Given $\theta \in\left(0, \frac{\pi}{2}\right)$

equation of hyperbola $\Rightarrow x^{2}-y^{2} \sec ^{2} \theta=10$

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

Hence eccentricity of hyperbola

$\left(\mathrm{e}_{\mathrm{H}}\right)=\sqrt{1+\frac{10 \cos ^{2} \theta}{10}} \ldots$.

$\left\{e=\sqrt{1+\frac{b^{2}}{a^{2}}}\right\}$

Now equation of ellipse $\Rightarrow x^{2} \sec ^{2} \theta+y^{2}=5$

$\Rightarrow \frac{\mathrm{x}^{2}}{5 \cos ^{2} \theta}+\frac{\mathrm{y}^{2}}{5}=1 \quad\left\{\mathrm{e}=\sqrt{1-\frac{\mathrm{a}^{2}}{\mathrm{b}^{2}}}\right\}$

Hence eccenticity of ellipse

$\left(e_{E}\right)=\sqrt{1-\frac{5 \cos ^{2} \theta}{5}}$

$\left(e_{E}\right)=\sqrt{1-\cos ^{2} \theta}=\sin \theta=\sin \theta \quad \cdots(2)$

$\left\{\because \theta \in\left(0, \frac{\pi}{2}\right)\right\}$

given $\Rightarrow \mathrm{e}_{\mathrm{H}}=\sqrt{5} \mathrm{e}_{\mathrm{e}}$

Hence $1+\cos ^{2} \theta=5 \sin ^{2} \theta$

$1+\cos ^{2} \theta=5\left(1-\cos ^{2} \theta\right)$

$1+\cos ^{2} \theta=5-5 \cos ^{2} \theta$

$6 \cos ^{2} \theta=4$

$\cos ^{2} \theta=\frac{2}{3}$

Now length of latus rectum of ellipse

$=\frac{2 \mathrm{a}^{2}}{\mathrm{b}}=\frac{10 \cos ^{2} \theta}{\sqrt{5}}=\frac{20}{3 \sqrt{5}}=\frac{4 \sqrt{5}}{3}$