Standard Form Ellipse Through Three Points and Parallel to X and Y Axes I want the general form $\frac{(x-x_0)^2}{a^2}+\frac{(y-y_0)^2}{b^2}=1$ for an ellipse with a specified eccentricity $e$ that passes through three (non-collinear) points $(x_1,y_1), (x_2, y_2), (x_3, y_3)$ and is parallel to the X and Y axes (i.e. major axis of ellipse parallel to the X axis and minor axis parallel to the Y axis).
I found this gem on Wikipedia:
$$
\frac{({\color{red}x} - x_1)({\color{red}x} - x_2) + {\color{blue}q}\;({\color{red}y} - y_1)({\color{red}y} - y_2)}
             {({\color{red}y} - y_1)({\color{red}x} - x_2) - ({\color{red}y} - y_2)({\color{red}x} - x_1)} =
        \frac{(x_3 - x_1)(x_3 - x_2) + {\color{blue}q}\;(y_3 - y_1)(y_3 - y_2)}
             {(y_3 - y_1)(x_3 - x_2) - (y_3 - y_2)(x_3 - x_1)}\ .
$$
where ${\color{blue}q} = \frac{a^2}{b^2} = \frac{1}{1 - e^2}$, which I think is supposed to work, but a) converting this equation to standard form is a bear (and maybe isn't doable?), and b) seems to introduce $xy$ terms which leads me to believe the ellipse will be tilted with respect to the X and Y axes.
Is this the right equation to be working with?  If so, is there a standard form of the equation?  Is there a different/better way to accomplish the task?
P.S. Having the standard form is pretty important: I'm going to use this with a graphics app where knowing $x_0, y_0, a,$ and $b$ is required.

Alternatively, the equation can be re-arranged in a compact form:
$$
\begin{vmatrix}
  (1-e^2)x^2+y^2 & x & y & 1 \\
  (1-e^2)x_1^2+y_1^2 & x_1 & y_1 & 1 \\
  (1-e^2)x_2^2+y_2^2 & x_2 & y_2 & 1 \\
  (1-e^2)x_3^2+y_3^2 & x_3 & y_3 & 1
\end{vmatrix}=0$$
where $e\ne 1$ and comparing with the general form
$$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$
Now,
\begin{align}
  A &= (1-e^2) C \\ \\
  B &= 0 \\ \\
  C &=
  \begin{vmatrix}
    x_1 & y_1 & 1 \\
    x_2 & y_2 & 1 \\
    x_3 & y_3 & 1
  \end{vmatrix} \\ \\
  D &= -
  \begin{vmatrix}
    (1-e^2)x_1^2+y_1^2 & y_1 & 1 \\
    (1-e^2)x_2^2+y_2^2 & y_2 & 1 \\
    (1-e^2)x_3^2+y_3^2 & y_3 & 1
  \end{vmatrix} \\ \\
  E &=
  \begin{vmatrix}
    (1-e^2)x_1^2+y_1^2 & x_1 & 1 \\
    (1-e^2)x_2^2+y_2^2 & x_2 & 1 \\
    (1-e^2)x_3^2+y_3^2 & x_3 & 1
  \end{vmatrix} \\ \\
  F &= -
  \begin{vmatrix}
    (1-e^2)x_1^2+y_1^2 & x_1 & y_1 \\
    (1-e^2)x_2^2+y_2^2 & x_2 & y_2 \\
    (1-e^2)x_3^2+y_3^2 & x_3 & y_3
  \end{vmatrix} \\
\end{align}
Re-arrange the equation as
$$A
\left( x+\frac{D}{2A} \right)^2+
C
\left( y+\frac{E}{2C} \right)^2=
\frac{D^2}{4A}+\frac{E^2}{4C}-F$$
implying the centre is
$$\left( -\frac{D}{2A}, -\frac{E}{2C} \right)$$
and the semi-axes
$$
(a,b)=
\left(
  \sqrt{\frac{D^2}{4A^2}+\frac{E^2}{4AC}-\frac{F}{A}},
  \sqrt{\frac{D^2}{4AC}+\frac{E^2}{4C^2}-\frac{F}{C}}
\right)$$
for $0 \le e<1$.
