The eccentricity of a hyperbola is $\sqrt{3}$ then the eccentricity of its conjugate hyperbola is

Given eccentricity of hyperbola $\left({\mathrm{e}}_1\right)=\sqrt{3}$ 

let eccentricity of conjugate hyperbola  be $e_2$

we know that $\frac{1}{{{\mathrm{e}}_1}^2}+\frac{1}{{{\mathrm{e}}_2}^2}=1$

$$\begin{gathered} \Rightarrow \frac{1}{3}+\frac{1}{{{\mathrm{e}}_2}^2}=1 \\ \Rightarrow \frac{1}{{{\mathrm{e}}_2}^2}=1-\frac{1}{3}=\frac{2}{3} \\ \Rightarrow {{\mathrm{e}}_2}^2=\frac{3}{2} \\ \Rightarrow {\mathrm{e}}_2=\sqrt{\frac{3}{2}} \end{gathered}$$