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I've seen many proofs online, but I can't really wrap my mind around it.

Being a generalization of the circle, I thought its equation would be as easy to understand as the circle's. Turns out I was wrong, or maybe I'm just too stupid to grasp the geometric intuition behind it.

Matt24
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    Clearly the change of variables $u = x/a$ and $v = y/b$ gives the equation of a circle. If you grasp that equation, perhaps you can think of the ellipse as being a circle "stretched" in the coordinate directions. – hardmath Oct 17 '15 at 22:13
  • try to see it as $x^2 + \frac{a^2}{b^2}y^2=a^2$ if this helps – Michael Medvinsky Oct 17 '15 at 22:13
  • @MichaelMedvinsky I think I see it now with this visual link and FarazMasroor's help below.

    If I'm correct, the ellipse equation would really be the equation of a circle but reexpressing one of the coordinates as a ratio between the major axis (or the original radius) and the minor axis, as can be seen in the link. Right?

     – Matt24 Oct 18 '15 at 01:56

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My geometric intuition, hopefully it works for you, is that if you have $x^2+y^2=1$ then you have a circle, but if you do x/a or y/b then you end up contracting or expanding the circle in the x or in the y direction. If you have the original circle, it intersects the x axis at 1 and -1, but in the new ellipse it intersects the x axis when x/a is 1 or -1, namely a and -a. Hopefully this helps you graph them too.

Faraz Masroor
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    Hmm, I see what you're driving at, but does this only work for the unit circle? I mean, how can I relate the ellipse to the more general circle equation (which has r^2 instead of 1)? – Matt24 Oct 18 '15 at 00:34
  • You could multiply by a^2 so then if y=0 then x^2=a^2 q – Faraz Masroor Oct 18 '15 at 00:35
  • I'm so lost, what does x^2=a^2 tell me about the ellipse?

    I'm sorry for being so dumb :/

     – Matt24 Oct 18 '15 at 00:49
  • @Matt24 it tells you the values of x where the ellips crosses the x axis (we set y=0 to get here) – Faraz Masroor Oct 18 '15 at 00:49
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    Oh, right, how did I not see it hahaha. Ok, I'm gonna explore this a bit more graphically, and if I'll tell you if I have another doubt. Thanks for your help :) – Matt24 Oct 18 '15 at 00:56
  • @Matt24 no problem. Graphs are always helpful :) – Faraz Masroor Oct 18 '15 at 00:56
  • If you'd like to check it out, I shared this link in a comment above. I think it continues your train of thought. What do you think? – Matt24 Oct 18 '15 at 13:28
  • @Matt24 exactly, it's like a stretching or compressing. – Faraz Masroor Oct 18 '15 at 13:31
  • And what would the coefficient (a/b) represent? I assumed it was the ratio between the major and minor axis, though I really don't quite get why this has to be so. I can definitely see that when y=b and x=0, then I get a^2=a^2, but it gets a bit more difficult to understand when y=anything else. – Matt24 Oct 18 '15 at 13:54
  • That's right, it's like a proportion because you're scaling things – Faraz Masroor Oct 18 '15 at 14:25
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It is the equation of the ellipse centered at the origin. Consider the ellipse centered at the origin and whose focal axis coincides with the axis X. The focus $F$ and $F'$ are on the x axis. $O$ as its center is a midpoint of FF' segment , the coordinates of F and F ' will for example $( c , 0 )$ and $( -c , 0 )$ , respectively , where c is a positive constant. Let p $( x , y)$ any point of the ellipse. By the definition of the curve, the point P must satisfy the condition. $$d(FP)+d(F´P)=2a$$ where $a$ is a positive constant greater than $c$, then $$d(FP)= \sqrt{(x-c)^2+y^2}$$ $$d(F´P)= \sqrt{(x+c)^2+y^2}$$ and remplace. $$\sqrt{(x-c)^2+y^2}+\sqrt{(x+c)^2+y^2}=2a$$ simplifying it. $$cx+a^2=a*\sqrt{(x+c)^2+y^2}$$, then it square $$c^2x^2+2a^2cx+a^4=a^2x^2+2a^2cx+a^2c^2+a^2y^2$$ $$(a^2-c^2)x^2+a^2y^2=a^2(a^2-c^2)$$ how $2a>2c$ and $a^2>c^2$ then $a^2-c^2$ is positive and get $b^2=a^2-c^2$ then $$b^2x^2+a^2y^2=a^2b^2$$ that finally it divide for $a^2b^2$ $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$

Yobamath
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