# Mean distance from the focus of an ellipse in polar coordinates

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.

• Different parameterizations yield different “weighting.” – amd Jun 13 at 0:04
• 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 – Saketh Malyala Jun 13 at 0:50
• what do you mean by weighting? surely there's only one true average distance from the focus? – Andres Klene-Sanchez Jun 14 at 21:20
• 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. – Ted Shifrin Jun 17 at 22:19
• @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? – Andres Klene-Sanchez Jun 18 at 21:28

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.