Find ellipse rotation angle and minor semi-axis by two points Given two points with coordinates, and major semi-axis (a), I have to draw an ellipse (with the center of (0, 0)), which intersects with these points.
So I have to find minor semi-axis (b), and ellipse rotation angle.
There is an equation of rotated ellipse:
$$\frac{(x\cos\theta + y\sin\theta)^2}{a^2} + \frac{(x\sin\theta - y\cos\theta)^2}{b^2}=1$$
I tried to make a system of equations, but still can't figure out how to solve my problem.
Edit: It looks like my problem, that I described above, can have more than one solutions.
What I need initially is to draw a curve like on Google Earth "measure distance" feature:

So I try to calculate an ellipse, which intersects two points, and has center in the (0, 0) point. But I'm not sure this is the right decision.
How can I achieve this?
 A: The equation of the ellipse, or in general the conic, that is centered at the origin is
$ r^T Q r = 1 $
where $r =[x,y]^T$ and
$Q = \begin{bmatrix} Q_{11} && Q_{12} \\ Q_{12} && Q_{22} \end{bmatrix} $
Since we're given two points then plugging these into the equation of the conic, we have the following linear equations in the elements of $Q$
$ Q_{11} x_1^2 + Q_{22} y_1^2 + 2 Q_{12} x_1 y_1 = 1 $
$ Q_{11} x_2^2 + Q_{22} y_2^2 + 2 Q_{12} x_2 y_2 = 1 $
We need a third equation, and this comes from the fact that
$ Q = R D R^T $
Since $R$ is a rotation matrix, then
$ \det(Q) = \det(D) = \dfrac{1}{a^2 b^2} $
and
$\text{Trace}(Q) = \text{Trace}(D) = \dfrac{1}{a^2} + \dfrac{1}{b^2} $
From the two above equations, we have our third equation relative the entries of matrix $Q$, which is
$ \dfrac{1}{b^2} = a^2 (Q_{11} Q_{22} - Q_{12}^2 ) = Q_{11} + Q_{22} - \dfrac{1}{a^2} $
Where the third equation is the equality on the right of this last equation.
Now we have a system of 2 linear equations, and 1 quadratic equation in the entries of matrix $Q$.  And these can be solved without much difficulty, because of the fact that we have two linear equations, which means that we can solve for $Q_{11}, Q_{22}$ in terms of $Q_{12}$, and then plug these two expressions into the third equation, which will give us a quadratic equation in one variable only which is $Q_{12}$.  Solving for $Q_{12}$ will generate two possible values, and by back substitution, we generate $Q_{11}$ and $Q_{22} $.
Having found two possible values for the matrix $Q$ we can immediately find $b$ from the above equations.  To find $\theta$ we need to diagonalize $Q$ into the form $R D R^T$.  The matrix $R$ has the form
$ R = \begin{bmatrix} \cos \theta && - \sin \theta \\ \sin \theta && \cos \theta \end{bmatrix} $
where matrix $D$ must be ordered as follows
$ D = \begin{bmatrix} \dfrac{1}{a^2} && 0 \\ 0 && \dfrac{1}{b^2} \end{bmatrix}$
As a numerical example, suppose $(x_1, y_1) = (2, 3)$ , $(x_2, y_2) = (-2, 1) $ and $a = 4$, then there will the following two ellipses as solutions.


If we use the same two points, but make $a = 1$, then there will be two solutions that are hyperbolas.


