If $e$ and $e^{\prime}$ are the eccentricities of the ellipse $5 x^2+9 y^2=45$ and the hyperbola $5 x^2-4…
If $e$ and $e^{\prime}$ are the eccentricities of the ellipse $5 x^2+9 y^2=45$ and the hyperbola $5 x^2-4 y^2=45$ respectively, then $e e^{\prime}$ is equal to
$1$
$4$
$5$
$9$
Solution
Equation of ellipse is
or
$
\begin{array}{r}
5 x^2+9 y^2=45 \\
\frac{x^2}{9}+\frac{y^2}{5}=1 \\
e=\sqrt{\frac{1-b^2}{a^2}}=\sqrt{\frac{1-5}{9}}=\frac{2}{3}
\end{array}
$
and equation of hyperbola is
$
5 x^2-4 y^2=45
$
or
$
\frac{x^2}{9}-\frac{y^2}{\left(\frac{45}{4}\right)}=1
$
$
e^{\prime}=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{45}{36}}=\sqrt{\frac{81}{36}}=\frac{9}{6}=\frac{3}{2}
$
Then, $e e^{\prime}=\frac{2}{3} \cdot \frac{3}{2}=1$