Question

Find the condition that the chord joining two points with eccentric angles $\theta_1$ & $\theta_2$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ subtend a right angle at its center.

Answer 1

The condition is $a^2\cos\theta_1\cos\theta_2 + b^2\sin\theta_1\sin\theta_2 = 0$. This can be simplified to $\tan\theta_1\tan\theta_2 = -\frac{a^2}{b^2}$ or $\cot\theta_1\cot\theta_2 = -\frac{b^2}{a^2}$.

Answer 2

The coordinates of the two points on the ellipse are $P_1(a\cos\theta_1, b\sin\theta_1)$ and $P_2(a\cos\theta_2, b\sin\theta_2)$. The center of the ellipse is the origin $O(0,0)$. The position vectors are $\vec{OP_1} = (a\cos\theta_1, b\sin\theta_1)$ and $\vec{OP_2} = (a\cos\theta_2, b\sin\theta_2)$. For the chord to subtend a right angle at the center, $\vec{OP_1} \cdot \vec{OP_2} = 0$. This gives $a^2\cos\theta_1\cos\theta_2 + b^2\sin\theta_1\sin\theta_2 = 0$. Dividing by $a^2\cos\theta_1\cos\theta_2$ yields $1 + \frac{b^2}{a^2}\tan\theta_1\tan\theta_2 = 0$, so $\tan\theta_1\tan\theta_2 = -\frac{a^2}{b^2}$.