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This question was motivated by Definition of an ellipsoid based on its focal points .

I'd like to avoid terms like ellipsoid, so I'll use terms like one-dimensional ellipse (a normal ellipse in the plane) and two-dimensional ellipse (an ellipsoid in space).

One description for a one-dimensional ellipse in the $x_1x_2$-plane captures the geometry that for any point on the ellipse, the sum of the distances to the foci is constant: $$E_1=\left\{\bar{x}\,\left|\,\sum_{i=1,2}\|\bar{x}-\bar{f}_i\|=c\right.\right\}$$ where $\bar{f_1}$ and $\bar{f}_2$ are the foci and $c$ is some constant. It occurred to me that the set $\{\bar{f}_1,\bar{f}_2\}$ is a zero-dimensional ellipse. So I thought to rephrase the description above as $$E_1=\left\{\bar{x}\,\left|\,\int_{\textrm E_0}\|\bar{x}-\bar{e}\|\,de=c\right.\right\}$$ where $E_0$ is the zero-dimensional ellipse $\{\bar{f}_1,\bar{f}_2\}$ and each point in $E_0$ has measure 1.

What do you get when you upgrade a dimension? In words, what surface is the collection of all points in space where for each of those points, the integrated distance between that point and a one-dimensional ellipse is constant? I.e. what surface is $S$ in $\mathbb{R}^3$ where $$S=\left\{\bar{x}\,\left|\,\int_{\textrm E_1}\|\bar{x}-\bar{e}\|\,de=c\right.\right\}$$ where $E_1$ is a one-dimensional ellipse in $\mathbb{R}^3$? Answers could be implicit equations in $x_1$, $x_2$, and $x_3$, or something else.

At first I thought it would be neat if this gave a generic two-dimensional ellipse, providing an answer to the question linked above, but I have convinced myself that this is not so.

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    If $C$ is small enough, in the special case that $E_1$ is a circle, I think you should get a torus. Does that sound right?2011-05-21
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    @Simon: That keeps a constant _minimum_ distance from points on the surface to the circle. I'd like to keep the integrated distance (ID) constant. A torus won't be the surface, since points on the outer rim have a greater ID from the circle than points on the inner rim. Some examples: the center of a circle of radius 1 has an ID of $2\pi$ ($\int_0^{2\pi} 1\,dt$) from the circle, and I'm pretty sure $2\pi$ is the minimal possible ID. A point **on** the circle has ID $4\pi$ ($\int_0^{\pi} 2\sin(t)\,dt$), so with $c=4\pi$ the surface would contain the circle, but also other points in space.2011-05-22
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    Also, you could divide integrated distance by the length of $E_1$ and have the more common _average_ distance. Keeping the average distance from $E_1$ constant would give the same shapes as keeping the integrated distance constant, just using different values of $c$. I prefer to use integrated distance, since that is the usual form of the equation for the one-dimensional ellipse.2011-05-22
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    What do you mean by an integral of the form $\int_{E_1} \ldots de\ $? Are you integrating over the interior or along the curve, or what?2011-05-22
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    @Christian: Along the curve, with respect to arc length.2011-05-22
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    I would guess this is highly dependent on the magnitude of $c$ in relation to the width of the one-dimensional ellipse $e$. In particular, if $c$ is too small, $S$ is empty. Is this really what you want?2012-06-11
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    @Glen The same issue applies to the classical definition of a regular ellipse. If two foci are fixed at $1$ unit apart and we take $c<1$, then $S$ will be empty. So yes, I think the question is still there as posed, but it's good to note what you have noted.2012-06-11

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