The eccentricity of the hyperbola -frac{x^2}{ | vedclass

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(D) The given equation of the hyperbola is $-\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,which can be rewritten as $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$.

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$e = \sqrt{\frac{a^2 + b^2}{b^2}}$

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(D) The given equation of the hyperbola is $-\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,which can be rewritten as $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$.This is a vertical hyperbola with the transverse axis along the $y$-axis.For a hyperbola of the form $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$,the relationship between the semi-axes $a, b$ and the eccentricity $e$ is given by $a^2 = b^2(e^2 - 1)$.Rearranging for $e^2$: $e^2 - 1 = \frac{a^2}{b^2} \implies e^2 = 1 + \frac{a^2}{b^2} = \frac{b^2 + a^2}{b^2}$.Taking the square root,we get $e = \sqrt{\frac{a^2 + b^2}{b^2}}$.

The eccentricity of the hyperbola $-\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is:

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