The sides AC and AB of a $∆\mathrm{ABC}$ touch the conjugate hyperbola of the hyperbola $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1.$ If the vertex A lies on the ellipse $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1,$ then the side BC must touch

$$\begin{gathered} \mathrm{let} \mathrm{A} (\mathrm{a} \mathrm{cos\theta }, \mathrm{b} \mathrm{sin\theta }) be the vertex of triangle \mathrm{which} \mathrm{lies} \mathrm{on} \mathrm{ellipse} \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1 \\ \mathrm{Given} \mathrm{AB} \mathrm{and} \mathrm{AC} \mathrm{are} \mathrm{tangents} \mathrm{to} \mathrm{the} \mathrm{hypernola} \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=-1 \\ \mathrm{Now} \mathrm{BC} \mathrm{is} \mathrm{the} \mathrm{chord} \mathrm{of} \mathrm{contact}. \\ \mathrm{Its} \mathrm{equation} \mathrm{is} \frac{x(a \cos \theta )}{a^2}-\frac{y(b \sin \theta )}{b^2}=-1 \\ \Rightarrow \frac{-\mathrm{x}}{\mathrm{a}}\mathrm{cos\theta }+\frac{\mathrm{y}}{\mathrm{b}}\mathrm{sin\theta }=1 \Rightarrow \frac{\mathrm{x}}{\mathrm{a}}\cos (\mathrm{\pi }-\mathrm{\theta })+\frac{\mathrm{y}}{\mathrm{b}}\sin (\mathrm{\pi }-\mathrm{\theta })=1 \\ \therefore \mathrm{BC} \mathrm{touches} \mathrm{the} \mathrm{given} \mathrm{ellipse} \mathrm{at} (\mathrm{a} \cos (\mathrm{\pi }-\mathrm{\theta }), \mathrm{bsin}(\mathrm{\pi }-\mathrm{\theta }) \end{gathered}$$