JEE Main 2026 — Mathematics Coordinate Geometry

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

Distance between directrices is $\dfrac{2a}{e} = \dfrac{8}{3}$e2a​=38​

Multiplying the two equations, we get $4a^2 = 16 \Rightarrow a^2 = 4$4a2=16⇒a2=4

Dividing the two equations, we get $e^2 = \dfrac{9}{4}$e2=49​

Using $b^2 = a^2(e^2 - 1)$b2=a2(e2−1), we get $b^2 = 4\left(\dfrac{9}{4} - 1\right) = 5$b2=4(49​−1)=5

The equation of the hyperbola is $\dfrac{x^2}{4} - \dfrac{y^2}{5} = 1$4x2​−5y2​=1

The line $x = \alpha$x=α intersects the hyperbola at $A$A and $B$B. Substituting $x = \alpha$x=α, we get $\dfrac{\alpha^2}{4} - \dfrac{y^2}{5} = 1 \Rightarrow y = \pm \dfrac{\sqrt{5(\alpha^2 - 4)}}{2}$4α2​−5y2​=1⇒y=±25(α2−4)​​

The coordinates of $A$A and $B$B are $\left(\alpha, \dfrac{\sqrt{5(\alpha^2 - 4)}}{2}\right)$(α,25(α2−4)​​) and $\left(\alpha, -\dfrac{\sqrt{5(\alpha^2 - 4)}}{2}\right)$(α,−25(α2−4)​​)

The length of the base $AB$AB is $\sqrt{5(\alpha^2 - 4)}$5(α2−4)​ and the height of the triangle $AOB$AOB from the origin is $|\alpha|$∣α∣

Area of $\Delta AOB = \dfrac{1}{2} \times AB \times |\alpha| = 4\sqrt{15}$ΔAOB=21​×AB×∣α∣=415​

$$\dfrac{1}{2} \sqrt{5(\alpha^2 - 4)} |\alpha| = 4\sqrt{15}$$21​5(α2−4)​∣α∣=415​

Squaring both sides, we get $\dfrac{1}{4} \times 5(\alpha^2 - 4) \alpha^2 = 240$41​×5(α2−4)α2=240

$$\alpha^4 - 4\alpha^2 - 192 = 0$$α4−4α2−192=0

$$(\alpha^2 - 16)(\alpha^2 + 12) = 0$$(α2−16)(α2+12)=0

Since $\alpha^2 > 0$α2>0, we have $\alpha^2 = 16$α2=16

Answer: $16$16