If $e$ and $e^{\prime }$ are the eccentricities of the ellipse $5x^2+9y^2=45$ and the hyperbola $5x^2-4y^2=45$ respectively, then $e{e}^{\prime }$ is equal to

The correct option is D $1$
Given equation of ellipse is $\frac{x^2}{9}+\frac{y^2}{5}=1$

$\therefore e=\sqrt{1-\frac{b^2}{a^2}}$

$\Rightarrow e=\sqrt{1-\frac{5}{9}}$

$\Rightarrow e=\frac{2}{3}$

and equation of hyperbola is

$\frac{x^2}{3^2}+\frac{y^2}{{\left(\sqrt{\frac{45}{4}}\right)}^2}=1$

$\therefore e^{\prime }=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{\frac{45}{4}}{9}}=\sqrt{\frac{81}{36}}=\frac{3}{2}$

Now

$e{e}^{\prime }=\frac{2}{3}\times \frac{3}{2}=1$