An ellipse passes through the foci of the hyperbola, 9 x ^ 2 - 4 y...

Equation of the hyperbola
$\dfrac{{x}^{2}}{4} - \dfrac{{y}^{2}}{9} = 1$



Focus of hyperbola $\left({a}_{1}{e}_{1}, 0\right) and ( - {a}_{1}{e}_{1}, 0 )$

${a}_{1} = 2, {e}_{1} = \sqrt{1 + \dfrac{9}{4}} = \dfrac{\sqrt{13}}{2}$

∴ Foci would be \left(+ \sqrt{13}, 0\right)$and \left(- \sqrt{13}, 0\right)$

Product of eccentricity would be

\dfrac{\sqrt{13}}{2} ⋅ {e}_{2} = \dfrac{1}{2}

∴ {e}_{2} = \dfrac{1}{\sqrt{13}} .

As the major & minor axis of the ellipse coincide with axis of hyperbola then the value of ${a}_{2}$ for ellipse would be $\sqrt{13}$ ,

${e}_{2} = \sqrt{1 - \dfrac{{b}_{2}^{2}}{{a}_{2}^{2}}}$

$\dfrac{1}{\sqrt{13}} = \sqrt{1 - \dfrac{{b}_{2}^{2}}{13}}$

${b}_{2}^{2} = 12$

∴ Equation of the ellipse would be

\dfrac{{x}^{2}}{13} + \dfrac{{y}^{2}}{12} = 1 .<br /><br />Option (1) \dfrac{39}{4 &sdot; ( 13 )} + \dfrac{3}{12} =1

Satisfies the $equation, hence it lies on the ellipse. <br /><br />Option (2) \dfrac{13}{4 ( 13 )} + \dfrac{3}{4.12} =1$

does not lie on the ellipse.

Option (3) $\dfrac{13}{2 ( 13 )} + \dfrac{6}{12} =1$ satisfy.

Option (4) $\dfrac{13}{13} +$0 = 1$satisfy.<br /><br />So, option \left(\dfrac{\sqrt{13}}{2}, \dfrac{\sqrt{3}}{2}\right) is$ the answer.