Equivalence between geometrical and analytical definitions of an ellipse

Geometric $\Longrightarrow$ Algebraic:

Up to translations and rotations, one can set the center of the ellipse to be $(0,0)$ and the foci coordinates to be $F_1=(-c,0)$ and $F_2=(c,0)$. Note that this won’t change the equation of the ellipse since it depends only on distances between points, which it is invariant under rotations and translations.

Now let $P=(x,y)$ be a point belonging to the ellipse. Hence it verifies the equation
$$MF_1+MF_2=2a,$$which is equivalent to
$$\sqrt{(x-(-c))^2+(y-0)^2}+\sqrt{(x-c)^2+(y-0)^2}=2a.$$The idea now is to write it as:
$$\sqrt{(x+c)^2+y^2}=2a-\sqrt{(x-c)^2+y^2}$$and then square it. After simplifications, this gives
$$cx=a^2-a\sqrt{(x-c)^2+y^2}.$$Now one only need to isolate the square root and square it again, we finally get
$$(a^2-cx)^2=a^2\left((x-c)^2+y^2\right),$$which can be rewritten as
$$\left(\frac{x}{a}\right)^2+\left(\frac{y}{\sqrt{c^2-a^2}}\right)^2=1$$So one defines $b:=\sqrt{c^2-a^2}$ the semi-minor axis and $c$ is the eccentricity.

By translating the center of the ellipse, one gets back the center coordinates.

Algebraic $\Longrightarrow$ Geometric:

Let $M(x,y)$ satisfy $\displaystyle\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. We show $MF_1 + MF_2 = 2a$.

Then we isolate $y^2$ :
$$y^2 = b^2\left(1 - \frac{x^2}{a^2}\right).$$

We define $F_1=(-c,0)$ and $F_2=(c,0)$ where $c^2=a^2-b^2$ and $c>0$.

Then one can compute explicitly $MF_1$ and $MF_2$. The computation leads to:
$$MF_1 = a + \frac{c}{a}x, \quad MF_2 = a - \frac{c}{a}x.$$Which can be summed in order to find
$$MF_1 + MF_2 = \left(a + \frac{c}{a}x\right) + \left(a - \frac{c}{a}x\right) = 2a.$$Thus, $M$ satisfies the focal definition.