Honoured SrSir

I returned but last week to Oxford, but whilst I was at in Northampton Shire I began to think on Mr Leibnitz's problem, and I think I have hit upon the Solution of it, it not being difficult to any that understand Fluxions, The method I used was this

Suppose Aa series of curves of the same nature which have all the same Axis $\mathrm{AB}$ and Vertex $\mathrm{A}$. Let $\mathrm{AI}$ be one of them and $\mathrm{BI}$ the curve which cuts it at right angles, Because the curve $\mathrm{AI}$ is given its subnormal $\mathrm{GD}$ will be given, But this subnormal is the subtangent of the curve $\mathrm{BI}$. therefore $\mathrm{FE}:\mathrm{EI}\colon\colon \mathrm{IG}:\mathrm{GD}$. that is calling $\mathrm{AG}$ $\mathrm{x}$ and $\mathrm{IG}$ $\mathrm{y}$: $\stackrel{·}{\mathrm{y}}:-\stackrel{·}{\mathrm{x}}\colon\colon \mathrm{y}:\mathrm{GD}$. (here $\stackrel{·}{\mathrm{x}}$ must be negative for when $\mathrm{y}$ increases $\mathrm{x}$ decreases, by this Analogy I got an equation and from the equation I obtain a value of the Latus Rectum. This value of the latus Rectum I put in its Room in the equation wchwhich expresses the nature of the curve proposed and I have a new exquationequation expressed by $\mathrm{x}$ $\mathrm{y}$ and their fluxions which gives the nature of the curve required. For example suppose the curve $\mathrm{AI}$ a Parabola whose equation is $2\mathrm{l}\mathrm{x}=\mathrm{y}\mathrm{y}$ then $\mathrm{GD}=\mathrm{l}$ and $\stackrel{·}{\mathrm{y}}:-\stackrel{·}{\mathrm{x}}\colon\colon \mathrm{y}\frac{-\stackrel{·}{\mathrm{x}}\mathrm{y}}{\mathrm{y}}=\mathrm{l}$ this value of $\mathrm{l}$ being put in the equation for the parabola gives $\frac{-2\mathrm{x}\stackrel{·}{\mathrm{x}}\mathrm{y}}{\stackrel{·}{\mathrm{y}}}=\mathrm{y}\mathrm{y}$ and $-2\mathrm{x}\stackrel{·}{\mathrm{x}}=\mathrm{y}\stackrel{·}{\mathrm{y}}$ hence ${\mathrm{a}}^2-{\mathrm{x}}^2=\frac{1}{2}{\mathrm{y}}^2$ or $2{\mathrm{a}}^2-2{\mathrm{x}}^2={\mathrm{y}}^2$ Hence the curve $\mathrm{BI}$ is an Ellipse whose Greater Axis is double in Power to its lesser. If the curves Proposed are Hyperbolas, whose center is $\mathrm{C}$ transverse Axis $\mathrm{AM}=2\mathrm{a}$, their equation is $\frac{2\mathrm{l}\mathrm{a}\mathrm{x}+\mathrm{l}{\mathrm{x}}^2}{2\mathrm{a}}={\mathrm{y}}^2$ and the Subnormal is $\frac{\mathrm{l}\mathrm{a}+\mathrm{l}\mathrm{x}}{2\mathrm{a}}={\mathrm{y}}^2$ and $\stackrel{·}{\mathrm{y}}:-\stackrel{·}{\mathrm{x}}\colon\colon \mathrm{y}:\frac{\mathrm{l}\mathrm{a}+\mathrm{l}\mathrm{x}}{2\mathrm{a}}$ hence $\mathrm{l}=-\frac{2\mathrm{a}\mathrm{y}\stackrel{·}{\mathrm{x}}}{\overline{\mathrm{a}+\mathrm{x}}\times \stackrel{·}{\mathrm{y}}}$ wchwhich value of $\mathrm{l}$ being put in the equation of the Hyperbola gives $\frac{-4{\mathrm{a}}^2\mathrm{y}\mathrm{x}\stackrel{·}{\mathrm{x}}-2\mathrm{a}\mathrm{y}{\mathrm{x}}^2\stackrel{·}{\mathrm{x}}}{2\mathrm{a}\times \overline{\mathrm{a}+\mathrm{x}}\times \stackrel{·}{\mathrm{y}}}={\mathrm{y}}^2$ and $\frac{-{\mathrm{x}}^2\stackrel{·}{\mathrm{x}}-2\mathrm{a}\mathrm{x}\stackrel{·}{\mathrm{x}}}{\mathrm{x}+\mathrm{a}}=\mathrm{y}\stackrel{·}{\mathrm{y}}$ that is $-\mathrm{x}\stackrel{·}{\mathrm{x}}-\mathrm{a}\stackrel{·}{\mathrm{x}}+\frac{{\mathrm{a}}^2\stackrel{·}{\mathrm{x}}}{\mathrm{a}+\mathrm{x}}=\mathrm{y}\stackrel{·}{\mathrm{y}}$ or $\frac{2{\mathrm{a}}^2\stackrel{·}{\mathrm{x}}}{\mathrm{a}+\mathrm{x}}=2\mathrm{x}\stackrel{·}{\mathrm{x}}+2\mathrm{a}\stackrel{·}{\mathrm{x}}+2\mathrm{y}\stackrel{·}{\mathrm{y}}$ and taking the Fluents we have Fluent $\frac{2{\mathrm{a}}^2\stackrel{·}{\mathrm{x}}}{\mathrm{a}+\mathrm{x}}$ $+$ a Given quantity $={\mathrm{x}}^2+2\mathrm{a}\mathrm{x}+{\mathrm{y}}^2$. From wchwhich we draw the followin construction. at the center $\mathrm{C}$ Asymptotes $\mathrm{CA}$ $\mathrm{CE}$ describe the hyperbola $\mathrm{LGH}$ whose rectangle inscribed is $=2{\mathrm{a}}^2$. make $\mathrm{BF}$ perpendicular to $\mathrm{BA}$ and $=$ to it. and make up the rectang.rectangle $\mathrm{MF}$ in $\mathrm{AB}$ take any point $\mathrm{N}$ and let $\mathrm{NO}$ be a mean proportional between $\mathrm{NA}$ and $\mathrm{NM}$, at the center $\mathrm{O}$ with the Radius $\mathrm{OP}$ whose Square is equal to the rectangle $\mathrm{MF}$ minus the Hyperbolick space $\mathrm{BNGR}$ describe the Arch $\mathrm{PQ}$ and let $\mathrm{NG}$ produced meet with it in $\mathrm{P}$ the point $\mathrm{P}$ will be the curve required.