Question

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

Answer 1

Eccentricity of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is$e=\sqrt{\frac{a^2+b^2}{a^2}}$
Eccentricity of conjugate hyperbola $e^{\prime }=\sqrt{\frac{a^2+b^2}{b^2}}$

Writing $x^2-3y^2=1$ in standard form,
$\frac{x^2}{1}-\frac{y^2}{\frac{1}{3}}=1$
$a^2=1;b^2=\frac{1}{3}$
$\Rightarrow e^{\prime }=\sqrt{\frac{1+\frac{1}{3}}{\frac{1}{3}}}=2$