Suppose you are given the two semi-axes $ a, b$, with $ a \gt b $ of an ellipse, as well as the angle $\theta$ the major axis makes with the positive $x$ axis. In addition, you are given two points $P_1(x_1, y_1), P_2(x_2,y_2)$ that lie on the ellipse. You want to find the center of this ellipse.
My attempt: The equation of the ellipse is
$$ (r - C)^T Q (r - C) = 1 $$
where $ r = [x, y]^T $ and $C = [C_x, C_y]^T $. And
$$ Q = R D R^T $$
with
$$ R = \begin{bmatrix} \cos \theta && - \sin \theta \\ \sin \theta && \cos \theta \end{bmatrix} $$
and
$$ D = \begin{bmatrix} \dfrac{1}{a^2} && 0 \\ 0 && \dfrac{1}{b^2} \end{bmatrix} $$
Since $a,b, \theta$ are known, then $Q$ is known.
Expanding the equation of the ellipse,
$ r^T Q r - 2 r^T Q C + C^T Q C - 1 = 0 $
Using $ r = P_1 $ and $ r = P_2 $ we obtain two equations
$ P_1^T Q P_1 - 2 P_1^T Q C + C^T Q C -1 = 0 $
and
$ P_2^T Q P_2 - 2 P_2^T Q C + C^T Q C - 1 = 0 $
To solve for $C$, subtract the two equations,
$ (P_1^T Q P_1 - P_2^T Q P_2) - 2 (P_1 - P_2)^T Q C = 0 $
which is linear in $C$, therefore, its solution is of the form
$ C = x_0 + t x_1 $, where $x_0$ and $x_1 $ are known and $t \in \mathbb{R} $
To determine $t$, plug this form into either of the two original equations involving $P_1$ and $P_2$, to get a quadratic equation in $t$, that can have $0, 1,$ or $2$ solutions. For example, if the two points separation is greater than $2 a$ then there will no solutions, while if the separation is $2 a$ or $2 b$ and the angle $\theta$ is chosen appropriately, then there will be one solution, while if the separation is less than $2 b$ then there will be $2$ solutions for $C$.
Alternative solutions, remarks, on this problem are highly appreciated.