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