JEE Main 2025 — Application of Derivatives Question with Solution

Equation of normal to parabola
$y^2=8x\text{ is }y=mx-4m-2m^3$
passes through $(0,-6)$ we get
$\begin{aligned}
& -6=-4 \mathrm{~m}-2 \mathrm{~m}^3 \\ & \Rightarrow \mathrm{~m}^3+2 \mathrm{~m}-3=0 \\ & \Rightarrow(\mathrm{~m}-1)\left(\mathrm{m}^2+\mathrm{m}+3\right)=0 \Rightarrow \mathrm{~m}=-1 \\ & \mathrm{P}=\left(\mathrm{am}^2,-2 \mathrm{am}\right)=(2,-4) \\ & \therefore \text { Shortest distance }=\mathrm{PC}-\mathrm{r} \\ & =(2 \sqrt{2}-1)
\end{aligned}$