If ${\mathrm{tan\theta }}_1\tan {\mathrm{\theta }}_2=\frac{-{\mathrm{a}}^2}{{\mathrm{b}}^2}$ then the chord joining two points ${\mathrm{\theta }}_1$ and ${\mathrm{\theta }}_2$ on the ellipse $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ will subtend a right angle at

$$\begin{gathered} Let end \mathrm{points} of chord be P(\mathrm{a} {\mathrm{cos\theta }}_1, \mathrm{b} {\mathrm{sin\theta }}_1) and \\ \mathrm{Q}({\mathrm{acos\theta }}_2, {\mathrm{bsin\theta }}_2) \\ \mathrm{centre} C(0, 0) \\ Now \mathrm{slope} of \overline{CP}=\frac{\mathrm{b} {\mathrm{sin\theta }}_1}{\mathrm{a} {\mathrm{cos\theta }}_1} \\ {\mathrm{m}}_1=\frac{\mathrm{b}}{\mathrm{a}} {\mathrm{Tan\theta }}_1 \\ \mathrm{slope} of \overline{CQ}=\frac{\mathrm{b} {\mathrm{sin\theta }}_2}{\mathrm{a} {\mathrm{cos\theta }}_2} \\ {\mathrm{m}}_2=\frac{\mathrm{b}}{\mathrm{a}} {\mathrm{Tan\theta }}_2 \\ Now m_1{m}_2=\frac{\mathrm{b}}{\mathrm{a}} {\mathrm{Tan\theta }}_1\frac{\mathrm{b}}{\mathrm{a}} {\mathrm{Tan\theta }}_2 \\ \Rightarrow m_1{m}_2=\frac{b^2}{a^2}{\mathrm{Tan\theta }}_1{\mathrm{Tan\theta }}_2 \\ =\frac{b^2}{a^2}\left(\frac{-a^2}{b^2}\right) \\ =-1 \\ \therefore chord makes right angle at centre \end{gathered}$$