A hyperbola passes through the foci of the ellipse $\frac{{\mathrm{x}}^2}{25}+\frac{{\mathrm{y}}^2}{16}=1$ and its transverse and conjugate axes coincide with major and minor axes of the ellipse respectively.If the product of their eccentricites is one,then the equation of the hyperbola is

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