How do I translate and rotate an Ellipse in Polar coordinates?

Question

So I'm trying to find the best ellipse that fits with a sample data, that is an easy task if the ellipses fallow the standard form: $$r = \frac{b}{\sqrt{- e^{2} \cos^{2}{\left(θ \right)} + 1}}$$

But when I transform my data from cartesian coordinates to polar coordinates, my data will not always be close to an ellipse as standardized as this one. I need to include a variable responsible to translation $(r_0)$ and one responsible to rotation of the axis $(\theta_0)$. Thus, by obtaining these variables, I can calculate the variational of the residual of my function $r(\theta)$ relative to them, in such a way that I optimize the residual $(data sample - approximation\ function)$. I don't know what is the "general" elliptic equation (which includes translation and rotation) in polar coordinates. I searched for this equation a lot and I tried to find it by myself but I didn't succeed. Does anyone know it?

Answer 1

Concerning the translation you have better to deduct $x_0, y_0$ from the "points" coordinates.
For the rotation you add a $\theta _0$ to $\theta$.

Answer 2

The given ellipse in Cartesian coordinates is of the form $$ \frac{x^2}{a^2}+ \frac{y^2}{b^2}=1;\; a=\frac{b}{\sqrt{1-e^2}} $$

and is very advantageous to retain in cartesian form until coefficients are evaluated by least squares.

Wlog we assume simplified conic form

$$ a x^2 + by^2 -1=0 $$

$$ a \Sigma x^2 + b\Sigma y^2 -n=0 \tag 1 $$

$$a \Sigma x^3 + b\Sigma x y^2 -\Sigma x =0 \tag 2 $$

Solving the above and priming as $(a',b')$ for new set the cluster of given points automatically decides an x- elongated ellipse whose eccentricity can be determined. As required we convert back to polar form:

$$ \frac{(r \cos \theta)^2}{a^{'2}}+ \frac{(r \sin \theta)^2}{b^{'2}}=1. \tag3 $$