Coordinate Geometry – Hyperbola – JEE Main 23 Jan 2026 Shift 2

Solution:

First, we use the formula for eccentricity to find $b^2$:
$$ e = \sqrt{1 + \frac{b^2}{a^2}} $$

Given $e = \sqrt{3}$ and $a^2 = 4$:
$$ \sqrt{3} = \sqrt{1 + \frac{b^2}{4}} \Rightarrow 3 = 1 + \frac{b^2}{4} $$

$$ \frac{b^2}{4} = 2 \Rightarrow b^2 = 8 $$

So, the equation of the hyperbola is:
$$ \frac{x^2}{4} – \frac{y^2}{8} = 1 $$


Let the coordinates of point P be $(2\sec\theta, 2\sqrt{2}\tan\theta)$.
Since triangle OPQ is equilateral and symmetric about the x-axis (as PQ is perpendicular to the x-axis), the angle $\angle POM = 30^\circ$ (where M is on the x-axis).

In $\Delta OMP$:
$$ \tan 30^\circ = \frac{PM}{OM} = \frac{y_P}{x_P} $$

$$ \frac{1}{\sqrt{3}} = \frac{2\sqrt{2}\tan\theta}{2\sec\theta} = \sqrt{2}\sin\theta $$

$$ \sin\theta = \frac{1}{\sqrt{6}} $$

Now, we calculate the area of $\Delta OPQ$.
$$ \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times (2 \times PM) \times OM = PM \times OM $$

$$ \text{Area} = (2\sqrt{2}\tan\theta)(2\sec\theta) = 4\sqrt{2} \frac{\sin\theta}{\cos^2\theta} $$

Using $\cos^2\theta = 1 – \sin^2\theta = 1 – \frac{1}{6} = \frac{5}{6}$:
$$ \text{Area} = 4\sqrt{2} \times \frac{1/\sqrt{6}}{5/6} = 4\sqrt{2} \times \frac{1}{\sqrt{6}} \times \frac{6}{5} $$

$$ = \frac{4\sqrt{2} \cdot \sqrt{6}}{5} = \frac{4\sqrt{12}}{5} = \frac{8\sqrt{3}}{5} $$

Ans. (2)