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

Question

For some $	heta \in (0,\frac{\pi }{2})$, 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 ${e}_{2}^{1}={e}_{2}^{2}({\sec }^{2}	heta +1)$, then $(\frac{{l}_{1}{l}_{2}}{{e}_{1}{e}_{2}}){	an }^{2}	heta $ is equal to

Answer: 8

Accepted Answer

Given hyperbola

${x}^{2}-{y}^{2}{\sec }^{2}	heta =8$

Convert it into standard form.

$\frac{{x}^{2}}{8}-\frac{{y}^{2}}{8{\cos }^{2}	heta }=1$

Eccentricity: ${e}_{2}^{1}=1+\frac{8{\cos }^{2}	heta }{8}=1+{\cos }^{2}	heta $ latus rectum

${l}_{1}=\frac{2	imes 8{\cos }^{2}	heta }{\sqrt{8}}=4\sqrt{2}{\cos }^{2}	heta $

Given ellipse

${x}^{2}{\sec }^{2}	heta +{y}^{2}=6$

Convert into standard form.

$\frac{{x}^{2}}{6{\cos }^{2}	heta }+\frac{{y}^{2}}{6}=1$

Since $	heta \in (0,\pi /2)\Rightarrow {\cos }^{2}	heta ${e}\_{2}\^{2}=1-\frac{6{\cos }\^{2}	heta }{6}={\sin }\^{2}	heta $

Latus rectum

${l}\_{2}=\frac{2	imes 6{\cos }\^{2}	heta }{\sqrt{6}}=2\sqrt{6}{\cos }\^{2}	heta $

It is given that

${e}\_{2}\^{1}={e}\_{2}\^{2}({\sec }\^{2}	heta +1)$

Substituting values of ${e}\_{2}\^{1}$ and ${e}\_{2}\^{2}$.

$1+{\cos }\^{2}	heta ={\sin }\^{2}	heta (\frac{1}{{\cos }\^{2}	heta }+1)$

$\Rightarrow$ $1+{\cos }\^{2}	heta ={\sin }\^{2}	heta (\frac{1+{\cos }\^{2}	heta }{{\cos }\^{2}	heta })$

$\Rightarrow$ ${	an }\^{2}	heta =1\Rightarrow 	heta =\frac{\pi }{4}$ as $	heta \in (0,\pi /2)$

$\Rightarrow$ $\frac{{l}\_{1}\cdot {l}\_{2}}{{e}\_{1}\cdot {e}\_{2}}=\frac{(4\sqrt{2}{\cos }\^{2}\frac{\pi }{4})(2\sqrt{6}{\cos }\^{2}\frac{\pi }{4})}{\sqrt{(1+{\cos }\^{2}\frac{\pi }{4})}(\sin \frac{\pi }{4})}$

$	herefore $ $

Solutions · 1