Let $P (4, 4 \sqrt{3})$ be a point on the parabola y² = 4ax 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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$\frac{343 \sqrt{3}}{8}$

$17 \sqrt{3}$

$\frac{263 \sqrt{3}}{8}$

$\frac{34 \sqrt{3}}{3}$

answer is C.

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Detailed Solution

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$(4, 4 \sqrt{3}) lies on y^{2} = 4 a x \begin{matrix} \Rightarrow 48 = 4 a .4 \\ 4 a = 12 \end{matrix}$

$\Rightarrow y^{2} = 12 x is equation of parabola$

$Now, parameter of P is t_{1} = \frac{2}{\sqrt{3}} \Rightarrow Parameters of Q$

$is t_{2} = - \frac{\sqrt{3}}{2} \Rightarrow Q (\frac{9}{4}, - 3 \sqrt{3})$

Area of trapezium PQNM

$\begin{array}{l} = \frac{1}{2} MN \cdot (PM + QN) \\ = \frac{1}{2} MN \cdot (PS + QS) \\ = \frac{1}{2} MN \cdot PQ \\ = \frac{1}{2} 7 \sqrt{3} \cdot \frac{49}{4} = (343) \frac{\sqrt{3}}{8} \end{array}$

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