Mathematics · Application of Derivatives

JEE Main 2025 — 3 April, Evening Shift — Question 33

The shortest distance between the curves $y^{2}=8 x$ and $x^{2}+y^{2}+12 y+35=0$ is:

Answer: C

Equation of normal: $y=m x-2 a m-a m^{3} \quad(a=2)$

$y=m x-4 m-2 m^{3}$

centre of circle: $c(0,-6)$ , radius $=1$

$-6=-4 m-2 m^{3}$

$\Rightarrow m=1$

$P\left(a m^{2},-2 a m\right)$

$=P(2,-4)$

Shortest distance: $C P-r$

$=\sqrt{4+4}-1$

$=2 \sqrt{2}-1$

Answer key and solution verified before publishing.

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