If the radii of director circles of $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ and $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}_{ }^2}=1$ are 2r and r respectively and ${\mathrm{e}}_1 \mathrm{and} {\mathrm{e}}_2$ be the eccentricities of above ellipse and  hyperbola respectively then
 

Equation of director circle of ellipse  $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is ${\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2+{\mathrm{b}}^2$

$$\begin{gathered} \Rightarrow 2\mathrm{r}=\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2} \\ \Rightarrow 4{\mathrm{r}}^2={\mathrm{a}}^2+{\mathrm{b}}^2 \to \left(1\right) \end{gathered}$$

Equation of director circle of hyperbola  $\frac{x^2}{a^2}-\frac{y^{2_{ }}}{b^2}=1$is 

$$\begin{gathered} {\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2-{\mathrm{b}}^2 \\ \mathrm{radius} \left(\mathrm{r}\right)=\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2} \\ \Rightarrow {\mathrm{r}}^2={\mathrm{a}}^2-{\mathrm{b}}^2 \\ \left(1\right)+\left(2\right)\Rightarrow {\mathrm{a}}^2=\frac{5{\mathrm{r}}^2}{2} \\ \left(1\right)-\left(2\right)\Rightarrow {\mathrm{b}}^2=\frac{3{\mathrm{r}}^2}{2} \end{gathered}$$

Now eccenticity of ellipse $$\begin{gathered} Now eccentricity of ellipse\left({\mathrm{e}}_1\right)=\sqrt{\frac{{\mathrm{a}}^2-{\mathrm{b}}^2}{{\mathrm{a}}^2}}=\sqrt{\frac{r^2}{{\displaystyle \frac{5r^2}{2}}}}=\sqrt{\frac{2}{5}} \\ \Rightarrow {e^2}_1=\frac{2}{5} \\ eccentricity of hyperbola \left(e_2\right)=\sqrt{\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{{\mathrm{a}}^2}}=\sqrt{\frac{4r^2}{{\displaystyle \frac{5r^2}{2}}}}=\sqrt{\frac{8}{5}} \\ \Rightarrow {{\mathrm{e}}_2}^2=\frac{8}{5} \\ \mathrm{Now} 4{{\mathrm{e}}_2}^2-{{\mathrm{e}}_1}^2=4\left(\frac{8}{5}\right)-\frac{2}{5} \\ =\frac{30}{5} \\ =6 \end{gathered}$$