Let H:x^2a^2-y^2b^2=1 be a hyperbola such that the distance between its foci is 6 and the …

Question

Let $H:\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ be a hyperbola such that the distance between its foci is $6$ and the distance between its directrices is $\dfrac83$. If the line $x=\alpha$ intersects the hyperbola $H$ at 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:

Accepted Answer

$2ae=6\Rightarrow ae=3$; $\dfrac{2a}{e}=\dfrac83\Rightarrow \dfrac{a}{e}=\dfrac43$. Multiplying: $a^2=4\Rightarrow a=2$, and $e^2=\dfrac{9}{4}$.

$$b^2=a^2(e^2-1)=4\left(\frac94-1\right)=5.$$

Hyperbola: $\dfrac{x^2}{4}-\dfrac{y^2}{5}=1$. At $x=\alpha$: $y=\pm\dfrac{\sqrt{5(\alpha^2-4)}}{2}$, so $AB=\sqrt{5(\alpha^2-4)}$ and the height from $O$ to the line $x=\alpha$ is $|\alpha|$.

$$4\sqrt{15}=\frac12\sqrt{5(\alpha^2-4)}\cdot|\alpha| \;\Rightarrow\; 240=\frac54\alpha^2(\alpha^2-4).$$

Let $t=\alpha^2$: $t^2-4t-192=0 \Rightarrow (t-16)(t+12)=0 \Rightarrow t=16$.

Solutions · 1