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If transverse and conjugate axes of a hyperbola are equal, then its eccentricity is
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The correct answer is:
$\sqrt{2}$
Hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. Here, transverse and conjugate axis of a hyperbola is equal.
i.e., $a=b \quad \therefore x^2-y^2=a^2$; which is a rectangular hyperbola. Hence, eccentricity
$e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{2}$
i.e., $a=b \quad \therefore x^2-y^2=a^2$; which is a rectangular hyperbola. Hence, eccentricity
$e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{2}$
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