Locus of intersection point of tangents at extremities of chords of $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Question

What is the locus of the point of intersection of tangents at the extremities of the chords of the ellipse $$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\;\;$$ subtending to a right angle at its center?

Answer 1

Let the endpoints of the chord $(a\cos\theta_1, b\sin\theta_1)$ and $(a\cos\theta_2, b\sin\theta_2)$. Then, the tangent lines at the endpoints are

$$\frac xa\cos\theta_1 + \frac yb\sin\theta_1 = 1\tag 1$$ $$\frac xa\cos\theta_2 + \frac yb\sin\theta_2 = 1\tag 2$$

Given that they also subtend a right angle at the center, their dot-product is zero,

$$a^2\cos\theta_1\cos\theta_2+b^2\sin\theta_1\sin\theta_2=0\tag 3$$

Take (1) - (2), $$\sin\frac{\theta_2-\theta_1}2 \left( \frac xa \sin\frac{\theta_2+\theta_1}2 - \frac yb \cos\frac{\theta_2+\theta_1}2\right) = 0$$

which leads to

$$\cos\frac{\theta_2+\theta_1}2 = \frac{bx}{\sqrt{b^2x^2+a^2y^2}}, \>\>\>\>\> \sin\frac{\theta_2+\theta_1}2 = \frac{ay}{\sqrt{b^2x^2+a^2y^2}} \tag 4$$ Take (1) + (2),

$$\cos\frac{\theta_2-\theta_1}2 \left( \frac xa \cos\frac{\theta_2+\theta_1}2 + \frac yb \sin\frac{\theta_2+\theta_1}2\right) = 1$$

Use (4) to get,

$$\cos\frac{\theta_2-\theta_1}2 = \frac{ab}{\sqrt{b^2x^2+a^2y^2}}\tag 5$$

Next, rewrite (3) as

$$(a^2+b^2)\cos(\theta_2-\theta_1)+(a^2-b^2)\cos(\theta_2+\theta_1)=0$$

Use the double-angle identity together with the results (4) and (5) to obtain the locus,

$$\frac{x^2}{\frac{a^2}{b^2}(a^2+b^2)} + \frac{y^2}{\frac{b^2}{a^2}(a^2+b^2)} =1$$

which is an ellipse.