JEE Main 2025 — Parabola Question with Solution

Normal at P $\begin{aligned} & y+ t x=2 t+t^3 \\ & \uparrow \\ &(\mathrm{a}, 0) \\ & \mathrm{at}= 2 \mathrm{t}+\mathrm{t}^3 \\ & \mathrm{a}= 2+\mathrm{t}^2 \\ & \mathbb{R}\left(2+\mathrm{t}^2, 0\right) \end{aligned}$ $\begin{aligned} & \mathrm{PR}=4 \Rightarrow 4+4 \mathrm{t}^2=16 \\ & 4 \mathrm{t}^2=12 \Rightarrow \mathrm{t}^2=3 \\ & \mathrm{a}=5 \mathbb{R}(5,0) \end{aligned}$ Focus $(1,0)$ $(1,0)\&(5,0)$ will be tha end pts. of diameter $\Rightarrow {\mathrm{Eg}}^{\mathrm{n}}$ of circle is $\begin{aligned} & (x-1)(x-5)+y^2=0 \\ & x^2+y^2-6 x+5=0 \end{aligned}$