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).

(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.

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.

equations of loci page