Theorem AS3
The affine symmetry group of P_{0} is
E_{P}(2) = {t : t(x,y) = (e^{2}x+2efy+f^{2},ey+f), e ≠ 0}.
Proof
We begin by noting that the parabola has parametric equations
x = t^{2}, y = t, t real
Now suppose that t is an affine transformation mapping
P_{0} to P_{0}.
As t is affine, t(x,y) = (ax+by+c,dx+ef+f), with aebd ≠ 0.
For any t, P = (t^{2},t) is on P_{0} so that t(P) must lie on
P_{0} so that
For all t, (dt^{2}+et+f)^{2} = at^{2}+bt+c.
Since this holds for all t, the coefficients on each side must be
identical. Comparing coefficients of t^{4}, d = 0. Using this, we get
e^{2}t^{2}+2eft+f^{2} =at^{2}+bt+c, so that
a = e^{2}, b = 2ef, and c = f^{2}.
Finally, aebd ≠ 0, so e ≠ 0.

