Let  $$H : \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ be a  hyperbola such that the distance between its foci equal to $$6$$ and distance between its directrices  is  $$\frac{8}{3}$$. If the line $$x = \alpha$$ intersects the hyperbola $$H$$ at $$A$$ and $$B$$, such that  the area of $$\triangle AOB$$ (where $$O$$ is the origin) is $$4\sqrt{15}$$, then $$\alpha^2$$ is equal to :

The hyperbola is $$\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$$ with centre at the origin and transverse axis along the $$x$$-axis.

Step 1: Use the distance between the foci.
For this hyperbola the foci are $$\bigl(\pm c,0\bigr)$$ where $$c^{2}=a^{2}+b^{2}$$.
Given distance between the foci $$=2c=6 \; \Longrightarrow \; c=3.$$p>

Step 2: Use the distance between the directrices.
For this hyperbola the directrices are $$x=\pm\dfrac{a}{e}$$ where the eccentricity $$e=\dfrac{c}{a}$$.
Hence the distance between the directrices is $$2\left(\dfrac{a}{e}\right)=\dfrac{2a^{2}}{c}.$$ Given distance $$=\dfrac{8}{3}$$, so $$\dfrac{2a^{2}}{c}=\dfrac{8}{3} \quad\Longrightarrow\quad \dfrac{2a^{2}}{3}=\dfrac{8}{3} \quad\Longrightarrow\quad 2a^{2}=8 \quad\Longrightarrow\quad a^{2}=4.$$

Step 3: Find $$b^{2}$$.
Using $$c^{2}=a^{2}+b^{2}$$ gives $$9=4+b^{2}\quad\Longrightarrow\quad b^{2}=5.$$ Thus the hyperbola is $$\dfrac{x^{2}}{4}-\dfrac{y^{2}}{5}=1.$$

Step 4: Coordinates of the intersection points with $$x=\alpha$$.
Substitute $$x=\alpha$$ in the hyperbola: $$\dfrac{\alpha^{2}}{4}-\dfrac{y^{2}}{5}=1 \;\Longrightarrow\; \dfrac{y^{2}}{5}=\dfrac{\alpha^{2}}{4}-1 \;\Longrightarrow\; y^{2}=5\left(\dfrac{\alpha^{2}}{4}-1\right) =\dfrac{5\alpha^{2}}{4}-5.$$ Hence $$A\bigl(\alpha,+\sqrt{\tfrac{5\alpha^{2}}{4}-5}\bigr),\; B\bigl(\alpha,-\sqrt{\tfrac{5\alpha^{2}}{4}-5}\bigr).$$

Step 5: Area of $$\triangle AOB$$.
Base $$AB$$ is a vertical segment of length $$2\sqrt{\tfrac{5\alpha^{2}}{4}-5}.$$ The perpendicular distance from the origin $$O(0,0)$$ to the line $$x=\alpha$$ is $$|\alpha|$$.
Therefore $$\text{Area}= \dfrac12\,( \text{base} )\,( \text{height} ) =\dfrac12\left[2\sqrt{\tfrac{5\alpha^{2}}{4}-5}\right]|\alpha| =|\alpha|\sqrt{\tfrac{5\alpha^{2}}{4}-5}.$$

The given area is $$4\sqrt{15}$$, so (taking $$\alpha\gt 0$$ because only $$\alpha^{2}$$ is needed) $$\alpha\sqrt{\tfrac{5\alpha^{2}}{4}-5}=4\sqrt{15}.$$ Square both sides: $$\alpha^{2}\left(\dfrac{5\alpha^{2}}{4}-5\right)=240.$$ Put $$\alpha^{2}=X$$: $$X\left(\dfrac{5X-20}{4}\right)=240 \;\Longrightarrow\; X(5X-20)=960 \;\Longrightarrow\; 5X^{2}-20X-960=0 \;\Longrightarrow\; X^{2}-4X-192=0.$$ Solve the quadratic: $$X=\dfrac{4\pm\sqrt{4^{2}+4\cdot192}}{2} =\dfrac{4\pm28}{2}\;\Longrightarrow\; X=16\; (\text{or }-12).$$ Since $$X=\alpha^{2}\gt 0$$, we take $$X=16.$$

Result.
$$\alpha^{2}=16.$$

Option B which is: $$16$$