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In classical geometry an ellipse is usually defined as the locus of points in the plane such that the distances from each point to the two foci have a given sum.

When we speak of an ellipse analytically, we usually describe it as a circle that has been squashed in one direction, i.e. something similar to the curve $x^2+(y/b)^2 = 1$.

"Everyone knows" that these two definitions yield the same family of shapes. But how can that be proved?

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Suppose we have a classical ellipse with its two foci and major axis given. We want to prove that it's a squashed circle.

Select a coordinate system with its origin in the center if the ellipse, the $x$-axis passing through the foci, and a scale such that the common sum-of-distances-to-the-foci is $2$. The foci have coordinates $(\pm c,0)$ for some $c\in[0,1)$.

By considering the nodes at the end of the minor axis we find that the semiminor axis of the ellipse must be $b=\sqrt{1-c^2}$. We must then prove that the equations $$ \tag{1} x^2 + \left(\frac{y}{\sqrt{1-c^2}}\right)^2 = 1 $$ $$ \tag{2} \sqrt{(x+c)^2+y^2} + \sqrt{(x-c)^2+y^2} = 2 $$ are equivalent.

Rearranging (1) gives $$\tag{1'} y^2 = (1-c^2)(1-x^2) $$ and therefore $(x+c)^2+y^2 = (1+xc)^2$ by multiplying out each side and collecting terms. Similarly, $(x-c)^2+y^2=(1-xc)^2$. So for points where (1) holds, (2) reduces to $(1+xc)+(1-xc)=2$, which is of course true.

Thus every solution of (1) is a solution of (2). On the other hand, for every fixed $x\in(-1,1)$, it is clear that the left-hand-side of (2) increases monotonically with $|y|$ so it can have at most two solutions, which must then be exactly the two solutions for $y$ we get from (1'). (And neither equation has solutions with $|x|>1$). So the equations are indeed equivalent.


Conversely, if we have a squashed circle with major and minor axes $2a$ and $2b$, we can find the focal distance $2c$ by $c^2+b^2=a^2$ and locate the foci on the major axis. Then the above argument shows that the classical ellipse with these foci coincides with our squashed circle.

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  • $\begingroup$ Awesome algebraic explanation! However, do you think there's a way to better see this connection geometrically? To somehow see that if we stretch a circle, there will be two points whose sum of the distances to any point on the ellipse will be a constant? I've been trying to come up with one and haven't been able to. Thank you! $\endgroup$ – Joshua Ronis Jan 5 at 18:51
  • $\begingroup$ @JoshuaRonis: If I had a good geometric proof, I wouldn't have needed to rely on mindless formula-crunching here. $\endgroup$ – Henning Makholm Jan 6 at 14:38
  • $\begingroup$ Haha okay I'm asking in a bunch of forums and trying to find one myself, I'll post here if I do $\endgroup$ – Joshua Ronis Jan 6 at 15:30
  • $\begingroup$ Do you know why, given a squashed circle, do the foci lie on the major axis? It's seems so obvious, but I cannot explain it rigorously... $\endgroup$ – Bman72 Jun 26 at 13:38

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