Find the locus. let $A$ & $B$ be two points on the ellipse $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$ such that they subtend $90^{0}$ 
on the centre of the ellipse. what is the locus of points of intersection of tangents at $A$ & $B$
 A: Let $\mathcal{E}$ be the ellipse $\left\{\; (u,v) \in \mathbb{R}^2 \;: \;\frac{u^2}{a^2} + \frac{v^2}{b^2} = 1\; \right\}$.
For any point $P = (x,y)$ outside $\mathcal{E}$, there are two tangents of $\mathcal{E}$ passing through $P$.
Let $A = (u_1,v_1)$ and $B = (u_2,v_2)$ be the intersections of these two tangents with $\mathcal{E}$.
For simplicity of argument, let us first assume $u_1, u_2 \ne 0$.
Let $Q = (u,v)$ be  either $A$ or $B$ and let $\displaystyle\;t = \frac{v}{u}\;$.
In order for $QP$ to be a tangent line of $\mathcal{E}$, $P$ and $Q$ satisfies
$$\frac{xu}{a^2} + \frac{yv}{b^2} = 1\iff u\left(\frac{x}{a^2}+\frac{yt}{b^2}\right) = 1 \implies u^2\left(\frac{x}{a^2}+\frac{yt}{b^2}\right)^2 = 1$$
Since $Q$ lies on $\mathcal{E}$,
$$\frac{u^2}{a^2} + \frac{v^2}{b^2} = 1 \iff u^2\left(\frac{1}{a^2} + \frac{t^2}{b^2}\right) = 1$$
This implies $t$ satisfies a quadratic equation.
$$\left(\frac{x}{a^2}+\frac{yt}{b^2}\right)^2 = \frac{1}{a^2} + \frac{t^2}{b^2}
\quad\iff\quad
\frac{t^2}{b^2}\left(1-\frac{y^2}{b^2}\right) - 2t\frac{xy}{a^2b^2} + \frac{1}{a^2}\left(1 - \frac{x^2}{a^2}\right) = 0
$$
It is clear $t_1 = \frac{v_1}{u_1}$ and $t_2 = \frac{v_2}{u_2}$ are the two roots of this polynomial. In order for  $A$, $B$ to subtend a right angle at center of $\mathcal{E}$, we need $$u_1u_2 + v_1v_2 = 0 \iff u_1u_2(1+t_1t_2) = 0 \iff t_1t_2 = -1$$
Apply Vieta's formula to 
above quadratic polynomial, the condition becomes
$$\frac{\frac{1}{a^2}\left(1 - \frac{x^2}{a^2}\right)}{\frac{1}{b^2}\left(1-\frac{y^2}{b^2}\right)} = t_1t_2 = -1
\quad\iff\quad \frac{x^2}{a^4} + \frac{y^2}{b^4} = \frac{1}{a^2}+\frac{1}{b^2}$$
This is the equation of another ellipse:
$$\mathcal{E}' = \left\{ \; (x,y) \in \mathbb{R}^2 \;:\; \frac{b^2x^2}{a^2(a^2+b^2)} + \frac{a^2y^2}{b^2(a^2+b^2)} = 1\; \right\}$$
When $u_1$ or $u_2$ is zero, it is easy to see the tangent lines through $A$, $B$ intersects at one of the 4 points $(\pm a,\pm b)$. Since $\mathcal{E}'$ also contains these 4 points, $\mathcal{E}'$ is the locus we seek.
