JEE Main 2025 — Parabola Question with Solution

$\begin{aligned} & (4,4 \sqrt{3}) \text { lies on } y^2=4 \mathrm{ax} \\ & \Rightarrow 48=4 \mathrm{a} \cdot 4 \\ & \quad 4 \mathrm{a}=12 \end{aligned}$ $\Rightarrow y^2=12x$ is equation of parabola Now, parameter of $P$ is $t_1=\frac{2}{\sqrt{3}}\Rightarrow$ Parameters of $Q$ is ${\mathrm{t}}_2=-\frac{\sqrt{3}}{2}\Rightarrow \mathrm{Q}\left(\frac{9}{4},-3\sqrt{3}\right)$ Area of trapezium PQNM $\begin{aligned} & =\frac{1}{2} \mathrm{MN} \cdot(\mathrm{PM}+\mathrm{QN}) \\ & =\frac{1}{2} \mathrm{MN} \cdot(\mathrm{PS}+\mathrm{QS}) \\ & =\frac{1}{2} \mathrm{MN} \cdot \mathrm{PQ} \\ & =\frac{1}{2} 7 \sqrt{3} \cdot \frac{49}{4}=(343) \frac{\sqrt{3}}{8} \end{aligned}$