Given an ellipsoid equation of the form \begin{equation} \textbf{v}^T \textbf{A} \textbf{v} \; = \; 1 \tag{1} \end{equation} where $ \textbf{A} \in \mathbb{R}^{n \times n} $ is positive definite and non-diagonal and $ \textbf{v} \in \mathbb{R}^{n \times 1} $. Let's consider $ n \equiv 3 $.
It is known that horizon of an ellipsoid from an arbitrary exterior center $ \textbf{v}_0 $ of projection is a plane section of the ellipsoid and it is an ellipse itself (possibly degenerate). Equation of the plane is:
\begin{equation} \textbf{v}^T \textbf{A} \textbf{v}_0 \; = \; 1 \tag{2} \end{equation}
The ellipse has semi-major and semi-minor axes. They form orthogonal basis in the section plane. Additionally there is a normal to the plane given by a coefficients of first-order terms in (2). All three of them form an orthogonal basis in $ 3 $-D space.
How to obtain the basis directly from $ \textbf{A} $ and $ \textbf{v}_0 $?
