Question

A curve is such that the length of the perpendicular from the origin on the tangent at any point P of the curve is equal to the abscissa of P. Prove that the differential equation of the curve is \[y^2 - 2xy\frac{dy}{dx} - x^2 = 0\], and hence find the curve.

Answer 1

Tangent  at P(x, y) is given by \[Y - y = \frac{dy}{dx}(X - x)\]
If p be the perpendicular from the origin, then 
\[p = \frac{x\frac{dy}{dx} - y}{\sqrt{\left[ 1 + \left( \frac{dy}{dx} \right)^2 \right]}} = x ............\left(\text{given}\right)\]
\[\Rightarrow x^2 \left( \frac{dy}{dx} \right)^2 - 2xy\frac{dy}{dx} + y^2 = x^2 + x^2 \left( \frac{dy}{dx} \right)^2 \]
\[ \Rightarrow y^2 - 2xy\frac{dy}{dx} - x^2 = 0 \]
Hence proved. 
\[\text{ Now, }y^2 - 2xy\frac{dy}{dx} - x^2 = 0 \Rightarrow \frac{dy}{dx} = \frac{y^2 - x^2}{2xy}\]
\[ \Rightarrow 2xy\frac{dy}{dx} - y^2 = - x^2 \]
\[ \Rightarrow 2y\frac{dy}{dx} - \frac{y^2}{x} = - x^{} \]
\[\text{ Let }y^2 = v\]
\[ \Rightarrow \frac{dv}{dx} - \frac{v}{x} = - x \]
Multiplying by the integrating factor \[e^{\int - \frac{1}{x}dx} = \frac{1}{x}\]
\[v . \frac{1}{x} = \int - x . \frac{1}{x}dx + c = - x + c\]
\[ \Rightarrow \frac{y^2}{x^2} = - x + c\]
\[ \Rightarrow x^2 + y^2 = cx\]