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Let $e_1$ be the eccentricity of a hyperbola for which distance between its focii is 2 times the distance between its directrices and $e_2$ be the eccentricity of another hyperbola for which the length of its transverse axis is twice the length of its the conjugate axis. Then $e_1 e_2=$
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Verified Answer
The correct answer is:
$\frac{\sqrt{10}}{2}$
Given $e_1$ is the eccentricity of hyperbola and eccentricity $\mathrm{e}_2$ of another hyperbola.
According to question,
$$
\begin{aligned}
& 2 \mathrm{a}_1 \mathrm{e}_1=2 \times \frac{2 \mathrm{a}_1}{\mathrm{e}_1} \\
& \mathrm{e}_1^2=2 \\
& \mathrm{e}_1= \pm \sqrt{2}
\end{aligned}
$$
Here, $e_1>1$ then $e_1=\sqrt{2}$
Now, $2 \mathrm{a}_2=4 \mathrm{~b}_2$
$$
\begin{aligned}
& \mathrm{a}_2=2 \mathrm{~b}_2 \\
& \mathrm{~b}_2=\frac{\mathrm{a}_2}{2}
\end{aligned}
$$
So, $c_2^2=a_2^2+b_2^2$
$$
\begin{aligned}
& \mathrm{c}_2^2=\mathrm{a}_2^2+\frac{\mathrm{a}_2^2}{4}=\frac{5 \mathrm{a}_2^2}{4} \\
& \mathrm{c}_2=\frac{\sqrt{5}}{2} \mathrm{a}_2 \\
& \frac{\mathrm{c}_2}{\mathrm{a}_2}=\frac{\sqrt{5}}{2} \\
& \mathrm{e}_2=\frac{\sqrt{5}}{2}
\end{aligned}
$$
Then, $\mathrm{e}_1 \mathrm{e}_2=\sqrt{2} \times \frac{\sqrt{5}}{2}=\frac{\sqrt{10}}{2}$
So, option (b) is correct.
According to question,
$$
\begin{aligned}
& 2 \mathrm{a}_1 \mathrm{e}_1=2 \times \frac{2 \mathrm{a}_1}{\mathrm{e}_1} \\
& \mathrm{e}_1^2=2 \\
& \mathrm{e}_1= \pm \sqrt{2}
\end{aligned}
$$
Here, $e_1>1$ then $e_1=\sqrt{2}$
Now, $2 \mathrm{a}_2=4 \mathrm{~b}_2$
$$
\begin{aligned}
& \mathrm{a}_2=2 \mathrm{~b}_2 \\
& \mathrm{~b}_2=\frac{\mathrm{a}_2}{2}
\end{aligned}
$$
So, $c_2^2=a_2^2+b_2^2$
$$
\begin{aligned}
& \mathrm{c}_2^2=\mathrm{a}_2^2+\frac{\mathrm{a}_2^2}{4}=\frac{5 \mathrm{a}_2^2}{4} \\
& \mathrm{c}_2=\frac{\sqrt{5}}{2} \mathrm{a}_2 \\
& \frac{\mathrm{c}_2}{\mathrm{a}_2}=\frac{\sqrt{5}}{2} \\
& \mathrm{e}_2=\frac{\sqrt{5}}{2}
\end{aligned}
$$
Then, $\mathrm{e}_1 \mathrm{e}_2=\sqrt{2} \times \frac{\sqrt{5}}{2}=\frac{\sqrt{10}}{2}$
So, option (b) is correct.
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