Orthogonal on an ellipse

Question

Consider two points $a \in \mathbb{R}^2$ and $b \in \mathbb{R}^2$ on the boundary of an ellipse such that $a^\top \Sigma^{-1} a = b^\top \Sigma^{-1} b$. I am trying to understand which points on the boundary of this ellipse satisfy $a^\top \Sigma^{-1} b = 0$. One possibility for this to happen is when $a \perp v_1$ and $b \perp v_2$, where $v_1$ and $v_2$ are eigenvectors of $\Sigma$, then $a^\top \Sigma^{-1} b = 0$. It is also clear that if $a$ and $b$ are valid, then $(-a, b), (a, -b) \text{ and } (-a, -b)$ are also valid solutions. Is it possible to find other points on the boundary of the ellipse such $a^\top \Sigma^{-1} b = 0$ or show that no other points besides these four exist, where this condition is satisfied ?

Answer 1

I assume that $\Sigma$ is symmetric postive definite. Let $R$ be the rotation $$R = \begin{pmatrix}0 & -1 \\[1 ex] 1 & 0 \end{pmatrix}.$$ Take any vector $b \neq 0$ and $$a = \Sigma ^{1/2} R \Sigma ^{-1/2} b.$$ Then $$a^{\mathrm t} \Sigma^{-1} a = b^{\mathrm t} \Sigma^{-1} b$$ and $$a^{\textrm t}\Sigma^{-1}b=0.$$

Answer 2

Start with the simplest case: ${\bf \Sigma} = {\bf I}$ (the identity matrix):

Here's what you get with ${\bf \Sigma} \neq {\bf I}$:

Clear?