TITLE: How to find a closed curve approximation to contour lines that look like ellipses?
QUESTION [6 upvotes]: I am finding it very difficult to find resources giving a overview of techniques that can be used to approximate a closed curve. For example, the wikipedia article on curve fitting has further links to linear/polynomial regression techniques that I am familiar with but their section for closed curve fitting is very brief and limited to mentions on fitting circles and ellipses. The references I can find here are also similar brief or specific:
Fitting of Closed Curve in the Polar Coordinate.
Fitting a closed curve on the roots of ${x \choose k}-c$
My specific problem is as follows. I have a function $f(x,y)$ for which I have plotted some contour lines:

Each contour looks like part of a closed ellipse to me. To solve for the contour lines I considered solving the equation, in polar coordinates, $f(r \sin(\theta), r \cos(\theta)) = k$. However, this is not solvable as confirmed by Sage math.
Thus, I am looking for general approaches that will let me find an approximate closed curve to these contour lines. The only thing I can think of doing is to write down a regression equation for $r$ in terms of $\theta$, and then solve the least squares problem: i.e. find coefficients of the equation such that $(f(x, y) - k)^2)$ is minimised.
EDIT: As requested in the comment, 
$$f(x,y) = e^{-x - y} \sum_{m = 0}^\infty \frac{x^my^m}{(m!)^2}$$
and contour lines are obtained by considering rate parameters $x$ and $y$ such that the probability of a Skellam distributed random variable $Z$ taking value zero is held constant. While solutions specific to this equation are appreciated (such as in one of the previously linked posts where Stirling's approximation is used), please note that more generic methods of fitting closed curves would be more instructive for me in the future.

REPLY [2 votes]: Perhaps they are ellipses with a focus at (0,0).
Then you have $$\sqrt{x^2+y^2}+\sqrt{(x-a)^2+(y-a)^2}=C\\
(x-a)^2+(y-a)^2=(C-\sqrt{x^2+y^2})^2$$
Tidy it up, leave just the square-root on one side, square again.
Note that $2x^2+2y^2=(x+y)^2+(x-y)^2$.
This converts to 
$$x-y=\pm\sqrt{g(x+y)}$$
for some quadratic function of $(x+y)$.
Then you can plot $(x,y)=((x+y)/2+(x-y)/2,(x+y)/2-(x-y)/2)$ and see if they fit.  
The question remains, what is the relation between $a$ and $C$?
Suppose an ellipse cuts the line $x=y$ at $(-b,-b)$ and $(d,d)$ where $d>a$.  Then the sum of their distances to the foci $(0,0)$ and $(a,a)$ is 
$$C = \sqrt{2}b+\sqrt{2}(a+b)=\sqrt{2}d+\sqrt{2}(d-a)$$
If you know $b$ and $d$, you can find $a$ and $C$.