QJEEM24Mabede: JEE Main 2024 Mathematics - Class 11 JEEM24Mabede Solution: Let the length of the focal chord PQ of the parabola be 15…

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

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

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

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

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

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

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