I mentioned in my previous answer that the "foci and string" form of the ellipse equation is rarely seen "except as the point of departure on an algebraic journey to the 'standard' form". I want to elaborate a little on that "algebraic journey".
Typically, the journey involves a lot of unenlightening, mechanical symbol-pushing to eliminate the square roots. Specifically, defining
$$d_i := \sqrt{(x-x_i)^2+(y-y_i)^2}$$
the argument tends to go something like this:
$$\begin{alignat}{2}\quad && d_1 + d_2 &= 2 a \qquad\text{(definition: sum of distances to foci is constant)} \tag{$\star$} \\ \to\quad && (d_1 + d_2)^2 &= (2 a)^2 \\[4pt]
\to\quad && d_1^2 + 2 d_1 d_2 + d_2^2 &= 4 a^2 \\[4pt]
\to\quad && 2 d_1 d_2 &= 4 a^2 - d_1^2 - d_2^2 \\[4pt]
\to\quad && (2d_1d_2)^2 &= ( 4 a^2-d_1^2-d_2)^2 \\[4pt]
\to\quad && 0 &= d_1^4+d_2^4+16a^4-2d_1^2d_2^2-8a^2d_1^2-8a^2d_2^2 \tag{$\star\star$}
\end{alignat}$$
so that $(\star\star)$ contains only even powers of the $d_i$, hence: no radicals. Mission accomplished! Replacing the $d_i$ (in particular, with $(x_i,y_i) = (\pm c,0)$, and defining $b^2 := a^2-c^2$), equation $(\star\star)$ simplifies (see below) to the origin-centered standard form equation we all know and love.
I believe OP is disappointed that, somewhere along the tedious journey from $(\star)$ to $(\star\star)$, we lose sight of $(\star)$.
However, it's still possible to catch a glimpse of $(\star)$ in $(\star\star)$, because $(\star\star)$ factors:
$$(d_1+d_2-2a)(d_1-d_2-2a)(-d_1+d_2-2a)(d_1+d_2+2a) = 0 \tag{$\star\star\star$}$$
(The reader might see a resemblance to Heron's formula in the above.)
Since $(\star)$ is right there in the first factor, the set of points satisfying $(\star\star\star)$ must include those satisfying $(\star)$, the (well, one) definition of the ellipse.
Note that the last factor of $(\star\star\star)$ contributes no points, since presumably $a > 0$ and $d_i \geq 0$.
Interestingly, the middle factors of $(\star\star\star)$ correspond to the relations
$$d_1 - d_2 = 2a \quad\text{or}\quad d_2 - d_1 = 2a \qquad\qquad\text{i.e.,}\quad |d_1-d_2| = 2a$$
which say precisely that the difference of distances to the foci is constant: the (well, one) definition of the hyperbola! (Each factor corresponds to an arm of the ostensible hyperbola.)
Consequently, $(\star\star)$ is simultaneously an ellipse equation and an hyperbola equation! Except, not exactly. The graph of the solution set is only one or the other, as determined by $a$'s relationship to the distance between the foci. To be specific, let's do the simplification hinted at earlier: take $(x_i,y_i) = (\pm c,0)$, so that $(\star\star)$ becomes
$$16 \left(\;a^2 (a^2 - c^2) - x^2(a^2-c^2) - a^2 y^2\;\right) = 0 \qquad\to\qquad \frac{x^2}{a^2} + \frac{y^2}{a^2-c^2} = 1$$
We see, then, that when $a > c$ ---so that the sum of the distances to the foci is bigger than the distance between the foci themselves--- the equation is that of an ellipse; in $(\star\star\star)$, the second and third factors cannot be zero. On the other hand, when $a < c$, the equation is that of an hyperbola; the first factor of $(\star\star\star)$ cannot be zero. (Exploring the degeneracies arising from $a=c$ is left as an exercise to the reader.)
Anyway, my point is this: We can get to $(\star\star)$ from $(\star)$ by plodding through sequence of algebraic steps that obscure the geometry; or, we can get to $(\star\star)$ by "rationalizing" $(\star)$ via multiplication by what one might call its "Heronic conjugate", the three factors of which are geometrically meaningful (although one is inherently extraneous). And we get the hyperbola equation for free ... because it's the same equation!
Kinda neat, that.
a circle equation can be understand very intuitivelyThen think of it the other way around. An ellipse is what a circle looks like if "dilated" along the axes. The canonical circle equation $\,x^2+y^2=r^2\,$ then becomes something like $\,\lambda x^2+\mu y^2=d^2\,$ i.e. the canonical ellipse equation. $\endgroup$ – dxiv Jun 23 '18 at 5:28