Question

Find the equation of the curve passing through the orgin if the middle point of the segment of its normal from any point of the curve to the x-axis lies on the parabola $2y^2=x$.

Answer 1

Find the equation of curve passing through the origin if the middle point of its normal from any point of the curve to the $x-$axis lies on the parabola $2y^2=x$

Equation of normal at point $p(x,y)$ is

$y-x=\frac{dy}{dx}(X-x)$

Normal meets at $x-$ axis at $Q(X,0)$

$y-x=\frac{dy}{dx}(X-x)$

$X=y\frac{dy}{dx}+xQ=(y\frac{dy}{dx}+x,0)$

Let middle point of $PQ$ is $R$ which is $(\frac{x+x+y\frac{dy}{dx}}{2},\frac{y+0}{2})$

$=(\frac{2x+y\frac{dy}{dx}}{2},\frac{y}{2})$

$R$ lies on parabola $2y^2=x$

Let $y^2=t..............\left(1\right)$

$2y\frac{dy}{dx}=dtt=2x+\frac{1}{2}\frac{dt}{dx}\frac{dt}{dx}-2t=-4xIF=e^{\int (-2)dx}=e^{-2x}t\left(IF\right)=\int (-4x)\left(IF\right)dx+Ct{e}^{-2x}=\frac{-4x}{-2}{e}^{-2x}-\int \frac{-4x}{-2}{e}^{-2x}+Ct{e}^{-2x}=2x{e}^{-2x}+e^{-2x}+Ct=2x++1+c{e}^{2x}$

From (1)

$y^2=2x+1+c{e}^{2x}$

If this passes through origin as given in $Q$

$C=-1$

Therefore the equation of required parabola is $y^2=2x+1-e^{2x}$