Let H: (x^2)/(a^2)-(y^2)/(b^2)=1 be a hyperbola such that the...

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Answer: $16$