The eccentricity of the conjugate hyperbola o | vedclass

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(A) The given hyperbola is ${x^2} - 3{y^2} = 1$,which can be written in the standard form $\frac{x^2}{1} - \frac{y^2}{1/3} = 1$.Here,${a^2} = 1$ and $

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$2$

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(A) The given hyperbola is ${x^2} - 3{y^2} = 1$,which can be written in the standard form $\frac{x^2}{1} - \frac{y^2}{1/3} = 1$.Here,${a^2} = 1$ and ${b^2} = \frac{1}{3}$.The conjugate hyperbola of $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = -1$,or $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$.The eccentricity $e'$ of the conjugate hyperbola is given by the formula $e' = \sqrt{1 + \frac{a^2}{b^2}}$.Substituting the values,$e' = \sqrt{1 + \frac{1}{1/3}} = \sqrt{1 + 3} = \sqrt{4} = 2$.

The eccentricity of the conjugate hyperbola of the hyperbola ${x^2} - 3{y^2} = 1$ is

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