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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
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Verified Answer
The correct answer is:
$1$
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, $e e^{\prime}=\frac{2}{3} \cdot \frac{3}{2}=1$
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, $e e^{\prime}=\frac{2}{3} \cdot \frac{3}{2}=1$
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