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 $a<c$ we can define $b^2=c^2-a^2$, yielding $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1.\qquad\text{(H)}$$ I also understand (by simply "looking" at the graphs of these two cases) that the former, (E), corresponds to an ellipse, and the latter, (H), to a hyperbola.

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 $$\sqrt{(x+c)^2+y^2} -\sqrt{(x-c)^2-y^2}=\pm 2a\qquad\text{(C)}$$ $$\frac{4xc}{\sqrt{(x+c)^2+y^2} +\sqrt{(x-c)^2-y^2}}=\pm 2a$$ $$\sqrt{(x+c)^2+y^2} +\sqrt{(x-c)^2-y^2}=\pm \frac{2cx}{a}$$ $$2\sqrt{(x+c)^2+y^2}=\pm 2\left(a+ \frac{xc}{a}\right)$$ $$x^2+2cx+c^2+y^2=a^2+ 2cx+ \frac{c^2x^2}{a^2}$$ $$\frac{x^2}{a^2}+\frac{y^2}{a^2-c^2}=1\qquad\text{(A)}$$

  • $\begingroup$ If I had to guess at an answer, it would run something like: (a) Squaring, which is done in several places in the derivations, is the problem; (b) I should not expect it to be obvious (not, perhaps, without some algebra) how the cases considered in "restoring" a distinction map to the distinction between the definitions (e.g., that $a>c$ is equivalent to "the sum" as opposed to "the difference"); (c) Furthermore, whenever one "collapses" information in this way, the correspondence between what restores it, at the end, and how it was stated, at the beginning, will not be obvious in general. $\endgroup$
    – orome
    Oct 7, 2011 at 22:40
  • $\begingroup$ And: (d) There's a name for what I am calling "loosing information" or "collapsing", neither of which terms is quite correct. $\endgroup$
    – orome
    Oct 7, 2011 at 22:40
  • 1
    $\begingroup$ If you have a line with equation $y=kx$, and increasing and decreasing lines would have different names in your languages, what would be the mystery about the fact that $y=kx$ is an equation for both types of lines? $\endgroup$
    – Phira
    Oct 7, 2011 at 22:45
  • $\begingroup$ @Phira: That's a very good point. But the core of the question remains: how we got from expressions that distinguished between "increasing" and "decreasing" to one that covers both — and the mapping between what we do to "split" that result back in two, and the way the split was originally stated. $\endgroup$
    – orome
    Oct 7, 2011 at 22:54
  • $\begingroup$ Well, there's a mistake in your argument: if you substitute $b^2 = a^2 - c^2$ into your equation (A) above, you do not get either (E) or (H) depending on the case; rather, you get (E) no matter what, but $b^2$ may be negative. Instead, what you are actually doing is taking $b^2 = |a^2 - c^2|$, and there's your answer -- the absolute value is collapsing the plus and minus cases together. $\endgroup$ Oct 7, 2011 at 22:58

2 Answers 2


This is the difference between real numbers, some of which are not squares of other real numbers, and complex numbers, all of which are squares of other complex numbers. Is the $a^2 - c^2$ in the denominator a negative number (i.e. one that is not a square) or a positive number (one that is a square) (the case where it's $0$, which also is a square, is not what we're examining here).

If you work with complex numbers rather than real numbers, then there's no difference between ellipses and hyperbolas.

(If you go a step further and work with projective spaces rather than affine spaces, then there's no difference between those and parabolas.)

  • $\begingroup$ Ah, I expected it would go that deep. So let me see if I have this much right: Though I thought of (A) as "one thing" it actually embodies two (well, three) distinct cases which must be distinguished. $\endgroup$
    – orome
    Oct 7, 2011 at 23:14
  • $\begingroup$ Three cases must be distinguished when you work with real numbers, and two when you work with complex numbers. $\endgroup$ Oct 7, 2011 at 23:19
  • $\begingroup$ Is it the case that (i) the initial formulations, (B) and (C), are equivalent (if one did the algebra) to $a>c$ and $a<c$ respectively, and that (ii) that distinction is "collapsed" (wrong word, I'm sure) by squaring (so that it has to be explicitly reconsidered later)? $\endgroup$
    – orome
    Oct 7, 2011 at 23:24

Notice that E and H have two constants, A has three. So A forms a superset of E and H and you should see graph of A to study how E and H are fitted together to form A in a certain composite geometrical pattern/relationship.

Because a certain manipulation on A can bring it into form of E or H, we can see that a three parameter plot contains subset graphs of E and H. In fact A is a set of orthogonally intersecting confocal E and H.


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