JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Parabola

Let

P (4, 43)

be a point on the parabola

y^{2} = 4 ax

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:

(A) $\frac{263 3}{8}$
(B) $173$
(C) $\frac{343 3}{8}$ ✓ Correct
(D) $\frac{34 3}{3}$

Answer: (C)

Solution

Since

P (4, 43)

lies on

y^{2} = 4 a x

48 = 16 a \Rightarrow a = 3,

so the parabola is

y^{2} = 12 x

, with focus

(3, 0)

and directrix

x = - 3

.

Write points in parametric form

(3 t^{2}, 6 t)

. For P:

6 t_{1} = 43 \Rightarrow t_{1} = \frac{2}{3}

(check:

3 t_{1}^{2} = 3 \cdot \frac{4}{3} = 4

, consistent).

For a focal chord,

t_{1} t_{2} = - 1

, so

t_{2} = - \frac{3}{2}

, giving

Q = (3 \cdot \frac{3}{4}, 6 \cdot (- \frac{3}{2})) = (\frac{9}{4}, - 33) .

M and N are the feet of perpendiculars from P and Q on the directrix

x = - 3

, so

M = (- 3, 43)

and

N = (- 3, - 33)

. PM and QN are both horizontal, hence parallel, so PQMN is a trapezium with parallel sides

P M = 4 - (- 3) = 7, QN = \frac{9}{4} - (- 3) = \frac{21}{4},

and height (distance between the parallel sides)

M N = 43 - (- 33) = 73 .

Area of trapezium:

\frac{1}{2} (P M + QN) \cdot M N = \frac{1}{2} (7 + \frac{21}{4}) \cdot 73 = \frac{1}{2} \cdot \frac{49}{4} \cdot 73 = \frac{343 3}{8} .

JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Single correct option

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JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Parabola

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