Question

lf $e$ 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

• A. $9$
• B. $5$
• C. $4$
• D. $1$

Answer 1

The correct option is D $1$

$\frac{x^2}{9}+\frac{y^2}{5}=1$ and $\frac{x^2}{9}-\frac{y^2}{\frac{45}{4}}$
$e=\sqrt{\frac{a^2-b^2}{a^2}}=\sqrt{\frac{9-5}{9}}=\frac{2}{3}$
$e^{{}^{\prime }}=\sqrt{\frac{a^2+b^2}{a^2}}=\sqrt{\frac{9+\frac{45}{4}}{9}}=\sqrt{\frac{81}{36}}=\sqrt{\frac{9}{4}}=\frac{3}{2}$
$\therefore e{e}^{{}^{\prime }}=\frac{2}{3}.\frac{3}{2}=1$