QJEEM25Ma3f58: JEE Main 2025 Mathematics - Class 11 JEEM25Ma3f58 Solution: Let the shortest distance from , to the parabola be 4 . Then…

$\begin{aligned}&\text { Normal at } \mathrm{P}\\&\begin{gathered}y+t x=2 t+t^3 \\\uparrow \\(a, 0)\end{gathered}\end{aligned}$

$\begin{aligned}& \mathrm{at}=2 \mathrm{t}+\mathrm{t}^3 \\& \mathrm{a}=2+\mathrm{t}^2 \\& \mathrm{R}\left(2+\mathrm{t}^2, 0\right) \\& \mathrm{PR}=4 \Rightarrow 4+4 \mathrm{t}^2=16\end{aligned}$

$\begin{aligned}& \quad 4 \mathrm{t}^2=12 \Rightarrow \mathrm{t}^2=3 \\& \mathrm{a}=5, \mathrm{R}(5,0)\end{aligned}$

Focus $(1,0)$

$(1,0) \&(5,0)$ will be the end points of diameter

$\Rightarrow E q^{\mathrm{n}}$ of circle is

$\begin{aligned}& (x-1)(x-5)+y^2=0 \\& x^2+y^2-6 x+5=0\end{aligned}$