If e and ${\mathrm{e}}^1$ are the eccentricities of the ellipse $5{\mathrm{x}}^2 +9{\mathrm{y}}^2 =45$ and the hyperbola $5{\mathrm{x}}^2 –4{\mathrm{y}}^2 =45$ then e ${\mathrm{e}}^1$ =

$$\begin{gathered} \mathrm{Given} \mathrm{ellipse} \frac{{\mathrm{x}}^2}{9}+\frac{{\mathrm{y}}^2}{5}=1.(\mathrm{a}>\mathrm{b}) \\ \Rightarrow \mathrm{e}= \sqrt{\frac{9-5}{9}}=\sqrt{\frac{4}{9}}=\frac{2}{3} \\ \mathrm{Given} \mathrm{Hyperbola} \frac{{\mathrm{x}}^2}{9}-\frac{{\mathrm{y}}^2}{{\displaystyle \frac{45}{4}}}=1 \\ \mathrm{Now} {\mathrm{e}}^1=\sqrt{\frac{9+\frac{45}{4}}{9}}=\sqrt{\frac{36+45}{36}}=\sqrt{\frac{81}{36}}=\frac{9}{6}=\frac{3}{2} \\ \mathrm{Now} \mathrm{e} {\mathrm{e}}^1=\left(\frac{2}{3}\right) \left(\frac{3}{2}\right)=1 \end{gathered}$$