Mathematics · Parabola

JEE Main 2024 — 27 January, Shift 1 — Question 7

If the shortest distance of the parabola $y^{2}=4 x$ from the centre of the circle $x^{2}+y^{2}-4 x-16 y+64=0$ is $d$ , then $d^{2}$ is equal to :

Answer: C

Equation of normal to parabola $\mathrm{y}=\mathrm{mx}-2 \mathrm{~m}-\mathrm{m}^{3}$

this normal passing through center of circle $(2,8)$

\begin{aligned}& 8=2 m-2 m-m^{3}, & m=-2\end{aligned}

So point P on parabola $\Rightarrow\left(\mathrm{am}^{2},-2 \mathrm{am}\right)=(4,4)$

And C = $(2,8)$

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

$d^{2}=20$

Answer key and solution verified before publishing.

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