The eccentricity of the conic section x^2 - 4 | vedclass

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(D) The given equation is $x^2 - 4y^2 = 1$. This is in the standard form of a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,where $a^2 = 1$ and $b

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$\frac{\sqrt{5}}{2}$

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(D) The given equation is $x^2 - 4y^2 = 1$. This is in the standard form of a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,where $a^2 = 1$ and $b^2 = \frac{1}{4}$. Here,$a = 1$ and $b = \frac{1}{2}$. The eccentricity $e$ of a hyperbola is given by the formula $e = \sqrt{1 + \frac{b^2}{a^2}}$. Substituting the values,we get $e = \sqrt{1 + \frac{1/4}{1}} = \sqrt{1 + \frac{1}{4}} = \sqrt{\frac{5}{4}}$. Therefore,$e = \frac{\sqrt{5}}{2}$.

The eccentricity of the conic section $x^2 - 4y^2 = 1$ is:

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