If tan θ 1 × tan θ 2 = - a 2 b 2 , then the chord joining 2 points θ 1 and θ 2 on…

Let the two points be $A\left(a\cos {\theta }_1,b\sin {\theta }_1\right) \& B\left(a\cos {\theta }_2,b\sin {\theta }_2\right)$

$O\left(0,0\right)$ is origin

Slope of $OA$$=\frac{b\sin {\theta }_1}{a\cos {\theta }_1}=\frac{b\tan {\theta }_1}{a}$

Slope of $OB$$=\frac{b\sin {\theta }_2}{a\cos {\theta }_2}=\frac{b\tan {\theta }_2}{a}$

Slope of $OA$$\times$Slope of $OB$$=\frac{b^2}{a^2}\tan {\theta }_1\tan {\theta }_2=-1$

Therefore, chord subtends the right angle at center.