Q.Show that the locus of the point of intersection of tangents at two points on an ellipse, whose eccentric angles differ by a constant, is an ellipse.

Let the ellipse be $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ and let the two points have eccentric angles $\theta_1,\theta_2$ with $\theta_1-\theta_2=2\alpha$ (a fixed constant).

The tangents at $\theta_1$ and $\theta_2$ are $\dfrac{x}{a}\cos\theta_1+\dfrac{y}{b}\sin\theta_1=1$ and $\dfrac{x}{a}\cos\theta_2+\dfrac{y}{b}\sin\theta_2=1$.

Solving these simultaneously (a standard identity) gives their intersection point as

Let $\varphi=\dfrac{\theta_1+\theta_2}{2}$ (which varies) while $\dfrac{\theta_1-\theta_2}{2}=\alpha$ is fixed, so $\cos\alpha$ is a constant. Then

$\dfrac{x\cos\alpha}{a}=\cos\varphi,\qquad \dfrac{y\cos\alpha}{b}=\sin\varphi.$ …

Recall (or derive) the coordinates of the intersection of two eccentric-angle tangents, substitute the fixed half-difference, and el …

Fixing the SUM of the eccentric angles instead of their difference — that case instead gives a straight …