The eccentricity of the hyperbola conjugate to $x^{2} - 3 y^{2} = 2 x + 8$ is

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Solution

The given equation reduces to $\frac{(x - 1)^{2}}{9} - \frac{y^{2}}{3} = 1$ . Its conjugate hyperbola is $\frac{y^{2}}{3} - \frac{(x - 1)^{2}}{9} = 1$ ; thus $a^{2} = 9, b^{2} = 3$
Using $a^{2} = b^{2} (e^{2} - 1)$ , we get
$9 = 3 (e^{2} - 1) \Rightarrow e = 2$ .

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