proof of theorem AC1

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 transformation t,
(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'), where
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
cases cannot arise.

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.

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