[alogo] 1. Degenerate conics

Conics are described in cartesian coordinates through equations:
                                                               f(x,y) = ax2 + 2bxy + cy2 + 2dx + 2ey + f = 0.
Degenerate conics are those for which the determinant of the corresponding symmetric matrix M is zero:


The matrix serves to represent the conic as a quadratic form in homogeneous coordinates.
Setting  X = (x,y,z) and denoting by Xt the transposed column-vector this is:


The equation f(x,y) =0 results by setting z=1: f(x,y) = F(x,y,1) =0. By the way, this representation justifies the use of "twos" in the first formula for f(x,y).
If f(x,y) is a product of lines  Ax+By+C, A'x+B'y+C', then it is easily seen that the resulting matrix M has determinant 0.
In fact M has then the form:

[0_0] [0_1]

The nature of the conic is determined by its three invariants: J1=a+c, J2=ac-b2 and J3=|M|, last denoting the determinant of the matrix. This is true for every conic. In particular for degenerate ones we have:
1)  J3=0 and J2 >0, then  two imaginary lines                   (example: x2+y2=0).
2)  J3=0 and J2 <0, then  two real intersecting lines         (example: x2-y2=0).
3)  J3=0 and J2 = 0, then two real parallel lines               (example: x2+x=0).

[alogo] 2. Determining the factors

Assuming the degenerate conic in the general form:
                                                               f(x,y) = ax2 + 2bxy + cy2 + 2dx + 2ey + f = 0,
(with |M|=0) the determination of the two lines from the coefficients involves a simple calculation,
which represents f(x,y) as a difference of squares  f(x,y) = m(x,y)2 - n(x,y)2, the two lines being then m(x,y)+n(x,y) and m(x,y)-n(x,y) (see Loney p. 95).
To achieve this multiply f by a:
                                                         a2x2+2abxy+acy2+2adx+2aey+af = 0.
Separate the terms involving y:
                                                         a2x2+2abxy +2adx  = -(acy2+2aey+af),
                                               <=>   a2x2+2ax(by+d)      = -(acy2+2aey+af).
Completing the square on the left:
                                                         a2x2+2ax(by+d)+(by+d)2  = (by+d)2-(acy2+2aey+af),
                                                 <=> (ax + by + d)2  =  y2[b2-ac]+2y[bd-ae]+[d2-af].
The condition |M|=0 is easily seen to be equivalent to the vanishing of the discriminant of the right-side quadratic polynomial in y. In this case the right side can be written in the form:
                                                         y2[b2-ac]+2y[bd-ae]+[d2-af] = (uy+v)2,
and the two required lines are                                                 (ax + by + d) + (uy + v) = 0 ,    (ax + by + d) - (uy + v) = 0.
u and v must satisfy  u2 = b2-ac,   v2 = d2-af   and   uv = bd-ae, later allowing the right selection of signs for u, v in terms of the given coefficients.

[alogo] 3. Example

As an example (Loney p.97) resolve   12x2+7xy-10y2+13x+45y-35 = 0  in two linear factors.
Doing the computations of the previous section it turns out that  |u|=23/2, |v| = 43/2 and  uv<0.
Thus the signs must be opposite, giving   u=23/2  and  v=-43/2  and consequently:
                                     (ax+by+d)+(uy+v) = (12x+(7/2)y+(13/2)) + ((23/2)y - (43/2)) = 0,
giving the factor           4x + 5y - 5 = 0.
Analogously                  (ax+by+d)-(uy+v) = (12x+(7/2)y+(13/2)) - ((23/2)y - (43/2)) = 0,
giving the factor           3x - 7y + 7 = 0.

[alogo] 4. Varying the constant

Starting with a degenerate conic and varying the constant f we obtain in general non-degenerate conics:
                                f(x,y) = ax2 + 2bxy + cy2 + 2dx + 2ey + f = k.
These are level-curves of the function f(x,y) and if the degenerate equation resolves to two intersecting real lines then the corresponding level surfaces are hyperbolas having these lines as asymptotes.

[0_0] [0_1] [0_2]
[1_0] [1_1] [1_2]

This results immediately from the fact that an equation of the form  (ax+by+c)(a'x+by'+c') = k (k non-zero) represents a hyperbola with respect to its asymptotes which are the two lines on the left side (see HyperbolaAsymptotics.html ).

There is a similar behavior for the other cases of degenerate conics. For instance in the case the degnerate conic represents two parallel lines (J2=0 in section 1) the level curves are again imaginary or real parallel lines. For example if f(x,y)=x2+x, depending on the discriminant of  x2+x-k, the equation f(x,y)=k represents two imaginary or two real parallel lines. Similarly in the case of J2>0 as is for f(x,y) = x2+y2, setting f(x,y) = k we obtain real or imaginary circles.

See Also



Loney, S. L. Coordinate Geometry. Delhi, AITBS Publishers, 1991, p. 97.

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