Value of Alpha Squared for Hyperbola Area Problem

To find the value of $\alpha^2$, we begin by analyzing the standard equation of the hyperbola given by: $H : \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

1. Find $a^2$ and $b^2$ using the given dimensions:

   • The distance between the foci is given as $6$: $2ae = 6 \implies ae = 3$
   • The distance between the directrices is given as $\frac{8}{3}$: $\frac{2a}{e} = \frac{8}{3} \implies \frac{a}{e} = \frac{4}{3}$

   Multiplying these two equations: $(ae) \cdot \left(\frac{a}{e}\right) = 3 \cdot \frac{4}{3} \implies a^2 = 4$

   The eccentricity squared $e^2$ is found by dividing the equations: $e^2 = \frac{ae}{a/e} = \frac{3}{4/3} = \frac{9}{4}$

   Now, we calculate $b^2$ using the standard identity $b^2 = a^2(e^2 - 1)$: $b^2 = 4 \left(\frac{9}{4} - 1\right) = 4 \cdot \frac{5}{4} = 5$

   Therefore, the equation of the hyperbola $H$ is: $\frac{x^2}{4} - \frac{y^2}{5} = 1$

2. Determine the coordinates of points $A$ and $B$: The line $x = \alpha$ intersects the hyperbola $H$ at points $A$ and $B$. Substituting $x = \alpha$ into the equation of the hyperbola: $\frac{\alpha^2}{4} - \frac{y^2}{5} = 1 \implies \frac{y^2}{5} = \frac{\alpha^2}{4} - 1 = \frac{\alpha^2 - 4}{4}$ $y^2 = \frac{5(\alpha^2 - 4)}{4} \implies y = \pm \frac{\sqrt{5(\alpha^2 - 4)}}{2}$

   Thus, the points of intersection are $A\left(\alpha, \frac{\sqrt{5(\alpha^2 - 4)}}{2}\right)$ and $B\left(\alpha, -\frac{\sqrt{5(\alpha^2 - 4)}}{2}\right)$.

3. Calculate the area of $\triangle AOB$:

   • The length of the base $AB$ (which is a vertical line segment) is: $AB = 2 \cdot \frac{\sqrt{5(\alpha^2 - 4)}}{2} = \sqrt{5(\alpha^2 - 4)}$
   • The height of $\triangle AOB$ from the origin $O(0,0)$ to the vertical line $x = \alpha$ is $|\alpha|$.

   The area of $\triangle AOB$ is given as $4\sqrt{15}$: $\text{Area}(\triangle AOB) = \frac{1}{2} \times \text{base} \times \text{height}$ $4\sqrt{15} = \frac{1}{2} \cdot \sqrt{5(\alpha^2 - 4)} \cdot |\alpha|$

4. Solve for $\alpha^2$: Multiplying both sides by $2$: $8\sqrt{15} = |\alpha|\sqrt{5(\alpha^2 - 4)}$

   Squaring both sides: $(8\sqrt{15})^2 = \alpha^2 \cdot 5(\alpha^2 - 4)$ $64 \cdot 15 = 5\alpha^2(\alpha^2 - 4)$ $960 = 5\alpha^2(\alpha^2 - 4)$

   Dividing by $5$: $\alpha^2(\alpha^2 - 4) = 192$ $\alpha^4 - 4\alpha^2 - 192 = 0$

   Factoring the quadratic equation in terms of $\alpha^2$: $(\alpha^2 - 16)(\alpha^2 + 12) = 0$

   Since $\alpha^2$ must be a positive real number, we take: $\alpha^2 = 16$

Correct Option: B