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Given a fixed distance $2a$, and two points $(F_1,F_2)$ in the Euclidean plane, one can define an ellipse as the set of points $E$ such that the sum of the distances $d(E,F_1) + d(E,F_2)$ is equal to $2a$.

How can one prove that those two foci $F_1$ and $F_2$ are unique or not? In other words, prove that for any other set of points $(F_1',F_2') \neq (F_1,F_2)$, and some distance $2a'$, there exists at least one point $P$ in the previously defined ellipse $E$, such that $d(P,F_1') + d(P,F_2') \neq 2a'$. It would be nice if there was a nice way of finding one or even all points P with that property. That means I wish the solution to be as constructive as possible, even though I will not discard a proof by contradiction.

A weak proof can be made using the Euclidean distance formula $d(A,B)=\sqrt{(x_a - x_b)^2 + (y_a - y_b)^2}$. A strong proof would require using a general distance formula using these properties.

Nighteen
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  • I just realized that my question might be rephrased as "Proof that the definition of an ellipse is unique". – Nighteen Jan 18 '19 at 01:03
  • Do you have the distance formula at your disposal? In other words, do you have the Cartesian equation for ellipses? Or must you stick to the definition you've given? – Clayton Jan 18 '19 at 01:12
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    The foci along with the distance $2a$ uniquely determine the major and minor axes of the ellipse. – Théophile Jan 18 '19 at 01:36
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    @Théophile You would have to define major and minor axes as the longest and shortest diameter. You would also have to prove that if two ellipses have different axes then they are different (have one point P that is in one and not in the other ellipse). – Nighteen Jan 18 '19 at 01:43
  • Isn't it immediate? If the axes are different, you can (very easily) determine a point on one but not the other. – Clayton Jan 18 '19 at 01:50
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    Try doing an indirect proof, aka proof by contradiction. Assume that the foci are not unique, and try to come up with a contradiction to that. – BadAtAlgebra Jan 18 '19 at 01:55
  • @Clayton I see now it is immediate, since if $d(P_1,P_2)=b_1$, with $b_1 < b_2$ being the minor axes values for both ellipses, then if $P_1$ is in the first ellipse, then $P_2$ cannot be in the second, since there is no point in the second ellipse with distance smaller than $b_2$. A trivial argument can be used if both $b_1$ and $b_2$ are equal in module but the axes have different endpoints. – Nighteen Jan 18 '19 at 02:04
  • @Clayton But I still cannot see how the foci and the distance $2a$ uniquely determines the major and minor axes, using the max/min diameter definition. In other words, how can one find the segments of the major and minor axes using the given definition of an ellipse. – Nighteen Jan 18 '19 at 02:04
  • The foci determine a unique line (the major axis); you should be able to use that to help. – Clayton Jan 18 '19 at 02:06
  • With your hint I can find the two points which SHOULD define major axes. I could also find the center point as the midpoint of the foci and, after drawing a perpendicular line passing through the center, I was also able to find the other two points that SHOULD define the minor axes. But I still cannot prove these points make the maximum and minimum distance in the set of points of the ellipse. – Nighteen Jan 18 '19 at 02:19
  • If the foci and the length of major axis $2a$ are fixed, the center is uniquely determined as the midpoint of $F_1F_2,;$ and $2b$ is uniquely obtained by Pythagorean theorem as $b^2=a^2-c^2.$ – user376343 Jan 18 '19 at 11:12

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