Thank you for reporting, we will resolve it shortly
Q.
The centroid of the triangle formed by the feet of the normals from the point $(h, k)$ to the parabola $y^{2}+4 a x$ $=0,(a>0)$ lies on
Conic Sections
Solution:
Co-ordinates of any point on the parabola $y^{2}=-4 a x$ are $\left(-a t^{2}, 2 a t\right)$
Equation of the normal at $\left(-a t^{2}, 2 a t\right)$ is
$y-x t=2 a t+a t^{3}$
If the normal passes through the point $(h, k)$, then
$k-t h=2 a t+a t^{3}$
or $ a t^{3}+(2 a+h) t-k=0$,
which is a cubic equation whose three roots $t_{1}, t_{2}, t_{3}$ are the parameters of the feet of the three normals.
$\therefore $ Sum of the roots $=t_{1}+t_{2}+t_{3}=-\frac{\text { Coefficient of } t^{2}}{\text { Coefficient of } t^{3}}=0$
$\therefore $ Centroid of the triangle formed by the feet of the normals
$=\left(-\frac{a}{3}\left(t_{1}^{2}+t_{2}^{2}+t_{3}^{2}\right), \frac{2 a}{3}\left(t_{1}+t_{2}+t_{3}\right)\right)$
$=\left(-\frac{a}{3}\left(t_{1}^{2}+t_{2}^{2}+t_{3}^{2}\right), 0\right)$
which, clearly, lies on the $x$-axis.