I'm reading through a derivation of the standard equation of a horizontal hyperbola, and while I can follow the the algebra, I'm hung up on an assumption it makes early on: that the vertices lie in between the foci rather than "outside"–or even directly on–them.

The derivation begins like this:

  • We define a hyperbola as the set of all points $(x, y)$ in the plane such that the absolute value of the difference of the distances between the foci and $(x, y)$ is a fixed value $d$.
  • Consider a hyperbola centered at the origin, whose foci lie on the $x$-axis at $(c,0)$ and $(-c,0)$, with $c>0$.
  • The vertices fall at $(a,0)$ and $(-a,0)$, with $a>0$.

The assumption that $a<c$ is reflected in the diagram (below), and is used explicitly later on. Since we're building from the definition, rather than observing hyperbolas "in the wild", I'd like to know how we know this is in fact the case.


I tried to prove it explicitly by considering the case where $a>c$, but didn't have much success: foci

Because $a>c>0$, $A$ is closer to $C_2$ than it is to $C_1$. So $\left| \overline{C_1A}\right| > \left| \overline{C_2A}\right|$, which means $\left|\overline{C_1A}\right|-\left|\overline{C_2A}\right|>0$. Therefore, $\left|\left|\overline{C_1A}\right|-\left|\overline{C_2A}\right|\right|=\left|\overline{C_1A}\right|-\left|\overline{C_2A}\right|$. Making use of the definition, we see that $\left|\overline{C_1A}\right|-\left|\overline{C_2A}\right|=d$.

We can calculate the above difference in distances in terms of $a$ and $c$, remembering that $a>c$ (and $a>-c$). $$(a-(-c))-(a-c)=d\\ (a+c)-(a-c)=d\\ 2c=d$$ I don't know if that's of any use, though I think it would suggest there's only one hyperbola for any two particular foci, which is certainly not true (though I wouldn't want to have to prove it). I haven't figured anything out for the case where $a=c$, but I supect my time would be better spent asking for clarification than to continue trying to push forward my own.

Any help?

Slight update: I see now (thanks for the help coffeemath!) that if $a>c$, then the difference in distances equals the gap between the two foci, $2c$, and that's true for every value of $a$ along the $x$-axis. So if we wanted to construct such a hyperbola using the above definition, we'd have to use $2c$ for our fixed value $d$. Though I still can't nail it down analytically, it looks like the closest you could get would the limit as $a\to c^-$.

I played around with some graphs to see what happens as $a\to c^-$, and it's cool to see the hyperbola actually approach rays along the $x$-axis. I graphed a series of hyperbolas to see this: $\left(\frac{x^2}{.5}-\frac{y^2}{.5}=1\right)$, $\left(\frac{x^2}{.75}-\frac{y^2}{.25}=1\right)$, $\left(\frac{x^2}{.9}-\frac{y^2}{.1}=1\right)$, $\left(\frac{x^2}{.99}-\frac{y^2}{.01}=1\right)$, $\left(\frac{x^2}{.999}-\frac{y^2}{.001}=1\right)$, $\left(\frac{x^2}{.9999}-\frac{y^2}{.0001}=1\right)$. The blue dots are the foci.







Update: Well, I think I got it nailed down. Kind of drawn out, I had to break things into a couple different cases and make use of the geometric interpretation of the triangle inequality. I'll post it if anyone's interested, otherwise I'm gonna keep moving :)

  • $\begingroup$ Yes, I thought the thing was a degenerate hyperbola as $d \to 2c^-.$ Note also that as $d \to 0^+$ it approaches another degenerate hyperbola, this time where both branches coincide with the $y$ axis. $\endgroup$ – coffeemath Mar 29 '13 at 10:16

Your calculation is OK, and shows that taking $a>c$ leads to $2c=d.$ But in this case, every point on the $x$ axis to the right of $c$ (or to the left of $-c$) will satisfy the locus definition, since the calculation will be (e.g. for $x>c$) $$|x+c|-|x-c|=(x+c)-(x-c)=2c=d.$$ We certainly don't want to call something a hyperbola which includes such rays of the $x$ axis.


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