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Question

Let $0<	heta<\dfrac\pi2$. If the eccentricity of the hyperbola $\dfrac{x^2}{\cos^2	heta}-\dfrac{y^2}{\sin^2	heta}=1$ is greater than $2$, then the length of its latus rectum lies in the interval:

Accepted Answer

Given hyperbola, $${{{x^2}} \over {{{\cos }^2}\theta }} - {{{y^2}} \over {{{\sin }^2}\theta }} = 1$$ here a = cos$$\theta $$ and b = sin$$\theta $$ We know, eccentricity of the hyperbola is, $$\sqrt {1 + {{{b^2}} \over {{a^2}}}} $$ $$ \therefore $$ Here eccentricity (e) = $$\sqrt {1 + {{{{\sin }^2}\theta } \over {{{\cos }^2}\theta }}} $$ Given that, $$\sqrt {1 + {{{{\sin }^2}\theta } \over {{{\cos }^2}\theta }}} > 2$$ $$ \Rightarrow $$ $$\sqrt {1 + {{\tan }^2}\theta } > 2$$ $$ \Rightarrow $$ 1 + tan2$$\theta $$ > 4 $$ \Rightarrow $$ tan2$$\theta $$ > 3 $$ \Rightarrow $$ tan$$\theta $$ > $$ \pm \sqrt 3 $$ As given $$\theta $$ $$ \in $$ $$\left( {0,{\pi \over 2}} \right)$$ possible value of tan$$\theta $$ > $$\sqrt 3 $$ So, $$\theta $$ can be in the range $${\pi \over 3} < \theta < {\pi \over 2}$$ We know latus ractum (LR) = $${{2{b^2}} \over a}$$ $$ \therefore $$ LR = $${{2{{\sin }^2}\theta } \over {\cos \theta }}$$ = 2 tan$$\theta $$ sin$$\theta $$ We know in the range $${\pi \over 3} < \theta < {\pi \over 2}$$ tan$$\theta $$ and sin$$\theta $$ both are increasing function. So, at $${\pi \over 3}$$ value of LR will be minimum and at $${\pi \over 2}$$ value of LR will be maximum. $$ \therefore $$ Minimum value of LR = 2tan$${\pi \over 3}$$ sin$${\pi \over 3}$$ = 2 $$ \times $$ $$\sqrt 3 \times {{\sqrt 3 } \over 2}$$ = 3 Maximum value of LR = 2tan$${\pi \over 2}$$ sin$${\pi \over 2}$$ = 2$$\left( \infty \right) \times 1$$ = $$\infty $$ $$ \therefore $$ Interval of LR = (3, $$\infty $$)

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