Number of points on the ellipse $\frac{{\mathrm{x}}^2}{50}+\frac{{\mathrm{y}}^2}{20}=1$from which pair of perpendicular tangents are drawn to the ellipse $\frac{{\mathrm{x}}^2}{16}+\frac{{\mathrm{y}}^2}{9}=1$ is

point on ellipse $\frac{{\mathrm{x}}^2}{50}+\frac{{\mathrm{y}}^2}{20}=1$ is $\left(\sqrt{50} \mathrm{cos\theta }, \sqrt{20} \mathrm{sin\theta }\right)$ point lies on director circle of ellipse $\frac{{\mathrm{x}}^2}{16}+\frac{{\mathrm{y}}^2}{9}=1$

Director circle is  ${\mathrm{x}}^2+{\mathrm{y}}^2=25$

$$\begin{gathered} 50 {\cos }^2\mathrm{\theta }+20 {\sin }^2\mathrm{\theta }=25 \\ 30{\cos }^2\mathrm{\theta }+20=25 \\ {\cos }^2\mathrm{\theta }=\frac{5}{30}=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right. \\ \mathrm{cos\theta }=\pm \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{6}$}\right. \\ director circle cuts ellipse at four distinct points \\ \therefore \mathrm{no}. \mathrm{of} \mathrm{points}=4 \end{gathered}$$