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.
On
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 $a^2b^2$ $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
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.