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
$$\begin{gathered} p=\frac{x{\displaystyle \frac{dy}{dx}}-y}{\sqrt{\left[1+{\left({\displaystyle \frac{dy}{dx}}\right)}^2\right]}}=x\left(given\right) \\ \Rightarrow x^2{\left(\frac{dy}{dx}\right)}^2-2xy\frac{dy}{dx}+y^2=x^2+x^2{\left(\frac{dy}{dx}\right)}^2 \\ \Rightarrow y^2-2xy\frac{dy}{dx}-x^2=0 \\ Henceproved. \\ Now,y^2-2xy\frac{dy}{dx}-x^2=0\Rightarrow \frac{dy}{dx}=\frac{y^2-x^2}{2xy} \\ \Rightarrow 2xy\frac{dy}{dx}-y^2=-x^2 \\ \Rightarrow 2y\frac{dy}{dx}-\frac{y^2}{x}=-x^{} \\ Let{y}^2=v \\ \Rightarrow \frac{dv}{dx}-\frac{v}{x}=-x \end{gathered}$$
Multiplying by the integrating factor $e^{\int -\frac{1}{x}dx}=\frac{1}{x}$
$$\begin{gathered} v.\frac{1}{x}=\int -x.\frac{1}{x}dx+c=-x+c \\ \Rightarrow \frac{y^2}{x^2}=-x+c \\ \Rightarrow x^2+y^2=cx \end{gathered}$$