proof of the classification theorem Theorem from Linear Algebra If M is a real symmetric 2x2 matrix, then there is an orthogonal 2x2 matrix P such that PTMP is the diagonal matrix D = diag(λ,μ). Remarks (1) As P is orthogonal, det(P) = ±1, so that λμ = det (D) = (det(P))2det(M) = det(M). (2) λ = μ = 0 if and only if D is the zero matrix, and then, as M = PDPT, M is also zero. If Q is an orthogonal matrix, then the transformation t(x) = Qx is euclidean, and is either reflection in a line through the origin or a rotation about the origin. If we write X = Qx, then, as QT = Q-1, we have x = PX, where P = QT, so that P is also orthogonal. The Classification Theorem The locus C = {(x,y) : f(x,y) = Ax2+Bxy+Cy2+Fx+Gy+H = 0}, with A, B, C not all zero is a circle, a plane conic, or a degenerate plane conic. If C has at least two points, and does not contain three collinear points, then it is a circle or a plane conic and is a circle or an ellipse if 4AC - B2 > 0, a parabola if 4AC - B2 = 0, a hyperbola if 4AC - B2 < 0. Note. It is a circle if and only if B = 0 and A = C. Proof The proof consists of three stages: (1) the application of a transformation t so that t(C) has an equation with no xy term, (2) the application of a translation to remove linear terms as far as possible, and (3) the classification of the simplified equations. (1) As noted, we can write the equation in matrix form xTMx + kx + H = 0. By the theorem from linear algebra (above), we can find an orthogonal matrix P such that PTMP = D = diag(λ,μ). If we apply the transformation t, with X = t(x) = PTx, then x = PX, so that t(C) has equation XTDX + kPX + H = 0, or in terms of XY-coordinates: λX2 + μY2 + αX + βY + H = 0 By an earlier remark, since A, B, C are not all zero, M is non-zero, so λ and μ are not both 0. (2) In the XY-equation, if λ is non-zero, we can rewite the terms involving X in the form λ(X + α/2λ)2 - (α)2/4λ Similarly, if μ is non-zero, the terms involving Y can be written as μ(Y+ β/2μ)2 - (β)2/4μ Provided that λ and μ are non-zero, we can now apply the translation u given by (x,y) = u(X,Y) = (X+α/2λ,Y+β/2μ) to get u(t(C)) with equation λx2 + μy2 = κ. Now suppose that μ is zero. Then λ is non-zero, and we can use u(X,Y) = (X+α/2λ,Y) to transform the locus to that with an equation of the form (after dividing by λ) x2 = γy + δ. If γ turns out to be non-zero, then repalcing y by (y +δ/γ) we get the form x2 = γy. If γ is zero, then the equation is already of the form x2 = δ. Finally, suppose that λ is zero. By a similar analysis, the locus t(C) may be translated to a locus with equation either y2 = γx, with γ ≠ 0, or y2 = δ. (3) We shall consider the first and third cases in (2), the second is similar to the third. The case λx2 + μy2 = κ, with λ and μ non-zero. κ = 0. If λ and μ have the same sign then it is easy to see that the only solution is (0,0). We have a one point degenerate conic. If λ and μ have opposite signs, then the equation has the form λ(x2 - b2y2) = 0, where b2 = -μ/λ. This gives the line pair x = by and x = -by. κ ≠ 0. We may divide through by κ so the equation becomes λ'x2 + μ'y2 = 1. It is easy to see that, if λ ad μ are positive, then we have the circle or ellipse x2/a2 + y2/b2 = 1, with a2= 1/λ and b2 = 1/μ. If λ and μ have opposite signs, then we have a hyperbola x2/a2 - y2/b2 = ±1. Finally, if λ and μ are negative, then there are no solutions, i.e. the locus is the empty set, a degenerate conic. the cases y2 = γx (γ ≠ 0) and y2 = δ The first is clearly a parabola, in standard form if γ > 0, and the mirror image otherwise. y2 = δ, δ < 0. There are no solutions, so we get the empty set. y2 = δ, δ = 0. We get the line y = 0. a degenerate conic. y2 = δ, δ > 0. Then we have the line pair, y = b and y = -b, where b2 = δ. This is again a degenerate conic. Summing up, an equation quadratic in x and y always gives a circle, a plane conic or a degenerate conic. In the course of our analysis, we saw that the locus is a circle, a conic, or is degenerate. In the last case, it consists of at most a single point, or contains a line. Thus, if a locus contains at least two points, but no three collinear points, it must be a circle or conic. Finally, from the analysis, we can see that, if the locus is non-degenerate, then we have λ, μ of the same sign, so λμ > 0, and we get a circle or an ellipse, we get a circle precisely when λ = μ. λ, μ of opposite signs, so λμ < 0, and we get a hyperbola, or one of λ and μ equal to zero, so λμ = 0 and we get a parabola. The last clause follows since λμ = det(diag(λ,μ)) = det(M) = AC - ¼B2. To see why the circle must have B = 0 and A = C, observe that the circle arises only when λ = μ, i.e. when the matrix D is diag(λ,λ) = λI. But D = PTMP, with P orthogonal, so M = PDPT = λI. Then B = 0 and A = C = λ.
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