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

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answer is 2.

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Detailed Solution

$\begin{array}{l} x^{2} - 3 y^{2} = 1 \\ \frac{x^{2}}{1} - \frac{y^{2}}{(\frac{1}{3})} = 1 \\ e_{1}^{2} = \frac{a^{2} + b^{2}}{a^{2}} = \frac{1 + (\frac{1}{3})}{1} = \frac{4}{3} \\ If are the eccentricities of two conjugate hyperbolas then \\ e_{1}^{2} + e_{2}^{2} = e_{1}^{2} e_{2}^{2} \end{array}$

$\begin{array}{l} \Rightarrow \frac{4}{3} + e_{2}^{2} = \frac{4}{3} e_{2}^{2} \\ \Rightarrow \frac{4}{3} = \frac{4}{3} e_{2}^{2} - e_{2}^{2} \\ \Rightarrow \frac{4}{3} = \frac{4 e_{2}^{2} - 3 e_{2}^{2}}{3} \\ \Rightarrow 4 = e_{2}^{2} \\ \Rightarrow e_{2} = 2 \end{array}$