JEE Main 2025 Jan 22 Shift 2, Mathematics Q3: Parabola: Chord, Tangent and Normal

JEE Main2025Jan 22, Shift 2Mathematics

Let $P (4, 43)$ be a point on the parabola $y^{2} = 4 a x$ and $P Q$ 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 $P QM N$ is equal to :

A
$\frac{263 3}{8}$
B
$173$
C
$\frac{343 3}{8}$
D
$\frac{34 3}{3}$

Solution

Since $P (4, 43)$ lies on $y^{2} = 4 a x$ : $(43)^{2} = 4 a (4) \Rightarrow 48 = 16 a \Rightarrow 4 a = 12$ . The parabola is $y^{2} = 12 x$ with $a = 3$ .

Parametric form: $P = (a t_{1}^{2}, 2 a t_{1}) = (3 t_{1}^{2}, 6 t_{1})$ . Matching with $(4, 43)$ : $3 t_{1}^{2} = 4$ gives $t_{1} = \frac{2}{3}$ .

For a focal chord, $t_{1} t_{2} = - 1$ , so $t_{2} = - \frac{3}{2}$ .

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

$P QM N$ is a trapezium with $P M ∥ QN$ (both perpendicular to the directrix). For a focal chord, $P M = P S = a (1 + t_{1}^{2})$ and $QN = QS = a (1 + t_{2}^{2})$ , so $P M + QN = P Q$ (focal chord length).

Using the standard result $P Q = a (t_{1} + \frac{1}{t _{1}})^{2}$ :

$t_{1} + \frac{1}{t _{1}} = \frac{2}{3} + \frac{3}{2} = \frac{4 + 3}{2 3} = \frac{7}{2 3}$ , so $P Q = 3 \cdot \frac{49}{12} = \frac{49}{4}$ .

Also $M N = ∣6 t_{1} - 6 t_{2} ∣ = 6 (\frac{2}{3} + \frac{3}{2}) = 6 \cdot \frac{7}{2 3} = \frac{21}{3} = 73$ .

Area of trapezium $= \frac{1}{2} \cdot M N \cdot (P M + QN) = \frac{1}{2} \cdot M N \cdot P Q = \frac{1}{2} \cdot 73 \cdot \frac{49}{4} = \frac{343 3}{8}$ .

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JEE Main 2025 Jan 22 Shift 2, Mathematics Q3: Parabola: Chord, Tangent and Normal

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