JEE Main 2026 — Hyperbola Question with Solution

Given distance between foci is $2ae=6\Rightarrow ae=3$

Distance between directrices is $\frac{2a}{e}=\frac{8}{3}$

Multiplying the two equations, we get $4a^2=16\Rightarrow a^2=4$

Dividing the two equations, we get $e^2=\frac{9}{4}$

Using $b^2=a^2(e^2-1)$, we get $b^2=4\left(\frac{9}{4}-1\right)=5$

The equation of the hyperbola is $\frac{x^2}{4}-\frac{y^2}{5}=1$

The line $x=\alpha$ intersects the hyperbola at $A$ and $B$. Substituting $x=\alpha$, we get $\frac{{\alpha }^2}{4}-\frac{y^2}{5}=1\Rightarrow y=\pm \frac{\sqrt{5({\alpha }^2-4)}}{2}$

The coordinates of $A$ and $B$ are $\left(\alpha ,\frac{\sqrt{5({\alpha }^2-4)}}{2}\right)$ and $\left(\alpha ,-\frac{\sqrt{5({\alpha }^2-4)}}{2}\right)$

The length of the base $AB$ is $\sqrt{5({\alpha }^2-4)}$ and the height of the triangle $AOB$ from the origin is $|\alpha |$

Area of $\Delta AOB=\frac{1}{2}\times AB\times |\alpha |=4\sqrt{15}$

$\frac{1}{2}\sqrt{5({\alpha }^2-4)}|\alpha |=4\sqrt{15}$

Squaring both sides, we get $\frac{1}{4}\times 5({\alpha }^2-4){\alpha }^2=240$

${\alpha }^4-4{\alpha }^2-192=0$

$({\alpha }^2-16)({\alpha }^2+12)=0$

Since ${\alpha }^2>0$, we have ${\alpha }^2=16$

Answer: $16$