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$$\begin{aligned} & A B=15 \Rightarrow\left(3\left(t^2-\frac{1}{t^2}\right)\right)+\left(6\left(t+\frac{1}{t}\right)\right)^2=225 \\ & \Rightarrow 9\left(t^2-\frac{1}{t^2}\right)+36\left(t+\frac{1}{t}\right)^2=225 \end{aligned}$$

$$ \begin{aligned} \Rightarrow & \left.\left.\left(t+\frac{1}{t}\right)^2 \right[\,\left(t-\frac{1}{t}\right)+4\right]=25 \\ & \left(t+\frac{1}{t}\right)^2\left(t+\frac{1}{t}\right)^2=25 \Rightarrow\left(t+\frac{1}{t}\right)^4=25 \\ \Rightarrow & t+\frac{1}{t}= \pm \sqrt{5} \Rightarrow\left(t-\frac{1}{t}\right)= \pm 1 \end{aligned}$$

Equation of $$A B:(y-6 t)=\left(\frac{2 t}{t^2-1}\right)\left(x-3 t^2\right)$$

$$\Rightarrow$$ Distance from $$y-6 t=m x-3 m t^2$$

$$\Rightarrow p=\frac{\left|3 m t^2-6 t\right|}{\sqrt{1+m^2}}=\frac{\left|\left(\frac{6 t}{t^2-1}\right)\right|}{\sqrt{5}}=\frac{6}{\sqrt{5}}$$

$$\left[ m=\frac{2 t}{t^2-1}=\frac{}{t-\frac{1}{t}}= \pm 2 \Rightarrow m^2=4\right]$$

$$\Rightarrow \quad 10 p^2=\frac{10 \times 36}{5} \Rightarrow 72$$

19:T631,

JEE Main 2024 (Online) 4th April Morning Shift Mathematics - Parabola Question 38 Hindi Explanation

$$ \begin{aligned} \Rightarrow & \left.\le

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