How to average ellipses? I would like to find the average among a number of ellipses.
The ellipses all have the same center, and the same major axis length, but they have different eccentricities and different orientations, which is what makes the problem challenging. 
Averaging two ellipses with the same eccentricity, but with an angle of $\frac{\pi}{2}$ between the major axes should result in a circle, as should the average between three same ellipses oriented at $\frac{\pi}{3}$ from one another. Averaging two ellipses with the same eccentricity, but with an angle of $\frac{\pi}{4}$ between the major axes should result an ellipse with a smaller major axis right in between the two axes.
Initially, it seemed that drawing a new ellipse through the intersection points between two ellipses would work, but that is not easily adapted to more than two ellipses, and it doesn't appear to work well if the eccentricity is different between ellipses. 
 A: One approach to "averaging" would be to write down the equations for each ellipse in the form $p_i(x,y)=0$ where $p_i$ is a second-degree polynomial normalized such that $p_i(0,0)=1$. Then you can take the average of the various polynomials (algebraically, that is just the same as averaging the coefficients position for position), the result will be an polynomial whose zeroes are another ellipse or circle:
$$ \frac 1n\Bigl(p_1(x,y)+\cdots+p_n(x,y)\Bigr) = 0 $$
A: The question I believe correctly states some key constraints, particularly that the average of an ellipse with itself rotated $\pi/2$ should be a circle.
Another important constraint on a method for computing the average of a set of ellipses is that the computed average must be independent of the ordering of the ellipses.
Developing a more concrete answer to this question requires more concise definition of the properties of the mean of a set of ellipses. One way to approach this is to define an error metric for the difference between one ellipse and another.
For this, I will assume that all the ellipses are centered on the origin, with semi-major axis $a$ oriented along the cartesian $x$-axis and semi-minor axis $b$ oriented along the cartesian $y$-axis, and each ellipse may have a rotation of angle $\alpha$ (positive counter-clockwise) from the $+x$ axis.
Let the eccentricity of the ellipse be $e = \sqrt{1 - \frac{b^2}{a^2}}$. Using the formulation
$$\left[ \frac{\cos^2(\alpha)}{a^2} + \frac{\sin^2(\alpha)}{b^2} \right] x^2 + 2\cos(\alpha)\sin(\alpha)\left[\frac{1}{a^2} - \frac{1}{b^2}\right]xy + \left[ \frac{\sin^2(\alpha)}{a^2} + \frac{\cos^2(\alpha)}{b^2} \right] y^2 = 1$$
from https://www.maa.org/external_archive/joma/Volume8/Kalman/General.html, if we then substitute 
\begin{align}
x \rightarrow r\cos(\theta) \\
y \rightarrow r\sin(\theta)
\end{align}
then working through all the tedious algebra yields an explicit formula in polar coordinates for the radius $r$ as a function of angle $\theta$:
$$r = \frac{b}{\sqrt{1 + e^2\left(\sin^2(\alpha) + \cos(2\alpha)\sin^2(\theta) - \frac{1}{2}\sin(2\alpha)\sin(2\theta) - 1 \right)}}  .$$
This generalized polar form for an arbitrary ellipse is a naturally suited coordinate system for evaluating the similarity between one ellipse and another. You could use the absolute area between two ellipses, computed in polar coordinates, as your chosen error metric for example:
$$A_{diff} = \int_0^{2\pi}\frac{1}{2}\|r^2_1 - r^2_0 \| d\theta$$
Probably a more practical error metric, since this ellipse mean in general will be computed for $N$ ellipses and most likely to be done numerically, might be (this is what I would use):
$$E = \sum_{i=1}^N E_i \\
E_i = \int_0^{2\pi} \left(r^2 - r_i^2 \right)^2 d\theta$$
and numerically find a discrete form of $r$ that minimizes $E$, then solve for the $e$ and $b$ that fit $r$.
Such a formulation in terms of functional minimization will ensure the mean is independent of the ordering of the set of input ellipses, and should ensure that the mean of ellipse with itself rotated by $\pi/2$ is a circle (though I haven't proven or tested this).
