The eccentricity of the hyperbola 2x^2 - y^2 | vedclass

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(D) The given equation of the hyperbola is $2x^2 - y^2 = 6$. Dividing both sides by $6$,we get $\frac{x^2}{3} - \frac{y^2}{6} = 1$. Comparing this wit

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$\sqrt{3}$

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(D) The given equation of the hyperbola is $2x^2 - y^2 = 6$. Dividing both sides by $6$,we get $\frac{x^2}{3} - \frac{y^2}{6} = 1$. Comparing this with the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we have $a^2 = 3$ and $b^2 = 6$. The eccentricity $e$ of a hyperbola is given by the formula $e = \sqrt{1 + \frac{b^2}{a^2}}$. Substituting the values,$e = \sqrt{1 + \frac{6}{3}} = \sqrt{1 + 2} = \sqrt{3}$. Thus,the eccentricity is $\sqrt{3}$.

The eccentricity of the hyperbola $2x^2 - y^2 = 6$ is

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