Question

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

Answer 1

One end of the chord is the vertex of parabola, which is $O(0,0)$.

Let the mid point of the chords be $M(h,k)$ and the other end of the chord be $P(a{t}^2,2at)$

Mid point of $OP$ is

$(\frac{a{t}^2+0}{2},\frac{2at+0}{2})\therefore M(\frac{a{t}^2}{2},at)\Rightarrow h=\frac{a{t}^2}{2}......\left(i\right)k=at\Rightarrow t=\frac{k}{a}$

Substituting $t$ in $\left(i\right)$, we get

$h=\frac{a{\left(\frac{k}{a}\right)}^2}{2}2h=\frac{k^2}{a}{k}^2=2ah$

Replacing $h$ by $x$ and $k$ by $y$

$y^2=2ax$

is the required locus of mid point of chords