Let the foci and length of the latus rectum of an ellipse x ^ 2 a ^...

Given,

Ellipse $\dfrac{{x}^{2}}{{a}^{2}} + \dfrac{{y}^{2}}{{b}^{2}} = 1$

And its focii $≡ \left(\pm 5, 0\right) and its latusrectum$\dfrac{2{b}^{2}}{a} = \sqrt{50}$</p><p>So,$ae = 5$and${b}^{2} = \dfrac{5\sqrt{2}a}{2}$</p><p>Now, using the formula eccentricity we get,</p><p>${b}^{2} = {a}^{2}\left(1 - {e}^{2}\right) = \dfrac{5\sqrt{2}a}{2}$</p><p>$⇒ a\left(1 - {e}^{2}\right) = \dfrac{5\sqrt{2}}{2}$\left\{\text{as}a \ne 0\right\}$

$⇒ \dfrac{5}{e}\left(1 - {e}^{2}\right) = \dfrac{5\sqrt{2}}{2}$

$⇒ \sqrt{2} - \sqrt{2}{e}^{2} = e$

$⇒ \sqrt{2}{e}^{2} + e - \sqrt{2} = 0$

$⇒ \sqrt{2}{e}^{2} + 2e - e - \sqrt{2} = 0$

$⇒ \sqrt{2}e\left(e + \sqrt{2}\right) - 1\left(1 + \sqrt{2}\right) = 0$

$⇒ \left(e + \sqrt{2}\right)\left(\sqrt{2}e - 1\right) = 0$

$∴ e \ne - \sqrt{2}; e = \dfrac{1}{\sqrt{2}}$

$⇒ a = 5\sqrt{2}$ and $b = 5$

Now, finding eccentricity of hyperbola,

$\dfrac{{x}^{2}}{{b}^{2}} - \dfrac{{y}^{2}}{{a}^{2}{b}^{2}} = 1$ 

We know that formula of eccentricity of hyperbola is given by, ${B}^{2} = {A}^{2}\left({{e}_{H}}^{2} - 1\right)$ where ${A}^{2} = {b}^{2} \text{ and } {B}^{2} = {a}^{2}{b}^{2}$

$⇒ {a}^{2}{b}^{2} = {b}^{2}\left({e}_{H}^{2} - 1\right)$

$⇒ {a}^{2} = \left({e}_{H}^{2} - 1\right)$

$⇒ 50 = \left({e}_{H}^{2} - 1\right)$

$⇒ {e}_{H}^{2} = 51$