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If the eccentricity of a hyperbola is $\sqrt{3}$; then the eccentricity of its conjugate hyperbola is :
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Verified Answer
The correct answer is:
$\sqrt{\frac{3}{2}}$
Let $e$ and $e^{\prime}$ are the eccentricities of a hyperbola and its conjugate hyperbola.
We know that
$\frac{1}{e^2}+\frac{1}{\left(e^{\prime}\right)^2}=1$
$\Rightarrow \quad \frac{1}{3}+\frac{1}{\left(e^{\prime}\right)^2}=1$
$\Rightarrow \quad \frac{1}{\left(e^{\prime}\right)^2}=\frac{2}{3} \Rightarrow e^{\prime}=\sqrt{\frac{3}{2}}$
We know that
$\frac{1}{e^2}+\frac{1}{\left(e^{\prime}\right)^2}=1$
$\Rightarrow \quad \frac{1}{3}+\frac{1}{\left(e^{\prime}\right)^2}=1$
$\Rightarrow \quad \frac{1}{\left(e^{\prime}\right)^2}=\frac{2}{3} \Rightarrow e^{\prime}=\sqrt{\frac{3}{2}}$
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