Let 0

Question

$\frac{{{x^2}}}{{{{\cos }^2}	heta }} - \frac{{{y^2}}}{{{{\sin }^2}	heta }} = 1$

$\because $ $e > 2$        (given)

Vedclass · Accepted Answer

$\left( {3,\infty } \right)$

Vedclass · Answer

$\frac{{{x^2}}}{{{{\cos }^2}	heta }} - \frac{{{y^2}}}{{{{\sin }^2}	heta }} = 1$

$\because $ $e > 2$        (given)

${e^2} > 4 \Rightarrow 1 + \frac{{{{\sin }^2}	heta }}{{{{\cos }^2}	heta }} > 4$

$ \Rightarrow 1 + {	an ^2}	heta  > 4$

$ \Rightarrow {	an ^2}	heta  > 3$

$\because $ $	heta  \in \left( {\frac{\pi }{3},\frac{\pi }{2}} ight)$

Latus ractum $ = 2\frac{{{{\sin }^2}	heta }}{{\cos 	heta }} = 2	an 	heta \sin 	heta $

$\because $ for $	heta  \in \left( {\frac{\pi }{3},\frac{\pi }{2}} ight),2	an 	heta \sin 	heta $ is increasing

Hence latus rectum $ \in \left( {3,\infty } ight)$

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

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