Question

The normal to a curve at $p(x,y)$ meets the axis at $G.$ if the distance of $G$ from the origin is twice the abscissa of $p,$ then the curve is

Answer 1

$\begin{array}{c}\left(Y-y\right)=-\frac{dx}{dy}\left(X-x\right)\\ 0-y=-\frac{dy}{dx}\left(G-x\right)\\ G=y\frac{dy}{dx}+x\\ \left|G\right|=2\left|x\right|\\ \left|x+y\frac{dy}{dx}\right|2\left|x\right|\\ x+y\frac{dy}{dx}=2xorx+y\frac{dy}{dx}=-2x\\ y\frac{dy}{dx}=xy\frac{dy}{dx}=-3x\\ \int ydy=\int xdxydy=-3x\\ \frac{y^2}{2}=\frac{x^2}{2}+C\frac{y^2}{2}=\frac{-3x^2}{2}+C\\ Hence,thecurveisHyperoblaandellipse\end{array}$