Theorem AS2
The affine symmetry group of H_{0} is
E_{H}(2) = {t : t(x,y) = (kx,y/k) or t(x,y) = (ky,x/k), k ≠ 0}.
Proof
We begin by noting that the hyperbola has parametric equations
x = t, y = 1/t, t ≠ 0
Now suppose that t is an affine transformation mapping H_{0} to H_{0}.
Since it must map O, the centre of H_{0} to itself, so must have the
form t(x,y) = (ax+by,dx+ey), with aebd ≠ 0.
For any t, P = (t,1/t) is on H_{0} so that t(P) must lie on
H_{0} so that
For all t, (at+b/t)(dt+e/t) = 1, i.e.
adt^{4} + (ae+bd)t^{2} + be = t^{2}.
Since this holds for all t, the coefficients on each side must be
identical, i.e. ad = be = 0, ae+bd = 1. As ad = 0, a or d = 0.
If a = 0, then 1 = 0e+bd, so b ≠ 0, and d = 1/b, and be = 0, so e = 0. This gives the
second form with k = b.
If d = 0, then 1 = ae+b0, so a ≠ 0, and e = 1/a, and be = 0,
so b = 0. This gives the first form with k = a.

