Question

Let the equation of the curve passing through the origin if the middle point of the segment of its normal from any point of the curve to the x - axis lies on the parabola $2y^2=x$ be $y^2=kx+m-e^{nx}$.Find $k+m+n$ ?

Answer 1

$Y-y=-\frac{1}{y^{\prime }}(X-x)$
At x axis,
$X=x+y.y^{\prime }$
mid point of PQ $=\left(\frac{2x+y{y}^{\prime }}{2},\frac{y}{2}\right)$
mid point lies on $2y^2=x$
$\therefore \frac{2y^2}{4}=\frac{2x+y{y}^{\prime }}{2}\Rightarrow \frac{ydy}{dx}-y^2=-2x$
Put $y^2=t$
$\therefore 2y\frac{dy}{dx}=\frac{dt}{dx}$
$\frac{1}{2}\frac{dt}{dx}-t=2x$
$\frac{dt}{dx}+(-2)t=-4x$
$\therefore t{e}^{-2x}=-4\int e^{-2x}.xdx+c$
$\therefore y^2{e}^{-2x}=e^{-2x}(2x+1)+c$
$x=0,y=0\Rightarrow c=-1$
$\therefore y^2=2x+1-e^{2x}$