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

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

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

Focus $(1,0)$

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

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

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

19:T4f8,

JEE Main 2025 (Online) 23rd January Evening Shift Mathematics - Parabola Question 25 Hindi Explanation

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

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

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

नाभि $(1,0)$

$(1,0)$ और $(5,0)$ व्यास के

Parabola

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