What is the eccentricity of the conjugate hyp | vedclass

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

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$2$

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(A) The given hyperbola is $x^2 - 3y^2 = 1$,which can be written as $\frac{x^2}{1} - \frac{y^2}{1/3} = 1$.Here,$a^2 = 1$ and $b^2 = 1/3$.The equation of the conjugate hyperbola is given by $-\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,which is $-x^2 + 3y^2 = 1$,or $\frac{y^2}{1/3} - \frac{x^2}{1} = 1$.For a hyperbola of the form $\frac{y^2}{B^2} - \frac{x^2}{A^2} = 1$,the eccentricity $e$ is given by $e = \sqrt{1 + \frac{A^2}{B^2}}$.Here,$A^2 = 1$ and $B^2 = 1/3$.Thus,$e = \sqrt{1 + \frac{1}{1/3}} = \sqrt{1 + 3} = \sqrt{4} = 2$.

What is the eccentricity of the conjugate hyperbola of the hyperbola $x^2 - 3y^2 = 1$?

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