The eccentricity of the conic {x^2} - 4{y^2} | vedclass

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(D) The given equation of the conic is ${x^2} - 4{y^2} = 1$,which can be written in the standard form of a hyperbola as $\frac{x^2}{1^2} - \frac{y^2}{

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

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

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

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