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Let $0 < \theta < \frac{\pi }{2}$. If the eccentricity of the hyperbola $\frac{{{x^2}}}{{{{\cos }^2}\,\theta }} - \frac{{{y^2}}}{{{{\sin }^2}\,\theta }} = 1$ is greater than $2$, then the length of its latus rectum lies in the interval
$\left( {3,\infty } \right)$
$\left( {\frac{3}{2},2} \right]$
$\left( {2,3} \right]$
$\left( {1,\frac{3}{2}} \right]$
Solution
$\frac{{{x^2}}}{{{{\cos }^2}\theta }} – \frac{{{y^2}}}{{{{\sin }^2}\theta }} = 1$
$\because $ $e > 2$ (given)
${e^2} > 4 \Rightarrow 1 + \frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }} > 4$
$ \Rightarrow 1 + {\tan ^2}\theta > 4$
$ \Rightarrow {\tan ^2}\theta > 3$
$\because $ $\theta \in \left( {\frac{\pi }{3},\frac{\pi }{2}} \right)$
Latus ractum $ = 2\frac{{{{\sin }^2}\theta }}{{\cos \theta }} = 2\tan \theta \sin \theta $
$\because $ for $\theta \in \left( {\frac{\pi }{3},\frac{\pi }{2}} \right),2\tan \theta \sin \theta $ is increasing
Hence latus rectum $ \in \left( {3,\infty } \right)$
