Find the equation of a curve such that the projection of its ordinate upon the normal is equal to its abscissa.

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Solution

$\begin{matrix} L e t s o l p e o f tan g e n t i s P tan θ = \frac{d y}{d x} P M = y = o r d i n a t e P S = M O = x = a b sec i i s a o j . o f P M o r n o r m a l : P M cos θ = x = y cos θ = x cos θ = \frac{x}{y} = \frac{b a s e}{h y p o .} tan θ = \frac{\sqrt{y^{2} - x^{2}}}{x} tan θ = \frac{d y}{d x} = \frac{\sqrt{y^{2} - 1^{2}}}{d} [H Y p o t e n o u s e = D G] y = v x \frac{d y}{d x} = v \div x \frac{d v}{d x} v + \frac{d y}{d x} = \frac{\sqrt{v^{2} x^{2} - x^{2}}}{x} v + x \frac{d y}{d x} = \frac{x \sqrt{v^{2} - 1}}{x} x \frac{d y}{d x} = \sqrt{v^{2} - 1} - v \int \frac{d v}{\sqrt{v^{2} - 1} - v} = \int \frac{d x}{x} = log x + C N o w, I = \int \frac{d v}{\sqrt{v^{2} - 1} - v} = \int \frac{- \sqrt{v^{2} - 1} - v}{(\sqrt{v^{2} - 1} - v) (- \sqrt{v^{2} - 1} - v)} d v = \int (- \sqrt{v^{2} - 1} - v) d v = - \int \sqrt{v^{2} - 1} d v - \int v d v L e t v = sec θ d v = sec θ + tan θ d θ = - \int tan θ + sec θ tan θ d θ - \frac{v^{2}}{x} [y = v x] = - \int {sec θ + tan}^{2} θ d θ - \frac{y^{2}}{2 x^{2}} = - \int {sec}^{3} θ d θ + \int sec θ d θ - \frac{y^{2}}{2 x^{2}} = - \frac{1}{2} tan (tan θ + sec θ) + \frac{sec θ tan θ}{2} + log (tan θ + sec θ) - \frac{y^{2}}{2 x^{2}} [b y s o l v i n g t h e i n t e r g r a t i o n b y p a r t s .] = \frac{1}{2} tan (t a n θ + sec θ) + \frac{sec θ tan θ}{2} - \frac{y^{2}}{x^{2}} = \frac{1}{2} log \sqrt{\frac{y^{2}}{x^{2}} - 1} + \frac{1}{x} + \frac{\frac{y}{x} \sqrt{\frac{y^{2}}{x^{2}} - 1}}{2} - \frac{y^{2}}{x^{2}} = 4 a_{1} x + C W h i c h i s t h e r e q u i r e d a n s w e r . \end{matrix}$

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