Let the lengths of the transverse and conjugate axes of a hyperbola in standard form be 2a…

Question

Let the lengths of the transverse and conjugate axes of a hyperbola in standard form be $2a$ and $2b$ respectively, and one focus and the corresponding directrix of this hyperbola be $(-5,0)$ and $5x+9=0$, respectively. If the product of the focal distances of a point $(\alpha,2\sqrt5)$ on the hyperbola is $p$, then $4p$ is equal to \_\_\_\_\_\_\_\_.

Answer: 189

Accepted Answer

Focus: $ae=5$; directrix: $\dfrac{a}{e}=\dfrac95$. Multiplying: $a^2=9\Rightarrow a=3,\ e=\dfrac53$.

$$b^2=a^2(e^2-1)=9\left(\frac{25}{9}-1\right)=16.$$

Hyperbola: $\dfrac{x^2}{9}-\dfrac{y^2}{16}=1$. At $y=2\sqrt5$: $\dfrac{\alpha^2}{9}=1+\dfrac{20}{16}=\dfrac94 \Rightarrow \alpha^2=\dfrac{81}{4}$.

$$p=e^2\alpha^2-a^2=\frac{25}{9}\cdot\frac{81}{4}-9=\frac{225}{4}-9=\frac{189}{4},\qquad 4p=189.$$

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