Given three points $A,B,C$, I want to identify the ellipse whose axes are parallel to the coordinate axes, and passing through these points
The equation of the ellipse is
$$ \dfrac{(x - x_0)^2}{a^2} + \dfrac{(y - y_0)^2}{b^2} = 1 $$
In addition to the three points, it is also given that $\dfrac{a}{b} = \alpha $, a known constant.
What is the procedure to follow to identify the parameters of this ellipse.
My effort:
Multiplying the given model by $a^2$ gives
$$ (x - x_0)^2 + \left( \dfrac{a}{b} \right)^2 ( y - y_0)^2 = a^2 $$
Expanding
$$ x^2 + \alpha^2 y^2 + (-2 x_0) x + (-2 y_0 \alpha^2 ) y + x_0^2 + \alpha^2 y_0^2 - a^2 = 0 $$
Define the unknown vector as $ u = -2 x_0 , v = - 2 y_0 \alpha^2 , w = x_0^2 + \alpha^2 y_0^2 - a^2 $, then the equation we want to identify becomes
$$ u x + v y + w = - x^{2} - \alpha^{2} y^2 $$
Substituting the three points we have $A(x_1, y_1), B(x_2, y_2), C(x_3, y_3)$, we obtain a linear system of three equations in the three unknowns $u, v, w$. Once we solve this linear system, it follows that
$$ x_0 = -\dfrac{u}{2} $$
$$y_0 = - \dfrac{v}{2 \alpha^2 } $$
$$ a^2 = x_0^2 + \alpha^2 y_0^2 - w $$
And finally, $ b^2 = \dfrac{a^2}{\alpha^2} $
My question is: Does the above make sense, and is there a better and more direct way, also, is the math correct, or are there any math errors in my formulation ? Thank you all very much.
Alternatively,
$$(x - x_0)^2 + (\alpha y - \alpha y_0)^2 = a^2 $$ is the equation of the circle of center $(x_0, \alpha y_0)$ and radius $a$ in a stretched plane $(x,\alpha y)$. This is a classical problem (solved by intersecting two bissectors).
https://stackoverflow.com/questions/62488827/solving-equation-to-find-center-point-of-circle-from-3-points