The purpose of this exercise is to find the equation of the ellipse whose axes are parallel to the coordinate axes, and passing through $4$ given points, where it is assumed that no $3$ of them are collinear.
My effort:
The equation of the above-described ellipse is
$$ \dfrac{(x - x_0)^2 }{a^2} + \dfrac{ (y - y_0)^2 }{b^2} = 1 $$
Define $A = \dfrac{1}{a^2}$ and $ B = \dfrac{1}{b^2} $, then the equation becomes
$$ A (x^2 - 2 x x_0 + x_0^2) + B (y^2 - 2 y y_0 + y_0^2) = 1 $$
Re-arranging
$$ A x^2 + B y^2 + C x + D y + E = 0 $$
where $ C = -2 x_0 A $ , $ D = - 2 y_0 B $ , $ E = -1 + A x_0^2 + B y_0^2 $
From the above definitions, it follows that
$ x_0 = - \dfrac{ C }{2 A} $ and $y_0 = - \dfrac{ D }{2 B} $
Therefore, $E = -1 + \dfrac{1}{4} \left( \dfrac{C^2}{A} + \dfrac{D^2}{B} \right) $
And this will lead to a nonlinear system in terms of $A,B,C,D$. To overcome this problem, and knowing that coefficient of $x^2$ cannot be zero, I divided the above ellipse equation by $A$, to obtain
$ x^2 + B' y^2 + C' x + D' y + E' = 0 $
where $ B' = \dfrac{B}{A}, C' = \dfrac{C}{A} , D' = \dfrac{D}{A}, E' = \dfrac{E}{A} $
Now there are only $4$ unknowns appearing linearly, so using the $4$ points we can find them all, by solving the $4 \times 4$ linear system:
$\begin{bmatrix} y_1^2 && x_1 && y_1 && 1 \\ y_2^2 && x_2 && y_2 && 1 \\ y_3^2 && x_3 && y_3 && 1 \\ y_4^2 && x_4 && y_4 && 1 \end{bmatrix} \begin{bmatrix} B' \\ C' \\ D' \\ E' \end{bmatrix} = \begin{bmatrix} - x_1^2 \\ - x_2 ^ 2 \\ - x_3^2 \\ - x_4^2 \end{bmatrix} $
It follows from the above analysis that
$ x_0 = - \dfrac{1}{2} C' $
$ y_0 = - \dfrac{1}{2} \dfrac{D'}{B'} $
and
$ E = -1 + A x_0^2 + B y_0^2 $
so that
$ E' = \dfrac{-1}{A} + x_0^2 + B' y_0^2 $
From this we can solve for $A$, and this also gives $B = A B' $
Now the ellipse is fully identified.
Question: Is the method described above correct, and is there a simpler and more direct way ? I look forward to any comments or suggestions or solutions. Thank you all.