If ${\mathrm{e}}_1,{\mathrm{e}}_2,{\mathrm{e}}_3$ are eccentricities of a parabola, ellipse and hyperbola respectively and ${\mathrm{e}}_1^2+{\mathrm{e}}_2^2+{\mathrm{e}}_3^2=\frac{46}{9}, {\mathrm{e}}_1^2-{\mathrm{e}}_2^2+{\mathrm{e}}_3^2=\frac{44}{9}$ then the eccentricity of conjugate hyperbola is

$$\begin{gathered} \mathrm{Given} {\mathrm{e}}_1=\mathrm{eccenturicity} \mathrm{of} \mathrm{parabola} =1 \\ Then \mathrm{given} \mathrm{equations} \mathrm{becomes} \\ 1+{{\mathrm{e}}_2}^2+{{\mathrm{e}}_3}^2=\frac{46}{9} \Rightarrow {{\mathrm{e}}_2}^2+{{\mathrm{e}}_3}^2=\frac{37}{9} — ① \\ \mathrm{and} 1-{{\mathrm{e}}_2}^2+{{\mathrm{e}}_3}^2=\frac{44}{9} \\ \Rightarrow -{{\mathrm{e}}_2}^2+{{\mathrm{e}}_3}^2=\frac{35}{9} — ② \\ \mathrm{Now} ① + ② \Rightarrow 2{{\mathrm{e}}_3}^2=\frac{37}{9}+\frac{35}{9} \\ \Rightarrow 2{{\mathrm{e}}_3}^2=8 \\ \Rightarrow {{\mathrm{e}}_3}^2 =4 \\ \Rightarrow {\mathrm{e}}_3=2 \\ let e_3{\text{'}}_{ } be eccentricity of conjugate hyperbola then \\ \frac{1}{{{\mathrm{e}}_3}^2} +\frac{1}{{\left({\mathrm{e}}_3^{\text{'}}\right)}^2} =1 \\ \Rightarrow \frac{1}{{\left({\mathrm{e}}_3^{\text{'}}\right)}^2} =1-\frac{1}{4}=\frac{3}{4} \\ \Rightarrow {\left({\mathrm{e}}_3^{\text{'}}\right)}^2=\frac{4}{3} \Rightarrow {\mathrm{e}}_3^{\text{'}}=\frac{2}{\sqrt{3}} \end{gathered}$$