proof of theorem AC1
If C has equation f(x,y) = 0, with f quadratic in x and y, and C is
not degenerate, then, for any affine
(1) t(C) has equation F(x,y) = 0, with F quadratic in x and y, and
(2) t(C) is not degenerate.
Recall that an affine transformation maps x to Ax + b, where A is an invertible 2x2 matrix
and b is a vector, and that the inverse is also an affine transformation. Working in terms
of coordinates, if t(x,y) = (X,Y), then (x,y) = t-1(X,Y) = (αX+βY+γ,α'X+β'Y+γ'),
α,β,γ,α',β',γ' are real numbers.
(1) Observe that
P(X,Y)εt(C) if and only if t-1(X,Y)εC
i.e. x = αX+βY+γ, and y = α'X+β'Y+γ' satisfy the equation of C,
i.e. f(αX+βY+γ,α'X+β'Y+γ') = 0.
Thus, t(C) has equation F(X,Y) = 0, with F at most quadratic in X and Y.
(2) Suppose that F is quadratic in X and Y, but that t(C) is degenerate.
Then t(C) is empty, consists of a single point, or of one or two lines.
The affine transformation t-1 maps any such set to another of the same type, so
we would have C = t-1(t(C)) degenerate. This would contradict our hypothesis
that C is not degenerate. Thus t(C) is not degenerate.
There remains the possibility that F is not actually quadratic in X and Y, i.e that
the coefficients of X2, Y2 and XY vanish. Then F is at most linear, so t(C) is a
line (if X or Y is present), the empty set (if the coefficients of X and Y also vanish),
or the whole plane if F(X,Y) is identically zero. Since C is not degenerate, these
Since an affine transformation has an invertible matrix A, we can show by algebra
that the pathological possibilities considered in the final paragraph do not occur.