Mathematics - Hyperbola Question with Solution | TestHub

For hyperbola $\frac{x^2}{10}-\frac{y^2}{10{\cos }^2\theta }=1$

The eccentricity of the hyperbola $e_H=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+{\cos }^2\theta }$

For ellipse $\frac{x^2}{5{\cos }^2\theta }+\frac{y^2}{5}=1$

The eccentricity of an ellipse $e_E=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-{\cos }^2\theta }=\sin \theta$

Given, $e_H=\sqrt{5} e_E$

$\Rightarrow 1+{\cos }^2\theta =5{\sin }^2\theta \Rightarrow {\cos }^2\theta =\frac{2}{3}$

Now, the length of the latus rectum of an ellipse $=\frac{2b^2}{a}=\frac{10{\cos }^2\theta }{\sqrt{5}}=\frac{20}{3\sqrt{5}}=\frac{4\sqrt{5}}{3}$