For some (0,2), let the eccentricity and the length of the latus rectum of the hyperbola x…

Question

For some $	heta\in\left(0,\dfrac{\pi}{2}ight)$, let the eccentricity and the length of the latus rectum of the hyperbola $x^2-y^2\sec^2	heta=8$ be $e_1$ and $l_1$ respectively, and let the eccentricity and the length of the latus rectum of the ellipse $x^2\sec^2	heta+y^2=6$ be $e_2$ and $l_2$ respectively. If $e_1^2=e_2^2(\sec^2	heta+1)$, then $\dfrac{l_1 l_2}{e_1 e_2}$ is equal to \_\_\_\_\_\_\_\_.

Answer: 8

Accepted Answer

Hyperbola: $\dfrac{x^2}{8}-\dfrac{y^2}{8\cos^2\theta}=1$, so $e_1^2=1+\cos^2\theta$ and $l_1=\dfrac{2(8\cos^2\theta)}{\sqrt8}=4\sqrt2\cos^2\theta$.

Ellipse: $\dfrac{x^2}{6\cos^2\theta}+\dfrac{y^2}{6}=1$; since $\cos^2\theta<1$, major axis is along $y$, so $e_2^2=1-\cos^2\theta=\sin^2\theta$ and $l_2=\dfrac{2(6\cos^2\theta)}{\sqrt6}=2\sqrt6\cos^2\theta$.

The condition $1+\cos^2\theta=\sin^2\theta\left(\dfrac{1}{\cos^2\theta}+1\right)$ simplifies (multiplying by $\cos^2\theta$) to $\cos^2\theta(1+\cos^2\theta)=\sin^2\theta(1+\cos^2\theta)$, so $\cos^2\theta=\sin^2\theta \Rightarrow \theta=\dfrac{\pi}{4}$.

At $\theta=\pi/4$: $\cos^2\theta=\tfrac12$, so $l_1=2\sqrt2,\ l_2=\sqrt6,\ e_1=\sqrt{3/2},\ e_2=1/\sqrt2$.

$$\frac{l_1l_2}{e_1e_2}=\frac{2\sqrt2\cdot\sqrt6}{\sqrt{3/2}\cdot1/\sqrt2}=\frac{4\sqrt3}{\sqrt3/2}=8.$$

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