The eccentricity of the hyperbola conjugate t | vedclass

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(A) For the given hyperbola $\frac{x^2}{4} - \frac{y^2}{12} = 1$,we have $a^2 = 4$ and $b^2 = 12$. The eccentricity $e_1$ of this hyperbola is given b

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$\frac{2}{\sqrt{3}}$

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(A) For the given hyperbola $\frac{x^2}{4} - \frac{y^2}{12} = 1$,we have $a^2 = 4$ and $b^2 = 12$. The eccentricity $e_1$ of this hyperbola is given by $e_1 = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{12}{4}} = \sqrt{1 + 3} = 2$. The conjugate hyperbola is $\frac{y^2}{12} - \frac{x^2}{4} = 1$. For a hyperbola and its conjugate,the eccentricities $e_1$ and $e_2$ satisfy the relation $\frac{1}{e_1^2} + \frac{1}{e_2^2} = 1$. Substituting $e_1 = 2$,we get $\frac{1}{4} + \frac{1}{e_2^2} = 1$. $\frac{1}{e_2^2} = 1 - \frac{1}{4} = \frac{3}{4}$. Therefore,$e_2^2 = \frac{4}{3}$,which implies $e_2 = \frac{2}{\sqrt{3}}$.

The eccentricity of the hyperbola conjugate to the hyperbola $\frac{x^2}{4} - \frac{y^2}{12} = 1$ is

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