$$ \mathbf x = \begin{bmatrix} x & y \end{bmatrix}^\mathsf T $$ $$ Q = \begin{bmatrix} A & B/2 \\ B/2 & C \end{bmatrix} $$
Eigendecomposition of $Q$: $ ~Q = \begin{bmatrix} \mathbf e_1 & \mathbf e_2 \end{bmatrix}^\mathsf T \begin{bmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{bmatrix} \begin{bmatrix} \mathbf e_1 & \mathbf e_2 \end{bmatrix} $
$$ 0 \lt \lambda_1 \le \lambda_2 $$ $$ \mathbf e_1 = \begin{bmatrix} e_1 & e_2 \end{bmatrix}^\mathsf T $$ $$ \alpha = \arctan \frac{e_2}{e_1} $$ $$ \mathbf T = \begin{bmatrix} h & k \end{bmatrix}^\mathsf T $$ $$ \mathbf x^\mathsf T Q \mathbf x + \mathbf T^\mathsf T \mathbf x = 1 \label{a}\tag{1} $$ Equation (1) is a 2nd degree polynomial equation whose geometry I can interpret and for which I can get a definite parameterization. If I am not mistaken, the ellipse defined in (1) is parameterized as: $$ \mathcal E = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \begin{bmatrix} \frac {1}{\sqrt \lambda_1} \cos t \\ \frac {1}{\sqrt \lambda_2} \sin t \end{bmatrix} + \begin{bmatrix} h \\ k \end{bmatrix} ~;~ 0 \le t \lt 2 \pi $$ Now let's consider another second degree polynomial equation. $$ \mathbf x^\mathsf T Q \mathbf x + \mathbf T^\mathsf T \mathbf x = 0 \label{b}\tag{2} $$ I can neither get a parameterization for, nor can I interpret the geometry of the curve defined by equation (2). Here is the question: How does the curve defined in equation (2) get parameterized and/or what are the geometric properties of that curve?