Let be a point on the parabola 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 : [2025]

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Solution

Q.
Let $P (4, 4 \sqrt{3})$ be a point on the parabola $y^{2} = 4 a x$ 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 : [2025]

1 $\frac{34 \sqrt{3}}{3}$

2 $\frac{343 \sqrt{3}}{8}$

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

4 $17 \sqrt{3}$

Ans.
(2)

Since, $P (4, 4 \sqrt{3})$ lies on $y^{2} = 4 a x$

$\Rightarrow 48 = 4 a (4) \Rightarrow 4 a = 12$

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

Let $(4, 4 \sqrt{3})$ be $(a t^{2}, 2 a t) \Rightarrow t = \frac{2}{\sqrt{3}}$

As PQ is focal chord

$∴ Q (\frac{a}{t^{2}}, \frac{- 2 a}{t}) = (\frac{9}{4}, - 3 \sqrt{3})$

Area of trapezium PQNM

$= \frac{1}{2} M N (P M + Q N)$

$= \frac{1}{2} M N (P S + Q S)$ [by definition of parabola]

$= \frac{1}{2} \times M N \times P Q$

$= \frac{1}{2} \times 7 \sqrt{3} (\frac{49}{4}) = (343) \frac{\sqrt{3}}{8}$ .

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