Statement1 : Ellipse $\frac{{\mathrm{x}}^2}{25}+\frac{{\mathrm{y}}^2}{16}=1$ and $12{\mathrm{x}}^2-4{\mathrm{y}}^2=27$ intersect each other at right angle

Statement 2 : Given ellipse and hyperbola have same foci

Then correct statement is

$$\begin{gathered} \mathrm{Given} \mathrm{ellipse} \frac{{\mathrm{x}}^2}{25}+\frac{{\mathrm{y}}^2}{16}=1 \\ \Rightarrow {\mathrm{a}}^2=25, {\mathrm{b}}^2=16 \\ \mathrm{e}=\sqrt{\frac{25-16}{25}}=\frac{3}{5} \\ \therefore \mathrm{foci}=(\pm \mathrm{ae}, 0)=(\pm 3, 0) \\ \mathrm{Given} \mathrm{Hyperbola} \frac{{\mathrm{x}}^2}{{\displaystyle \frac{9}{4}}}-\frac{{\mathrm{y}}^2}{{\displaystyle \frac{27}{4}}}=1 \\ \mathrm{e}=\sqrt{\frac{{\displaystyle \frac{9}{4}}+{\displaystyle \frac{27}{4}}}{{\displaystyle \frac{9}{4}}}}=2 \\ \therefore \mathrm{foci}=(\pm \mathrm{ae}, 0)=(\pm 3, 0) \\ \therefore \mathrm{foci} \mathrm{of} \mathrm{ellipse}=\mathrm{foci} \mathrm{of} \mathrm{hyperbola}. \\ \mathrm{Since} \mathrm{both} \mathrm{are} \mathrm{confocal} \mathrm{they} \mathrm{inter}se\mathrm{ct} \mathrm{orthog}o\mathrm{nally}. \\ \therefore \mathrm{Statement}-\mathrm{I}, \mathrm{Statement}-\mathrm{II} \mathrm{are} \mathrm{ture} and \mathrm{II} \mathrm{is} \mathrm{correct} \mathrm{explanation} \mathrm{of} \mathrm{I}. \end{gathered}$$