JEE Main 2026 — Jan 23, Shift 2 — Mathematics solution

JEE Main2026Jan 23, Shift 2Mathematics

Let $P Q$ be a chord of the hyperbola $\frac{x ^{2}}{4} - \frac{y ^{2}}{b ^{2}} = 1$ , perpendicular to the x-axis such that $O P Q$ is an equilateral triangle, $O$ being the centre of the hyperbola. If the eccentricity of the hyperbola is $3$ , then the area of the triangle $O P Q$ is :

A
$23$
B
$\frac{8 3}{5}$
C
$\frac{11}{5}$
D
$\frac{9}{5}$

Solution

Eccentricity: $e^{2} = 1 + \frac{b ^{2}}{4} = 3 \Rightarrow b^{2} = 8$ .

Let $P = (2 sec θ, 22 tan θ)$ and $Q = (2 sec θ, - 22 tan θ)$ , with $M$ the midpoint of $P Q$ on the x-axis: $M = (2 sec θ, 0)$ .

$O P Q$ equilateral means $∠ P O M = 3 0^{\circ}$ , so $tan 3 0^{\circ} = \frac{P M}{O M} = \frac{2 2 tan θ}{2 sec θ} = 2 sin θ$ .

$2 sin θ = \frac{1}{3} \Rightarrow sin θ = \frac{1}{6}$ , hence $cos^{2} θ = \frac{5}{6}$ .

Area $= 2 \cdot \frac{1}{2} \cdot O M \cdot P M = 2 sec θ \cdot 22 tan θ = 42 \cdot \frac{sin θ}{cos ^{2} θ} = 42 \cdot \frac{1/ 6}{5/6} = \frac{8 3}{5}$ .

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