<s>[INST] A geometric reason why the square of the focal length of a hyperbola is equal to the sum of the squares of the axes? When I teach conics, I give a simple geometric argument for the Pythagorean relation for ellipses. (By "Pythagorean relation" I mean the fact that the square of the focal length $c$ is the difference of the squares of the semimajor and semiminor axes, i.e. $c^2=|a^2-b^2|$ for the ellipse defined by $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.) The picture argument goes as follows: just draw the obvious isosceles triangle from the two foci to one of the two points at the end of the minor axis. We see two right triangles; in each, one leg is the focal length $c$ and the other leg is the semiminor axis, while the hypotenuse is the semimajor axis. The result follows from the Pythagorean theorem applied to this right triangle.
But when it comes to the Pythagorean relation for hyperbolas, I can't think of an equally convincing picture to draw. Instead I have to give a rather inelegant algebraic argument for the relation $c^2=a^2+b^2$ (for the hyperbola defined by $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, with focal length $c$), directly from the definition of the hyperbola. The messy details are below.*
Now, obviously there is an asymmetry between the argument for the ellipse and the argument for the hyperbola: the ellipse argument is almost trivial and doesn't involve working straight from the locus definition, but the hyperbola argument is tedious and does require working straight from the definition. So, my question:

Is there an equally persuasive picture I could draw to convince high
  school students of the Pythagorean relation for hyperbolas -- without having to get technical with points at infinity, say?

The natural triangle to draw is the right triangle with legs from the center to a vertex and from that vertex to an asymptote, which has legs of length $a$ and $b$; but then the problem becomes why the length along the asymptote should be equal to $c$. I doubt there is a purely (Euclidean) geometrical reason because this triangle lies on the asymptote, not on the hyperbola itself; I suspect some sort of limiting procedure will be required.

[*] The tedious algebraic details:

The foci lie at $(\pm c,0)$.
  Say $a>b$. Then hyperbola is the locus of points $(x,y)$ such that 
$$\left|\sqrt{(x+c)^2+y^2}-\sqrt{(x-c)^2+y^2}\right|=2a$$
Assume for simplicity the expression between the absolute value bars
  is positive. (The algebra is even more tedious if we don't make such a
  simplifying assumption.) Then moving the right root to the other side
  and squaring gives
$$(x+c)^2+y^2=4a^2+(x-c)^2+y^2+4a\sqrt{(x-c)^2+y^2}$$
Simplifying gives
$$xc-a^2=a\sqrt{(x-c)^2+y^2}$$
and upon squaring again we have
$$x^2c^2-2xca^2+a^4=a^2(x^2-2xc+c^2+y^2)=a^2x^2-2xca^2+c^2a^2+a^2y^2$$
which reduces to
$$x^2(c^2-a^2)-y^2a^2=a^2(c^2-a^2)$$
Defining $b^2:=c^2-a^2$ brings the equation into the form
  $x^2/a^2+y^2/b^2=1$, and the Pythagorean relation follows from the
  definition of $b$.

 [/INST] Since the hyperbola is defined as the locus where the difference of the two distances is constant, I don't see any analogous argument. But certainly a limit argument does work: As $(x,y)\to \infty$ (with $x>0$), we have $\left|\frac yx\right|\to\frac ba$, and so
\begin{align*}
 \sqrt{(x+c)^2+y^2}-\sqrt{(x-c)^2+y^2}&=\\\left(\sqrt{(x+c)^2+y^2}-\sqrt{(x-c)^2+y^2}\right)&\frac{\sqrt{(x+c)^2+y^2}+\sqrt{(x-c)^2+y^2}}{\sqrt{(x+c)^2+y^2}+\sqrt{(x-c)^2+y^2}}\\
&=\frac{4c}{\sqrt{(x+c)^2+y^2}+\sqrt{(x-c)^2+y^2}}\\&=\frac{4c}{\sqrt{(1+c/x)^2+(y/x)^2}+\sqrt{(1-c/x)^2+(y/x)^2}}\\
&\to \frac{2ac}{\sqrt{a^2+b^2}}\,.
\end{align*}
</s>