For some $\theta \in \left( {0,{\pi \over 2}} \right)$, if the eccentricity of the
hyperbola, x²–y²sec²$\theta$ = 10 is $\sqrt 5$ times the
eccentricity of the ellipse, x²sec²$\theta$ + y² = 5, then the length of the latus rectum of the ellipse, is :

Given equation of hyperbola $\Rightarrow {x^2} - {y^2}{\sec ^2}\theta = 10$<br><br> $\Rightarrow {{{x^2}} \over {10}} - {{{y^2}} \over {10{{\cos }^2}\theta }} = 1$<br><br> Hence eccentricity of hyperbola<br><br> $\left( {{e_H}} \right) = \sqrt {1 + {{10{{\cos }^2}\theta } \over {10}}}$&nbsp;&nbsp;&nbsp;...(i)<br><br> &nbsp;&nbsp;&nbsp;&nbsp;$\left\{ { \because \,\, e = \sqrt {1 + {{{b^2}} \over {{a^2}}}} } \right\}$<br><br> Now equation of ellipse $\Rightarrow {x^2}{\sec ^2}\theta + {y^2} = 5$<br><br> $\Rightarrow {{{x^2}} \over {5{{\cos }^2}}} + {{{y^2}} \over 5} = 1\,$&nbsp;&nbsp;&nbsp;$\,\left\{ {e = 1 - {{{a^2}} \over {{b^2}}}} \right\}$<br><br> Hence eccenticity of ellipse<br><br> $\left( {{e_E}} \right) = \sqrt {1 - {{5{{\cos }^2}\theta } \over 5}}$<br><br> $\left( {{e_E}} \right) = \sqrt {1 - {{\cos }^2}\theta }$&nbsp;&nbsp;&nbsp;...(ii)<br><br> given ${e_H} = \sqrt 5 {e_e}$<br><br> Hence $$\sqrt {1 + {{\cos }^2}\theta } = \sqrt 5 \times \left( {\sqrt {1 - {{\cos }^2}\theta } } \right)$$<br><br> Squaring both sides<br><br> $1 + {\cos ^2}\theta = 5\left( {1 - {{\cos }^2}\theta } \right)$<br><br> $1 + {\cos ^2}\theta = 5 - 5{\cos ^2}\theta$<br><br> $6{\cos ^2}\theta = 4$<br><br> ${\cos ^2}\theta = {2 \over 3}$&nbsp;&nbsp;&nbsp;...(iii)<br><br> Now length of latus rectum of ellipse = $$ = {{2{a^2}} \over b} = {{10{{\cos }^2}\theta } \over {\sqrt 5 }} = {{20} \over {3\sqrt 5 }} = {{4\sqrt 5 } \over 3}$$