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

 $$\begin{gathered} \mathrm{eccentricity} \mathrm{of} \mathrm{ellipse}={\mathrm{e}}_1=\sqrt{\frac{25-16}{25}}=\frac{3}{5} \\ \mathrm{eccentricity} \mathrm{of} \mathrm{hyperbola}\Rightarrow {\mathrm{e}}_2=\frac{5}{3}(\because e_1{e}_2=1) \\ Now foci of ellipse=(0,\pm be)(\because a<b) \\ =(0,\pm 3) \\ let hyperbola be \frac{x^2}{a^2}-\frac{y^2}{b^2}=-1 \\ it passes through (0,3) then b^2=9 \\ Now a^2=b^2(e^2-1)=16 \end{gathered}$$

 equation of hyperbola is $\frac{{\mathrm{x}}^2}{16}-\frac{{\mathrm{y}}^2}{9}=-1$