I understand how to get from the definitions of a hyperbola — as the set of all points on a plane such that the absolute value of the difference between the distances to two foci at $(-c,0)$ and $(c,0)$ is constant, $2a$ — and an ellipse — as the set of all points for which the sum of these distances is constant — to the equation $\frac{x^2}{a^2}+\frac{y^2}{a^2-c^2}=1,\qquad\text{(A)}$ and I also understand if $a>c$ we can define $b^2=a^2-c^2$, yielding $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\qquad\text{(E)}$ and if $a
However, it seems that the general equation, (A), obscures a distinction that was specified in the definitions: if two different definitions produce the same equation, hasn't something been lost in the process? At some point the derivations must have taken a step that wiped out the some feature of the equations — (B) and (C) below — that distinguishes the definitions. I see that one can "restore" a distinction by considering the relationship between $a$ and $b$, as above, but how that distinction maps back to the distinction between the definitions is obscure to me.
What steps in the derivations of (A) from the respective definitions, (B) and (C), is obscuring information that distinguishes those definitions? Is something going on here that can be generalized?
The derivations I'm referring to are pretty standard, they appear in many texts and also in several places on this site, but are repeated here for reference.
From Spivak's Calculus (p. 66): a point $(x,y)$ is on an ellipse if and only if $\sqrt{(x+c)^2+y^2} +\sqrt{(x-c)^2+y^2}= 2a\qquad\text{(B)}$ $\sqrt{(x+c)^2+y^2}= 2a-\sqrt{(x-c)^2+y^2}$ $x^2+2cx+c^2+y^2=4a^2-4a\sqrt{(x-c)^2+y^2}+x^2-2cx+c^2+y^2$ $4(cx-a^2)=-4a\sqrt{(x-c)^2+y^2}$ $c^2x^2-2cxa^2+a^4=a^2(x^2-2cx+c^2+y^2)$ $(c^2-a^2)x^2-a^2y^2=a^2(c^2-a^2)$ $\frac{x^2}{a^2}+\frac{y^2}{a^2-c^2}=1\qquad\text{(A)}$
From a related post: a point $(x,y)$ is on a hyperbola if and only if $\sqrt{(x+c)^2+y^2} -\sqrt{(x-c)^2-y^2}=\pm 2a\qquad\text{(C)}$ $\frac{4xc}{\sqrt{(x+c)^2+y^2} +\sqrt{(x-c)^2-y^2}}=\pm 2a$ $\sqrt{(x+c)^2+y^2} +\sqrt{(x-c)^2-y^2}=\pm \frac{2cx}{a}$ $2\sqrt{(x+c)^2+y^2}=\pm 2\left(a+ \frac{xc}{a}\right)$ $x^2+2cx+c^2+y^2=a^2+ 2cx+ \frac{c^2x^2}{a^2}$ $\frac{x^2}{a^2}+\frac{y^2}{a^2-c^2}=1\qquad\text{(A)}$