Theorem AS3
The affine symmetry group of P0 is
EP(2) = {t : t(x,y) = (e2x+2efy+f2,ey+f), e ≠ 0}.
Proof
We begin by noting that the parabola has parametric equations
x = t2, y = t, t real
Now suppose that t is an affine transformation mapping
P0 to P0.
As t is affine, t(x,y) = (ax+by+c,dx+ef+f), with ae-bd ≠ 0.
For any t, P = (t2,t) is on P0 so that t(P) must lie on
P0 so that
For all t, (dt2+et+f)2 = at2+bt+c.
Since this holds for all t, the coefficients on each side must be
identical. Comparing coefficients of t4, d = 0. Using this, we get
e2t2+2eft+f2 =at2+bt+c, so that
a = e2, b = 2ef, and c = f2.
Finally, ae-bd ≠ 0, so e ≠ 0.
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