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Topic: ellipse, mean distance from point on perimeter to a focus ?
Replies: 8   Last Post: Feb 12, 2010 5:56 PM

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Rob Johnson

Posts: 1,771
Registered: 12/6/04
Re: ellipse, mean distance from point on perimeter to a focus ?
Posted: Feb 12, 2010 5:22 AM
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In article <20100211111323.200$>,
David W. Cantrell <> wrote:
> (Rob Johnson) wrote:
>> In article
>> <>, G
>> Patel <> wrote:

>> >Is the mean distance from a point on perimeter to a focus equal to the
>> >semi major axis length?

>> Yes.

>Well, as Achava correctly pointed out, it depends on the variable with
>respect to which the mean is calculated.

Indeed. I realized that soon after I posted. As soon as I read the
original post, the picture of the "string and pins" construction
popped into my head and I went with it. Any weight symmetric with
respect to the foci will give the semi-major axis as the mean.

>> Recall the "string and pins" method of drawing an ellipse, which
>> depends on the fact that the sum of the distances from a point on the
>> ellipse to the two foci of that ellipse is the major axis of that
>> ellipse. Since the sum of those two distances is always the major
>> axis, the sum of their means is the major axis. By symmetry, the
>> means of these distances are the same. Therefore, each is equal to
>> the semi-major axis.

>That's a simple and excellent argument that _a_ mean is the semi-major axis
>length. Perhaps in some sense, that is the most useful mean.
>But I initially (before seeing Achava's post) calculated the mean distance
>with respect to theta, thinking of the ellipse in polar coordinates as
>given by
>r = a (1 - e^2) / (1 - e cos(theta))
>Doing that, we find instead
>mean distance = a sqrt(1 - e^2) = b, the semi-minor axis length.

I get that as well. Furthermore, averaging with respect to time,
using equal area in equal time, I get a mean of a(1+e^2/2).

Rob Johnson <>
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