Question
Let $H:\frac{x^{2}}{a^{2}} - \frac{y^{2}}{{\text{ }b}^{2}} = 1$ be a hyperbola such that the distance between its foci is 6 and the distance between its directrices is $\frac{8}{3}$. If the line $x = \alpha$ intersects the hyperbola H at the points A and B such that the area of the triangle AOB is $4\sqrt{15}$, where O is the origin, then $\alpha^{2}$ equals
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Solution
Sol. [Key Concept: $2ae = 6$ and $\frac{2a}{e} = \frac{8}{3}$ fix $a^{2},b^{2}$; the area of $triangle\ AOB$ at $x = \alpha$ gives $\alpha^{2}$.]
$$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 $$
Given:
$${2ae = 6 \Rightarrow ae = 3 }{\frac{2a}{e} = \frac{8}{3} \Rightarrow \frac{a}{e} = \frac{4}{3}}$$
$$\begin{matrix} a^{2} = 4\ e = \frac{3}{2}\ b^{2} = 5 & \\ \frac{x^{2}}{4} - \frac{y^{2}}{5} = 1 & \text{~}\text{solve with}\text{~}x = \alpha \\ y = \pm \sqrt{\frac{5\left( \alpha^{2} - 4 \right)}{4}} & \\ A\left( \alpha,\sqrt{\frac{5\left( \alpha^{2} - 4 \right)}{4}} \right) & B\left( \alpha, - \sqrt{\frac{5\left( \alpha^{2} - 4 \right)}{4}} \right) \end{matrix}$$
Ar. of $\bigtriangleup OAB = \frac{1}{2}|\alpha|\sqrt{5\left( \alpha^{2} - 4 \right)} = 4\sqrt{15}$
$${\Rightarrow 16 \times 15 = \frac{1}{4}\alpha^{2}(5)\left( \alpha^{2} - 4 \right) }{\Rightarrow \alpha^{2} = 16 }$$
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