loci defined by quadratic equations
We know that every plane conic, defined by the focus-directrix property, has an equation
which is quadratic in x and y, i.e. of the form
f(x,y) = Ax2+Bxy+Cy2+Fx+Gy+H = 0,
where A, B and C are not all zero. But such equations can give other kinds of curve. We give
some examples. Later, we shall see that the locus f(x,y) = 0 is always either a plane conic,
a circle, or one of these exceptional types, called degenerate plane conics.
type | equation | locus |
the empty set |
x2+y2+1=0 |
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a single point |
x2+y2 = 0 |
the point {(0,0)} |
a single line |
x2 = 0 |
the line x = 0 |
a pair of lines |
xy = 0 |
the lines x = 0 and y = 0 |
Note that a degenerate plane conic either has at most one point or contains a line. It is
useful to characterize the loci which contain a line.
Lemma
If the locus C = {(x,y) : f(x,y) = Ax2+Bxy+Cy2+Fx+Gy+H = 0} contains three collinear points,
the it contains the entire line containing these points.
Proof
Suppose first that L is the line y = mx + c. Then, where L meets C, f(x,mx+c) = 0. This gives
an equation of the form px2+qx+r = 0. This either has p=q=r=0 or has at most two real roots.
Thus, if L meets C in three points, we must lhave the former case, so all points of the line L lie on
the locus C.
The remaining cases, where L has equation x = k, are similar.
Observe that, when the locus contains a line ax+by+c=0, f(x,y) must have a factor (ax+by+c).
Since f(x,y) is quadratic, the other factor must also be linear, i.e. of the form (dx+ey+f). Thus,
the locus consists of the two lines ax+by+c = 0 and dx+ey+f = 0. Of course, these lines may
coincide, giving a locus consisting of a single line.
the classification of loci
Although it is possible to classify the loci using xy-coordinates, a more elegant approach uses
vectors and matrices. This does, however require some knowledge
of linear algebra.
Observe that f(x,y) = Ax2+Bxy+Cy2+Fx+Gy+H may be written in matrix form as
xTMx + kx + H,
where xT = (x,y), k = (F,G), and M is the symmetric matrix given on the right
Note that, as A, B and C are not all zero, M is a non-zero matrix.
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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 of the classification theorem Note. It is lengthy!
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