JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Hyperbola

The eccentricity fixes $b^2$: $e^2=1+\frac{b^2}{4}=3$ gives $b^2=8$, so the curve is $\frac{x^2}{4}-\frac{y^2}{8}=1$. The chord $PQ$ is vertical, so with foot $M=(x_0,0)$ the vertices are $P=(x_0,y_0)$ and $Q=(x_0,-y_0)$, while $O$ and $M$ lie on the axis. In the equilateral triangle the line $OP$ makes $30^{\circ }$ with $OM$, so $\tan 30^{\circ }=\frac{y_0}{x_0}=\frac{1}{\sqrt{3}}$, i.e. $x_0=\sqrt{3}{y}_0$. Substituting into the hyperbola, $\frac{3y_0^2}{4}-\frac{y_0^2}{8}=1\Rightarrow \frac{5y_0^2}{8}=1\Rightarrow y_0^2=\frac{8}{5}$. The area is $\frac{1}{2}\cdot PQ\cdot OM=\frac{1}{2}(2y_0)(x_0)=\sqrt{3}{y}_0^2=\sqrt{3}\cdot \frac{8}{5}=\frac{8\sqrt{3}}{5}$.