Question

Find the locus of the mid - points of the chords of the parabola $y^2=4ax$ which subtend
a right angle at the vertex.

• A. $y^2=2a(x-4a)$
• B. $y^2=a(x-4a)$
• C. $y^2=a/4(x-5a)$
• D. $y^2=2a(x-5a)$

Answer 1

The correct option is A $y^2=2a(x-4a)$

Let $P$ be $t_1$ and $Q$ be $t_2$

Since, $PQ$ subtends a right angle at the vertex $O(0,0)$ therefore as in $\left(c_1\right)$

$t_1{t}_2=-4.$ If $(h,k)$ be the mid - point, then

$2h=a(t_1^2+t_2^2)$ and $2k=2a(t_1+t_2)$

$2h=a[{(t_1+t_2)}^2-2t_1{t}_2]$

$2h=a[\frac{k^2}{a^2}+8]$ or $2a(h-4a)=k^2$

$\therefore$ Locus is $y^2=2a(x-4a)$