I'm deriving the standard equation for an ellipse from two arbitrary points \(F_1\) and \(F_2\).
An ellipse around these two focal points is a set of points \(P\) for which the sum of the distances to \(F_1\) and \(F_2\) is a constant \(s\), where \(s\) may be any value greater \(\overline{F_1 F_2}\).
$$ \overline{F_1 P} + \overline{F_2 P} = s $$
Let's sketch this out with a little SVG (luckily, SVG has an ellipse element).
We have our focal points \(F_1 = (-c, 0)\) and \(F_2 = (c, 0)\). The midpoint of \(\overline{F_1 F_2}\) is the point of origin and center of the ellipse \(O = (0,0)\). The points \(V_1 = (a, 0)\) and \(V_2 = (0, b)\) are the vertices of the ellipse. Parameter \(a\) is the semi-major axis (maximal radius of the ellipse), \(b\) the semi-minor axis (minimal radius of the ellipse), and \(c\) the linear eccentricity (distance from the center to any of the two foci).
The sum of the distances from \(F_1\) and \(F_2\) to \(V_1\) is \(2a\), so in general \(s = 2a\).
$$
\overline{F_1 V_1} + \overline{F_2 V_1} = (c + a) + (a - c) = 2a
$$
In case you don't believe me, consider the points \(P\), \(F_1\) and \(Q = (-c, y)\).
If \(-c = x\), then the distance obviously is \(\sqrt{(x+c)^2 + y^2} = y\).
Otherwise, those three points form a right triangle with the hypotenuse \(\overline{P F_1}\).
Line segment \(\overline{P Q}\) is parallel to the x-axis and has length \(x + c\).
Line segment \(\overline{F_1 Q}\) is parallel to the y-axis and has length \(y\).
The Pythagorean theorem says \(\overline{P F_1}^2 = (x + c)^2 + y^2 \).
The distance from any point \(P = (x,y)\) of the ellipse to \(F_1\) is \(\sqrt{(x+c)^2 + y^2}\), accordingly to \(F_2\) it is \(\sqrt{(x-c)^2 + y^2}\).
Now we can express the definition of the ellipse with the following equation:
$$\sqrt{(x-c)^2 + y^2} + \sqrt{(x+c)^2 + y^2} = 2a .$$
At this point we use the Pythagorean theorem to substitute \(c^2 = a^2 - b^2\).
$$
\begin{align*}
\frac{(a^2 - b^2) x^2}{a^2} + \cancel{a^2} &= x^2 + \cancel{a^2} - b^2 + y^2 \\
\cancel{a^2 x^2} - b^2 x^2 &= \cancel{a^2 x^2} - a^2 b^2 + a^2 y^2\\
-x^2 &= -a^2 + \frac{a^2 y^2}{b^2} \\
\frac{x^2}{a^2} &= 1 - \frac{y^2}{b^2}
\end{align*}
$$
We use a bit of algebraic manipulation and the Pythagorean theorem to derive the standard ellipse equation...
$$
\begin{align*}
\sqrt{(x-c)^2 + y^2} + \sqrt{(x+c)^2 + y^2} &= 2a \\
\sqrt{(x-c)^2 + y^2} &= 2a - \sqrt{(x+c)^2 + y^2}\\
\cancel{x^2} - 2cx + \cancel{c^2} + \cancel{y^2} &= 4a^2 - 4a \sqrt{(x+c)^2 + y^2} \\
&\hphantom{=} + \cancel{x^2} + 2cx + \cancel{c^2} + \cancel{y^2}\\
\cancel{-4} (cx + a^2) &= \cancel{-4}a \sqrt{(x+c)^2 + y^2}\\
c^2x^2 + 2 c x a^2 + a^4 &= a^2 (x^2 + 2cx + c^2 + y^2) \\
\frac{c^2 x^2}{a^2} + \cancel{2cx} + a^2 &= x^2 + \cancel{2xc} + c^2 + y^2 \\
\end{align*}
$$
If \(F_1 = F_2\) (there is only one focal point) and thus \(a = b = r\), we have the special case of a circle with radius \(r\) around the center \(O\). $$ x^2 + y^2 = r^2 $$