If ${\mathrm{e}}_1$ is the eccentricity of the ellipse $\frac{{\mathrm{x}}^2}{16}+\frac{{\mathrm{y}}^2}{25}=1$ and ${\mathrm{e}}_2$ is the eccentricity of the hyperbola passing through the foci of the ellipse and ${\mathrm{e}}_1$${\mathrm{e}}_2$=1 then the equation of the hyperbola is

$$\begin{gathered} Given ellipse \frac{x^2}{16} +\frac{y^2}{25}=1 \\ \mathrm{for} \mathrm{ellipse} {\mathrm{a}}^2=16, {\mathrm{b}}^2=25(\because a>b) \\ \mathrm{Now} {\mathrm{e}}_1=\sqrt{\frac{25-16}{25}}=\frac{3}{5} \\ \mathrm{Given} {\mathrm{e}}_1{\mathrm{e}}_2=1 \Rightarrow {\mathrm{e}}_2=\frac{5}{3} \\ \mathrm{foci} \mathrm{of} \mathrm{ellipse} =(0, 3) \\ \mathrm{let} \mathrm{hyperbola} \mathrm{be} \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=-1 \\ \mathrm{It} \mathrm{passes} \mathrm{through} (0, 3) \mathrm{then} \\ \Rightarrow \frac{-9}{{\mathrm{b}}^2}=-1 \Rightarrow {\mathrm{b}}^2=9 \\ \mathrm{Now} {\mathrm{a}}^2={\mathrm{b}}^2({\mathrm{e}}^2-1) \\ =9\left(\frac{16}{9}\right)=16 \\ \therefore \mathrm{Hyperbola} \mathrm{is} \frac{{\mathrm{x}}^2}{16}-\frac{{\mathrm{y}}^2}{9}=-1 \end{gathered}$$