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I've learned two separate ways to define an ellipse:

  1. It's a stretched circle. We get the formula for a unit circle, $X^2 + Y^2 = 1$, and stretch it by dividing the terms like so: $(\frac{X}{a})^2 + (\frac{Y}{b})^2 = 1$. In order to satisfy the same equation, for every $Y$ we previously had, $X$ must get stretched by a factor of $a$, and for every $X$ we previously had, $Y$ must be stretched (multiplied) by a factor of $b$.

  2. An ellipse is the set of all points whose sum of the distances from two points, the foci, is a constant. We can represent this with the equation $\sqrt{(x+f)^2 + y^2} + \sqrt{(x-f)^2 + y^2} = c^2$, where $c = 2a$ from the previous equation, $f$ is the distance from a focus to the origin, and $x$ and $y$ are the variables.

But as far as I've learned, it all seems like everything having to do with the foci in the first definition comes from first accepting that the foci DO indeed exist from the second definition. I can't seem to find a connection between them (the first definition doesn't even have an $f$ in it!!!).

How can we deduce that such "foci" exist from the first explanation? Every explanation of ellipses I've seen seems to use the stretched circle equation, and then go on to deduce the rest of the characteristics of the ellipse but only ASSUMING that the foci exist in the first place.

Thank you!

joshuaronis
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    There's algebra that predicts where the foci should be starting from $a$ and $b$ in your first definition. Then you can check that those two points when used in the second definition give you the same ellipse. Then anything you derive about the ellipse from either definition will be valid when you use the other. – Ethan Bolker Jan 03 '19 at 16:06
  • Somewhat related: Geometric proof of equivalence between two constructs of ellipse. The "two constructs" in that problem are the "two-foci" and "two-circle". "Two-foci" is the constant-sum-of-distances definition. "Two-circle" amounts to the parametric definition, which is equivalent to a "stretched circle" definition. My solution proves equivalence by deducing a common algebraic form, which I admit is unsatisfying. I still think there's a more-geometric argument, but I've set the pursuit aside. – Blue Jan 03 '19 at 17:10
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    3Blue1Brown recently posted a video giving a beautiful geometric proof of the relationship between the 3 most common descriptions of ellipses. – Paul Sinclair Jan 04 '19 at 00:38
  • @PaulSinclair that video and that channel is amazing. However, in it, he talks about the relationship between the "constant-sum" and "sliced-cone" definition more than anything. Do you know of any resource similar to that video, or perhaps an intuitive explanation you know yourself, as to how to deduce the foci from the stretched circle equation, without assuming them in the first place? – joshuaronis Jan 06 '19 at 11:50
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    He talks about it starting at roughly 9:40, though he leaves the details to you (they are very similar to Dandelin's proof). Stretching a circle is easily seen to be equivalent to intersecting a cylinder with a slanted plane. Put two spheres in the cylinder tangent to both the cylinder and to the plane, and the foci are the points of tangency between the spheres and the plane, and just as in Dandelin's proof for the cone, the distances from them to the ellipse are the same as to the two circles of tangency to the cylinder, whose total distance is constant. – Paul Sinclair Jan 08 '19 at 00:56
  • @PaulSinclair Okay, I see how to get the ellipse from slicing the cylinder now. In fact, its even easier to picture than from slicing the cone, because in the cylinder situation the spheres will both be the exact same size and I know which part of them will be touching the edge of the cylinder (the widest part). However, I'm still confused as to how this connects to a stretched circle...like let's say I get every X component of a stretched circle and double it. What would that correspond to in slicing the cylinder, or the cone? Thanks! – joshuaronis Jan 09 '19 at 02:43
  • It would correspond to rotating the plane about the $y$ axis by an angle that makes the hypotenuse exactly twice the length of the radius of the cylinder. I.e. $\cos \theta=1/2$ and so $\theta=π/3$. The plane normal will make a 60 degree angle with the z axis. It would help if I could draw you a picture, but since this question is closed, I can't. It might help to view the plane as fixed and slant the cylinder instead, – Paul Sinclair Jan 10 '19 at 18:10

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