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?
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.
(Stretched circle $\Rightarrow$ Ellipse with focis)
It's easy to show that projection of a diagram at an angle is the same as stretching. Let $E$ be a stretched circle, then it is not hard to find the corresponding angle s.t. $E$ is the projection of a circle $C$ at that angle.
Let $C_0$ be the unqiue cyclinder s.t. it contains the circle and it is normal to the plane containing the circle. Clearly $E$ is the intersection of $C_0$ with the plane containing $E$.
And there is a classical proof showing that the intersection of a cyclinder with a plane is an ellipse with focis, see @Element118 's answer.
(Ellipse with focis $\Rightarrow$ Stretched circle)
Let $E$ be an ellipse with focis $F_1,F_2=(\pm c,0,0)$, center $O=(0,0,0)$, major axis $2a$ and minor axis $2b$ sitting on the plane $z=0$. Clearly $a^2=b^2+c^2$.
Consider two spheres of radius $b$ and centers respectively $(c,0,b)$ and $(-c,0,-b)$ as the following diagram.
There exists an unique cyclinder $C_0$ inscribing both of the spheres. Clearly $C_0\cap \{z=0\}$ is an ellipse $E_1$ with focis $F_1$ and $F_2$ due to the classical proof mentioned above.
To prove that $E_1=E$, it reduces to showing that $OM=a$, which is equivalent to showing that $\angle OAM=\angle OMA$. Since $\angle OAM = \angle AMJ$, it reduces to showing $\angle OMA=\angle AMJ$ which is clearly true. Thus $E_1=E$.
And we clearly have that $E$ is the projection of a circle, which is the same as a stretched circle. The result follows.
From 3Blue1Brown, at about 9:40, expanding the comment from @Paul Sinclair into an answer.
A squashed circle with major axis $2a$ and minor axis $2b$ can sit inside tangent to a cylinder with radius $b$. By constructing spheres tangent to the cylinder and the squashed circle, they would be tangent to the squashed circle at the foci, let them be $F_1$ and $F_2$.
The spheres are tangent to the cylinder, touching the cylinder at a circle each. These circles are perpendicular to the axis of the cylinder.
Pick a point $P$ on the perimeter of the squashed circle. $F_1P$ is tangent to a sphere, and another line $l$ from $P$ tangent to the sphere at $Q_1$ runs along parallel to the axis of the cylinder. Likewise, $F_2P$ is tangent to the other sphere, and there is another line $m$ from $P$ tangent to the sphere $Q_2$ running along, parallel to the axis of the cylinder.
Since $l$ and $m$ are parallel with common point $P$, they are the same line. It follows that $F_1P+F_2P=Q_1P+Q_2P=Q_1Q_2$. It follows that no matter what point $P$ we picked, $Q_1Q_2$ is of a constant length, which implies that $F_1$ and $F_2$ are indeed foci and the squashed circle is indeed an ellipse by the "sum of distances" definition.
I know this question is 10 years old, but here is a completely pre-algebraic solution.
I will start with the Foci- Formula $$ \sqrt{(x-e)^2 + y^2} + \sqrt{(x+e)^2 + y^2} = 2.$$
To get rid of lefthand root, we shift the righthand root to the right and square. We obtain \begin{align} (x-e)^2 + y^2 &= (2 - \sqrt{(x+e)^2 + y^2})^2 \\ &= 4 - 4 \sqrt{(x+e)^2 + y^2} + (x+e)^2+y^2. \end{align} By eliminating $y^2$ and shifting $(x+e)^2$ to the left we get $$ -4xe = 4(1-\sqrt{(x+e)^2 + y^2}) $$ Now dividing by 4, shifting $1$ to the left and multiplying by $-1$ yields $$ xe + 1 = \sqrt{(x+e)^2 + y^2} $$ Now we are able to eliminate the second root by squaring and get $$ x^2e^2 + 1 = x^2+e^2+y^2. $$ Observe the identity $b^2 = 1-e^2$, eliminate all $e^2$ and rearrange: $$ x^2(1-b^2)+b^2 = x^2+y^2. $$ $x^2$ cancels and by dividing by $b^2$ we get $$ -x^2+1 = \frac{y^2}{b^2}, $$ which is obviously equivalent to the Squashed-Circle-Formula $$ x^2 + \left(\frac{y}{b}\right)^2 = 1. $$ Of course all steps are indeed equivalences and so the equivalence between the two formulas is proven.
