r x y} = v$ .

^[108] And though y^e angle ebg made by intersection of x & y is not determined {illeg}|w|hither {sic} it acute obtuse or a right one, yet may y^e line bg bee found as easily after y^e same manner w^ch determines y^e position of y^e tang^nt eg . For suppose {illeg} $= t$ . cb {illeg} $cb = y$ , $fc = z$ , $ab = x$ , & y^t $z ∥ y$ . Then (supposing y^e distance of fc & eb to be {infinit}ely little) it is, $t ∶ y ∷ t + o ∶ y + \frac{o y}{t} = z$ . Now if y^e nature of y^e line is $p + q y$ {illeg} $y^{3} =$ {0 } Then is $- q y x - r q^{2} x - s y^{3} x (= p x) + \ddot{p} o + q x z + \ddot{q} o z + r x z^{2} + \ddot{r} o z^{2} + s x z^{3}$ {illeg} And by putting y^e valor of z into its stead in those terms in w^ch {illeg} results $\ddot{p} o + \frac{q x o y}{t} + \ddot{q} o z + \frac{2 r x o y^{2}}{t} + \ddot{r} o z^{2} + \frac{3 s x o y^{3}}{t} + \ddot{s} o z^{3} = 0$ . Or $t =$ {illeg}

Soe y^t y^e variation of y^e angle ebg makes no variation.

Note y^t th{illeg}|e| foundacon of this operacon of y^t {illeg} {illeg} pag 131) {illeg} tangents {illeg} But since {illeg} Equation is the sa{illeg} {illeg}stons y^t it would bee if {illeg} {illeg}

<50r>

To draw perpendiculars to crooked lines in all other cases.

Although y^e unknone {sic} quantitys x & y are not related to one another as in the precedent rules (y^t^{illeg} that is soe y^t y move upon x in a given angle), y{illeg}|et| may there be drawne tangents to them by y^e same metho{illeg}|d|.

^[109] As if efm is an Elipsis {sic} described by y^e thred aen (as is usuall) Then make $gb = v$ . $en = y$ . $an = a$ . $x + y =$ {illeg} And let y^e relation twixt x & y be $x + y = b$ . $ab = z$ . Then is $bn = a - z$ . ${eb}^{2} = x x - z z = y y - a a + 2 a z - z z$ . And consequently $\frac{x x - y y + a a}{2 a} = z = ab$ And. And $x x \frac{- x^{4} + 2 x x y y - 2 x x a a - y^{4} + 2 a a y y - a^{4}}{4 a a} = {eb}^{2} = \frac{2 x x y y + 2 x x a a + 2 a a y y - x^{4} - y^{4} - a^{4}}{4 a a} = {ae}^{2} - {ab}^{2}$ . Againe suppose $af = s$ , $fn = t$ , $bc = o$ . Then is $ac = x + o = o + \frac{x x - y y + a a}{2 a} = \frac{s s - t t + a a}{2 a}$ . That is $2 a o + x x - y y + t t = s s$ . Alsoe $v ∶ g ∷ v + o ∶ g + \frac{o g}{v} = fc$ . And ${fc}^{2} = g g + \frac{2 g^{2} o}{v} = \frac{2 s s t t + 2 s s a a + 2 t t a a - s^{4} - t^{4} - a^{4}}{4 a a}$ {illeg} Making $eb = g$ . And consequently $\frac{2 x x y y + 2 x x a a + 2 a a y y - x^{4} - y^{4} - a^{4}}{4 a a} + \frac{2 g^{2} o}{v} = \frac{2 s s t t + 2 s s a a + 2 t t a a - s^{4} - t^{4} - a^{4}}{4 a a}$ Or $2 x x y y + 2 x x a a + 2 a a y y - x^{4} - y^{4} - a^{4} + \frac{8 a a g^{2} o}{v} = 4 t t a o + 2 t t x x - 2 t t x x - 2 t t y y + t^{4} + 4 a^{3} o^{3} + 2 a a x x - 2 a a y y + 4 a a t t - a^{4} - s^{4}$ . ^[110] Or $4 a o x x - 4 a o y y + \frac{8 a a g g o}{v} + 4 a a y y - t^{4} - 4 a^{3} o - 4 a a t t = 0$ . Lastly by y^e nature of y^e line $b - t = s$ . And $b a - 2 b t + t t = s s = 2 a o + x x - y y + t t$ . Or $\frac{- 2 a o - x x + y y + b b}{2 b} = t$ . And { $4 t t = \frac{4 a o x x - 4 a o y^{2} - 4 a b b o + Q: - x x + y y + b b :}{2 b b}$ } w^ch valor substituted into y^e Equation A, the result is { $- t^{4} - 8 x^{3} o + 6 a x y y - x^{4} - y^{4} - a^{4} + 2 a x y y - 2 x x a a = 0$ } $\frac{8 a a g g o}{v} - x^{4} - y^{4} - a^{4} + 2 x x y y - 2 x x a a + 2 a a y y = 0$ .
{ $\frac{8 a a g g o}{v} \frac{+ 4 a^{3} b b o - 4 a^{3} b b o - x^{4} a^{2} + 2 x x y y a a + 2 a a b b x x - a a y^{4} + 2 a a b b y y - b^{4} a}{b b}$ }.

^[111] As if

^{[Editorial Note 2]}

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^[9] October 1664

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^[26] For y^e 5^t

^[27] For y^e 6^t &c

^[28] For y^e first Equation of y^e seacond Sort

^[29] For y^e seacond

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^[51] This line is a streight one y^e equation being divisible by $b = y = 0$

^[52] Endeavor not to find y^e quantity d in these cases, but suppose it given^{[Editorial Note 1]}

^{[Editorial Note 1]} There is a line connecting the end of this note to the following one

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^[101] March \May/ 20^th 1665

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^[103] Theoreme 1^st Observation 1^st

^[104] Observacon 2^d

^[105] An universall theorem for tangents to crooked lines, when $y ⊥ x$ .

^[106] See Des Cartes his Geometry. booke 2^d, pag 42, 46, 47. Or thus, $\begin{matrix} x^{3} & - & b x x & - & c d x & + & d y x & + & b c d \\ 2 & 1 & 0 & 0 & - 1 \end{matrix}$ { $\frac{2 x x y - b x y}{- d x} \frac{+ b c d y}{d x x} = v$ }. And $\frac{b c y}{x x} + \frac{b y}{d} - \frac{2 x y}{d} = v$ .

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^[108] An universall theo{illeg}|r|em for drawing tangents to crooked lines when x & y intersect at any determined angle

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^{[Editorial Note 2]} The rest of the page is damaged.

Newton's Waste Book (Part 2)

How to find ye axes vertices Diamiters, \Centers,/ or Asymptotes of any Crooked Line supposeing it have them.

[9] Haveing {an} the nature of a crooked line expresed in al{illeg}|g|ebraicall termes to find its axi|e|s if {i {sic}}{i} {shave} { anq }

A generall rule to find ye axes of any line.

But since there is noe use of those termes in wch ϩ is of eaven dimensions ye Calculation will bee much abre{illeg}|v|iated by this following table.

T{illeg}|h|e use of ye precedent table in fin{illeg}ding ye Axes of crooked Lines, declared by Examples.

[39] To find ye Axis or Diame{illeg}|t|er of any crooked Line supposeing it have ym.

[54] Haveing ye Diameter to find ye Vertex of ye line.

[66] To find ye Quantity of crookednesse in lines.

To find ye Quantity of crookednesse in lines[71]