How to find ellipse with two points, angle and radii I have the coordinates of two different points on the edge of an ellipse, and I have the $x$ and $y$ radius of the ellipse, and I have the angle of rotation for the ellipse. I'm fairly certain there are two possible resulting ellipses, but how do I go about finding their centers?
 A: HINT.- You have as data the equation $\frac{(x’)^2}{a^2}+\frac{(y’)^2}{b^2}=1$ where $a$ and $b$ are the radius of your ellipse, $x’$ and $y’$ being the new coordinates; you have also the angle of rotation $\alpha$ besides of two points of the ellipse  $(x_1,y_1)$ and $(x_2,y_2)$.
►Put $m=\tan \alpha$ and let $(h,k)$ the coordinates of the center.
The new axes are the lignes
$$\begin{cases}OX': y-k=m(x-h)\iff y-mx+mh-k=0\\[2ex]OY':y-k=\frac{-(x-h)}{m}\iff my+x-h-mk=0\end{cases}$$
►Let $P=(x,y)$ be a generic point in the ellipse; the corresponding coordinates $(x’,y’)$ in the new system are given by the distances of $P$ to the two new axes (the above lines).
It follows $$x'=\frac{my+x-h-mk}{\sqrt{1+m^2}}$$ $$y’=\frac{y-mx+mh-k}{\sqrt{1+m^2}}$$
►So you have $$\frac{(my+x-h-mk)^2}{(1+m^2)a^2}+\frac{(y-mx+mh-k)^2}{(1+m^2)b^2}=1\qquad (*)$$
►Now you can get two equations from $(*)$ by using the given points $(x_1,y_1)$ and $(x_2,y_2)$ of the ellipse. This allows the calculation of the coordinates $(h,k)$ of the center.
