Let the length of the focal chord PQ of the parabola $y^2=12 x$ be 15 units. If the distance of…
Solution

$\begin{aligned} & \text { length of focal chord }=4 a \operatorname{cosec}^2 \theta=15 \\ & 12 \operatorname{cosec}^2 \theta=15 \\ & \sin ^2 \theta=\frac{4}{5} \\ & \tan ^2 \theta=4 \\ & \tan \theta=2 \\ & \text { equation } \frac{y-0}{x-3}=2 \\ & y=2 x-6 \\ & 2 x-y-6=0 \\ & P=\frac{6}{\sqrt{5}} \\ & 10 p^2=10 \cdot \frac{36}{5}=72\end{aligned}$
Asked in: JEE Main 2024 (04 Apr Shift 1)