JEE Main 2024 — Parabola Question with Solution

Given: Equation of parabola is $y^2=4x$  and the equation of circle is $x^2+y^2-4x-16y+64=0$.

We know that, equation of normal to parabola is given by, $y=mx-2m-m^3$.

Also, the centre of given circle is $\left(2,8\right)$.

The normal of parabola is passing through center of circle.

$\Rightarrow 8=2m-2m-m^3$

$\Rightarrow m=-2$

So point $P$ on parabola is given by,

$\left({\mathrm{am}}^2,-2\mathrm{am}\right)=(4,4)$

And $\mathrm{C}=(2,8)$

$\Rightarrow PC=\sqrt{4+16}=\sqrt{20}$

${\mathrm{d}}^2=20$.