My question is about the average distance from the focus of an ellipse.

If we let the equation of an ellipse in polar coordinates (centred at the focus) be $$r = \frac{\ell}{1+\varepsilon\cos{\theta}}$$ where $\ell$ is the semi-latus rectum and $\varepsilon$ is the eccentricity, then the mean distance should be the limit of the sum of $n$ radii divided by $n$. Taking this from $-\pi$ to $\pi$ gives the following integral, for which $d\theta$ is the limit of $\frac{2\pi}{n}$, \begin{align} \overline{r} &= \int_{-\pi}^{\pi} \frac{\ell}{1+\varepsilon\cos{\theta}} \frac{d\theta}{2\pi} \\ &= \frac{\ell}{\pi} \int_0^{\pi} \frac{1}{1+\varepsilon\cos{\theta}}d\theta \\ &= \frac{\ell}{\pi} \int_0^{\infty} \frac{1}{1+\varepsilon\frac{1-t^2}{1+t^2}} \frac{2}{1+t^2}dt \tag{t = $\tan{\frac{\theta}{2}}$}\\ &= \frac{2\ell}{\pi} \int_{0}^{\infty} \frac{1}{(t\sqrt{1-\varepsilon})^2 + (\sqrt{1+\varepsilon})^2}dt \\ &= \frac{2\ell}{\pi} \frac{1}{\sqrt{1-\varepsilon^2}} \arctan{\left(\frac{t \sqrt{1-\varepsilon}}{\sqrt{1+\varepsilon}}\right)} \Big|_0^\infty \\ &= \frac{\ell}{\sqrt{1-\varepsilon^2}}\end{align} This is the semiminor axis. However, the most commonly quoted 'average distance' from the focus I see is the semimajor axis, which is equal to $$a= \frac{\ell}{1-\varepsilon^2}$$ What am I doing wrong? Thank you for your time in reading this question.

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    $\begingroup$ Different parameterizations yield different “weighting.” $\endgroup$ – amd Jun 13 at 0:04
  • $\begingroup$ in one parametrization you could be sweeping across the ellipse by angle, and in another one by x coordinate. in the ellipse case, you'd be giving less weight to the major axis if you used a polar representation $\endgroup$ – Saketh Malyala Jun 13 at 0:50
  • $\begingroup$ what do you mean by weighting? surely there's only one true average distance from the focus? $\endgroup$ – Andres Klene-Sanchez Jun 14 at 21:20
  • $\begingroup$ You should be integrating with respect to arclength on the ellipse, not with respect to $\theta$. So make that adjustment. "One true average"? That's the whole point of probability — what is the measure with respect to which we're counting? You get all sorts of paradoxes if you don't realize that. In particular, check out Bertrand's paradox. $\endgroup$ – Ted Shifrin Jun 17 at 22:19
  • $\begingroup$ @TedShifrin the Bertrand paradox was amazing to read about, so thank you for that link. Could you provide more detail as to the specific integral/technique, perhaps with an answer? $\endgroup$ – Andres Klene-Sanchez Jun 18 at 21:28

When I try to do the arclength integral, I get something I do not know how to integrate. But here is the easy argument people have in mind, I think. You're averaging the distance from one focus. The distance from the other focus has the same average (by symmetry). But since the sum of the distances is always twice the semi-major axis, the average of their sum is twice the semi-major axis. Thus, the average of one of the distances is the semi-major axis.

Note that here it's important that I average over arclength, so as to have the symmetry argument I claim. Using $d\theta$ as a measure surely won't work, as $\theta$ is not at all a natural variable to use for distance from the other focus.


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