The eccentricity of the conic section $4 (x^{2} - y^{2}) = 1$ is

\sqrt{2}

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2

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4

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\frac{1}{4}

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Solution

The correct option is A $\sqrt{2}$
$4 (x^{2} - y^{2}) = 1$
$\frac{x^{2}}{1 / 4} - \frac{y^{2}}{1 / 4} = 1$
$\Rightarrow a^{2} = 1 / 4 = b^{2}$
Thus eccentricity of the given hyperbola is $e = \sqrt{1 + \frac{b^{2}}{a^{2}}} = \sqrt{2}$
Note: Given hyperbola is Rectangular and any rectangular hyperbola's eccentricity is $\sqrt{2}$

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