For the hyperbola H: x ^ 2 - y ^ 2 = 1 and the ellipse E: x ^ 2 a ^...

Given $H: {x}^{2} - {y}^{2} = 1, E: \dfrac{{x}^{2}}{{a}^{2}} + \dfrac{{y}^{2}}{{b}^{2}} = 1$

${e}_{H} = \sqrt{2} \text{ and } {e}_{E} = \dfrac{1}{{e}_{H}} = \dfrac{1}{\sqrt{2}}$

For hyperbola ${e}^{2} = 1 - \dfrac{{b}^{2}}{{a}^{2}} = \dfrac{1}{2} ⇒ \dfrac{{b}^{2}}{{a}^{2}} = \dfrac{1}{2}$

Also given that the common tangent of $H \text{ and } E$ is $y = \sqrt{\dfrac{5}{2}}x + k$ \left(i . e . m = \sqrt{\dfrac{5}{2}}\right)$</p><p>We know that the condition for common tangent of ellipse$\dfrac{{x}^{2}}{{a}^{2}} + \dfrac{{y}^{2}}{{b}^{2}} = 1$and hyperbola$\dfrac{{x}^{2}}{{A}^{2}} - \dfrac{{y}^{2}}{{B}^{2}} = 1$is${a}^{2}{m}^{2} + {b}^{2} = {A}^{2}{m}^{2} - {B}^{2}$</p><p>Now, for common tangency:$\dfrac{5}{2}{a}^{2} + {b}^{2} = \dfrac{5}{2} - 1$</p><p>$\dfrac{5}{2} + \dfrac{{b}^{2}}{{a}^{2}} = \dfrac{3}{2{a}^{2}} ⇒ {a}^{2} = \dfrac{1}{2}$</p><p>$∴ {a}^{2} + {b}^{2} = \dfrac{1}{2} + \dfrac{1}{4} = \dfrac{3}{4}$</p><p>$⇒ 4\left({a}^{2} + {b}^{2}\right) = 3