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If $e_1$ is the eccentricity of the ellipse $\frac{x^2}{16}+\frac{y^2}{25}=1$ and $e_2$ is the eccentricity of a hyperbola passing through the foci of the given ellipse and $e_1 e_2=1$, then the equation of such a hyperbola among the following is
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Verified Answer
The correct answer is:
$\frac{y^2}{9}-\frac{x^2}{16}=1$
Equation of ellipse $\frac{x^2}{16}+\frac{y^2}{25}=1$
Foci $\quad=(0, \pm 3)$
$\Rightarrow \quad e_2=\sqrt{1-\frac{16}{25}}=\frac{3}{5}$ given $e_2 e_2=1$
(where $e_2$ is eccentricity of hyperbola)
Let equation of hyperbola $-\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
its passes through $(0, \pm 3)$
$$
\begin{array}{ll}
\therefore & b^2=9 \\
\Rightarrow & e_2=\sqrt{1+\frac{a^2}{b^2}} \\
\Rightarrow & e_2^2=1+\frac{a^2}{b^2} \\
\Rightarrow & \frac{1}{e_2^2}=1+\frac{a^2}{9} \Rightarrow \frac{25}{9}=1+\frac{a^2}{9} \\
\Rightarrow & a^2=16
\end{array}
$$
Hence, the equation of hyperbola is $\frac{y^2}{9}-\frac{x^2}{16}=1$
Foci $\quad=(0, \pm 3)$
$\Rightarrow \quad e_2=\sqrt{1-\frac{16}{25}}=\frac{3}{5}$ given $e_2 e_2=1$
(where $e_2$ is eccentricity of hyperbola)
Let equation of hyperbola $-\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
its passes through $(0, \pm 3)$
$$
\begin{array}{ll}
\therefore & b^2=9 \\
\Rightarrow & e_2=\sqrt{1+\frac{a^2}{b^2}} \\
\Rightarrow & e_2^2=1+\frac{a^2}{b^2} \\
\Rightarrow & \frac{1}{e_2^2}=1+\frac{a^2}{9} \Rightarrow \frac{25}{9}=1+\frac{a^2}{9} \\
\Rightarrow & a^2=16
\end{array}
$$
Hence, the equation of hyperbola is $\frac{y^2}{9}-\frac{x^2}{16}=1$
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