Let $\mathrm{P}(4,4 \sqrt{3})$ be a point on the parabola $y^2=4 \mathrm{a} x$ and PQ be a focal chord of the parabola. If M and $N$ are the foot of perpendiculars drawn from $P$ and $Q$ respectively on the directrix of the parabola, then the area of the quadrilateral PQMN is equal to :

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Aicharya · Accepted Answer

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

Let $\mathrm{P}(4,4 \sqrt{3})$ be a point on the parabola $y^2=4 \mathrm{a} x$ and PQ be a focal chord of…

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