## Quadratic Forms, Hyperbolic Mirror

### Hyperbolic Mirror

Recall that an elliptical mirror reflects light from one focus to the other.
In contrast, a hyperbolic mirror reflects light from one focus *away* from the other.
After the light has been reflected,
pull it back through the mirror, and it converges at the other focus.
Put a light bulb at the front focus,
and it looks like it is shining from the back focus, behind the mirror.
These mirrors are used in x-ray telescopes,
where grazing angles must be extremely shallow.

Once again, the angle of incidence must equal the angle of reflection.
The incident angle is determined by the tangent line
at p, and the line joining p to the front focus.
The reflected angle is determined by the tangent line
and the line joining p to the back focus.
These angles must agree, for every point p on the hyperbola.

If you haven't done so already,
I'll ask you to review the
previous proof.
This one is practically the same.
In fact the proofs are identical up
to this equation.

(v2-u2)xy +uv(x2-y2) - f2uv = 0

At this point we need a parameterization for the curve,
and sine and cosine just won't do.
Instead, use secant and tangent.
For notational convenience, let s = sec(θ) and let t = tan(θ).
Now x = as, y = bt, u = ast, and v = bss.
Substitute and expand, and clear out the common factor abs3t,
yielding the following equation.

a2×(s2-t2) + b2×(s2-t2) - f2 = 0

Since s2-t2 = 1, and f2 = a2+b2, this equation is always true.
That completes the proof.