The eccentricity of the conjugate hyperbola of the hyperbola x² - 3y² =1 is

 The given hyperbola is 

 $\frac{{\mathrm{x}}^2}{1}-\frac{{\mathrm{y}}^2}{1/3}=1$

Its eccentricity e is given by

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

Hence, eccentricity e' of the conjugate hyperbola is given by

$$\begin{gathered} \frac{1}{{e_1}^2}=1-\frac{3}{4} \\ \Rightarrow {e^2}_1=4 \\ \Rightarrow e_1=2 \end{gathered}$$