Let the hyperbola H: x ^ 2 a ^ 2 - y ^ 2 = 1 and the ellipse E: 3 x...

Given

$E ≡ 3{x}^{2} + 4{y}^{2} = 12 ⇒ \dfrac{{x}^{2}}{4} + \dfrac{{y}^{2}}{3} = 1$

Now, ${e}_{E} = \sqrt{1 - \dfrac{{b}^{2}}{{a}^{2}}} = \sqrt{1 - \dfrac{3}{4}} = \dfrac{1}{2}$

Length of L.R. $= \dfrac{2{b}^{2}}{a} = \dfrac{2 \times 3}{2} = 3$

Now $H ≡ \dfrac{{x}^{2}}{{a}^{2}} - \dfrac{{y}^{2}}{1} = 1$

Length of L.R.$= \dfrac{2{b}^{2}}{a} = \dfrac{2 \times 1}{a}$

Given length of L.R. of $E$ and $H$ are equal

So $\dfrac{2}{a} = 3 ⇒ a = \dfrac{2}{3}$

Now ${e}_{H} = \sqrt{\dfrac{{b}^{2}}{{a}^{2}} + 1} = \sqrt{\dfrac{1}{{\left(\dfrac{2}{3}\right)}^{2}}} + 1 ⇒ {e}_{H} = \sqrt{\dfrac{9}{4} + 1} = \sqrt{\dfrac{13}{4}}$

So $12\left({e}_{H}^{2} + {e}_{E}^{2}\right) = 12\left(\dfrac{13}{4} + \dfrac{1}{4}\right) = 12\left(\dfrac{14}{4}\right) = 14 \times 3 = 42$