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
P^{T}MP is the diagonal matrix D = diag(λ,μ).
Remarks
(1) As P is orthogonal, det(P) = ±1, so that λμ = det (D) = (det(P))^{2}det(M) = det(M).
(2) λ = μ = 0 if and only if D is the zero matrix, and then, as M = PDP^{T}, 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 Q^{T} = Q^{1}, we have x = PX, where P = Q^{T}, so that P is also orthogonal.
The Classification Theorem
The locus C = {(x,y) : f(x,y) = Ax^{2}+Bxy+Cy^{2}+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  B^{2} > 0,
 a parabola if 4AC  B^{2} = 0,
 a hyperbola if 4AC  B^{2} < 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 x^{T}Mx + kx + H = 0.
By the theorem from linear algebra (above), we can find an orthogonal matrix P such that
P^{T}MP = D = diag(λ,μ). If we apply the transformation t, with
X = t(x) = P^{T}x, then x = PX,
so that t(C) has equation X^{T}DX + kPX + H = 0,
or in terms of XYcoordinates:
λX^{2} + μY^{2} + αX + βY + H = 0
By an earlier remark, since A, B, C are not all zero, M is nonzero, so λ and μ are not both 0.
(2) In the XYequation, if λ is nonzero, we can rewite the terms involving X in the form
λ(X + α/2λ)^{2}  (α)^{2}/4λ
Similarly, if μ is nonzero, the terms involving Y can be written as
μ(Y+ β/2μ)^{2}  (β)^{2}/4μ
Provided that λ and μ are nonzero, 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
λx^{2} + μy^{2} = κ.
Now suppose that μ is zero. Then λ is nonzero, 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)
x^{2} = γy + δ.
If γ turns out to be nonzero, then repalcing y by (y +δ/γ) we get the form x^{2} = γy.
If γ is zero, then the equation is already of the form x^{2} = δ.
Finally, suppose that λ is zero. By a similar analysis, the locus t(C) may be translated to
a locus with equation either y^{2} = γx, with γ ≠ 0, or y^{2} = δ.
(3) We shall consider the first and third cases in (2), the second is similar to the third.
The case λx^{2} + μy^{2} = κ, with λ and μ nonzero.
 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(x^{2}  b^{2}y^{2}) = 0,
where b^{2} = 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'x^{2} + μ'y^{2} = 1.
It is easy to see that, if λ ad μ are positive, then we have the circle or ellipse
x^{2}/a^{2} + y^{2}/b^{2} = 1, with a^{2}= 1/λ and b^{2} = 1/μ.
If λ and μ have opposite signs, then we have a hyperbola
x^{2}/a^{2}  y^{2}/b^{2} = ±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 y^{2} = γx (γ ≠ 0) and y^{2} = δ
The first is clearly a parabola, in standard form if γ > 0, and the mirror image otherwise.
 y^{2} = δ, δ < 0.
There are no solutions, so we get the empty set.
 y^{2} = δ, δ = 0.
We get the line y = 0. a degenerate conic.
 y^{2} = δ, δ > 0.
Then we have the line pair, y = b and y = b, where b^{2} = 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 nondegenerate, 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  ¼B^{2}.
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 = P^{T}MP,
with P orthogonal,
so M = PDP^{T} = λI. Then B = 0 and A = C = λ.

