Question

Prove that the locus of the middle points of all chords of the parabola $y^2=4ax$ passing through the vertex is the parabola $y^2=2ax$.

Answer 1

$y^2=4ax$

let the mid-point of chord be $(h,k)$ , So chord with mid-point given,

=>$ky-2ah=k^2-4ah+2ax=>ky-2ax=k^2-2ah\to \left(ii\right)$

$\left(ii\right)$ passes through vertex of $\left(i\right)$ $(0,0)$=>$k\left(0\right)-2a\left(0\right)=k^2-2ah=>k^2-2ah$

For locus, replace $h\stackrel{}{\to }x,k\stackrel{}{\to }y=>y^2=2ax$ is a Parabola.