# 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 $pi/2$ between the major axes should result in a circle, as should the average between three same ellipses oriented at $pi/3$ from one another. Averaging two ellipses with the same eccentricity, but with an angle of $pi/4$ 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.

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The term eccentricity already has another meaning, namely $\sqrt{1-(m/M)^2}$ where $m$ is the minor axis and $M$ is the major axis. –  André Nicolas Jun 25 '12 at 21:18
@AndréNicolas: I don't particularly care about the definition of the eccentricity (even though results may somewhat differ in the end). I have updated my question to hopefully avoid confusion. –  Jonas Jun 25 '12 at 21:23
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## 1 Answer

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: $$\frac 1n\Bigl(p_1(x,y)+\cdots+p_n(x,y)\Bigr) = 0$$

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This sounds like an excellent idea. Now how do I get the second-degree polynomials from the semi-axes and the angle again? –  Jonas Jun 25 '12 at 22:03
The equation for an ellipse with the major axis along the $x$ axis is $1 - \frac {x^2}{a^2} - \frac{y^2}{b^2} = 0$. Now substitute $x\mapsto \cos(\theta) x + \sin(\theta) y$ simultaneously with $y\mapsto \cos(\theta) y - \sin(\theta) x$. I don't remember the details of going in the other direction with the final equation, but they are out there somewhere. –  Henning Makholm Jun 25 '12 at 22:11
I accept the answer, since it almost works. The problem is that for at least the first three coefficients of the polynomial, you need to average the inverse square root, which works badly when any of the parameters are zero. Consequently, I ended up plotting the ellipses and performing a least-squares fit through the points. –  Jonas Jun 29 '12 at 18:59
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