The Singular Point of an Algebraic Cubic

Marco Paluszny, Richard Patterson and Francisco Tovar

Abstract. Given a singular algebraic cubic, tangent at two points to prescribed lines, formulas are given for the coordinates of its singular point. The coordinates of the singular point are expressed as rational functions of degree at most five in the coefficients of the cubic.

§1. Introduction

Singular algebraic cubics and rational Bézier curves of degree three are the same family of curves, which is widely used in Computer Aided Geometric Design (CAGD).

For curves that have already been parametrized, such as Bézier curves, the problem of determining the singular point(s) has been of interest in CAGD. The problem was studied for polynomial curves by Su Buchin and Dingyuan [9]. Stone and De Rose [8] did a geometric analysis for polynomial cubics. For rational cubics a solution is given in Patterson [3], another geometric analysis was provided by Pottmann and DeRose [4]. The problem of locating cusps in rational curves is discussed by Manocha and Canny in [1]. The problem of approximating double points was solved with an interactive procedure by Sederberg in [7].

But all of the papers mentioned above for finding double points require a parametrization to be given first, and yet given a singular algebraic curve, the usual method for finding a parametrization requires the double point to be known in advance.

The problem of determining the singularity of an algebraic curve of degree n, with rational coefficients, was also studied by Sakkalis and Farouki in [5]. Their method consists in isolating the coordinates of singular points in intervals and approximating them numerically. One of their formulas, in the case of cubics, can be extended to yield formulas similar to the ones given here, but of higher degree.

The goal of this paper is to provide explicit formulas for the coordinates of the singular point of a given implicit singular cubic in terms of its coefficients.

Mathematical Methods for Curves and Surfaces II 1 Morten Dæhlen, Tom Lyche, Larry L. Schumaker (eds.), pp. 1–8. Copyright o^{c} 1998 by Vanderbilt University Press, Nashville, TN. ISBN 1-xxxxx-xxx-x. All rights of reproduction in any form reserved.

§2. Expression of the Cubic

A point P (x, y, z) is a singular point of the curve F (x, y, z) = 0 if F_{x}(P) = F_{y}(P) = F_{z}(P) = 0, where F_{x}, F_{y} and F_{z} denote the partial derivatives of F. Otherwise, the point P is called regular. Every cubic is known to have at most one singular point. At a regular point P, the tangent line is given by F_{x}(P)x + F_{y}(P)y + F_{z}(P)z = 0.

Suppose that P_{0}(x_{0}, y_{0}, z_{0}) and P_{2}(x_{2}, y_{2}, z_{2}) are two regular points of F = 0 and P_{1}(x_{1}, y_{1}, z_{1}) is the intersection point of the tangent lines at P_{0} and P_{2}.

Let (s, t, u) be the new system of homogeneous coordinates given by

(x, y, z) = (s, t, u) ( \begin{matrix}x_{0} & y_{0} & z_{0} \\ x_{1} & y_{1} & z_{1} \\ x_{2} & y_{2} & z_{2} \end{matrix} ) .

We then have barycentric coordinates if we also require that s + t + u = 1, but for our purposes this is only necessary in Section 4. In the new coordinate system the expression of the cubic is

F (s, t, u) = as^{2}u + bsu^{2} − cst^{2} − dt^{2}u + estu − ft^{3}.  (2.1)

Here the minus signs are incorporated in some coefficients to make later formulas simpler.

If a = 0 in (2.1), then P_{0} is a singularity. Likewise, if b = 0, then P_{2} is a singularity. Thus the problem of finding the singular point in these cases is solved, and consequently in the rest of the paper we make the standing assumption that a and b are different from zero. With this assumption, a singular point of F = 0 can not lie on any of the lines s = 0, t = 0, or u = 0. This is because each of these lines already has two points in common with the curve, so if one of the lines also passed through a singular point it would then have four-point contact with the curve, which is impossible by Bezout’s theorem (see [10] p. 59).

§3. Singular Point of the Cubic

In this section we will see how to determine if a cubic has a singular point, and find formulas for the singular point when there is one. The following formula in the coefficients of F can be used to determine singularity.

Theorem 3.1. The cubic F (s, t, u) = 0 given by equation (2.1) is singular or reducible if and only if Δ(a, b, c, d, e, f) = 0, where

Δ = − 16a^{2}cd^{3}ef − 16a^{2}c^{2}d^{4} − 36ab^{2}cef^{3} + 16b^{3}c^{3}f^{2}  (3.1)
+ 16a^{3}d^{3}f^{2} − 8bc^{3}d^{2}e^{2} − 8bc^{2}de^{3}f − 24ab^{2}c^{2}df^{2}
− 36a^{2}bdef^{3} + 8a^{2}d^{2}e^{2}f^{2} − 8ac^{2}d^{3}e^{2} + 27a^{2}b^{2}f^{4}
− 24a^{2}bcd^{2}f^{2} + 32abc^{3}d^{3} + 64abc^{2}d^{2}ef + 46abcde^{2}f^{2}
− c^{2}d^{2}e^{4} − cde^{5}f + cbe^{4}f^{2} − 16b^{2}c^{3}def − 8acd^{2}e^{3}f
− 16b^{2}d^{2}c^{4} + ade^{4}f^{2} − abe^{3}f^{3} + 8b^{2}c^{2}e^{2}f^{2}.

Proof: See [6] pages 189-192 or the discussion in [2].

The following lemma gives a sufficient condition for the cubic F to be reducible.

Lemma 3.2. If the coefficients c, d and f of the cubic F of (2.1) are different from zero and satisfy the condition \frac{a}{c} = \frac{b}{d} = \frac{e}{f}, then the cubic is reducible.

Proof: Let α be the ratio in the Lemma. Substituting a = cα, b = dα and e = fα we find F = (αsu − t^{2})(cs + ft + du).

Theorem 3.3. If the cubic F (s, t, u) = 0 given by (2.1) is singular and irreducible, then the homogeneous coordinates of the singular point (\overline{s}, 1, \overline{u}) are given by the following formulas:

1) If S_{1} and U_{1} are not zero,

\overline{s} = \frac{G_{1}(a, b, c, d, e, f)}{S_{1}(a, b, c, d, e, f)}  (3.2)

\overline{u} = \frac{G_{1}(a, b, c, d, e, f)}{U_{1}(a, b, c, d, e, f)}  (3.3)

2) If S_{1} = 0 or U_{1} = 0,

\overline{s} = \frac{G_{2}(a, b, c, d, e, f)}{S_{2}(a, b, c, d, e, f)}  (3.4)

\overline{u} = \frac{G_{2}(a, b, c, d, e, f)}{U_{2}(a, b, c, d, e, f)}  (3.5)

The explicit expressions of G_{1}, S_{1}, U_{1}, G_{2}, S_{2} and U_{2} are

G_{1}(a, b, c, d, e, f) = 7cde^{2}f − 8adef^{2} + 8c^{2}d^{2}e − 4acd^{2}f
− 8bcef^{2} + 9abf^{3} − 4bc^{2}df  (3.6)
S_{1}(a, b, c, d, e, f) = −12acdfe + 12a^{2}df^{2} − 8ac^{2}d^{2} + ce^{3}f
− ae^{2}f^{2} + 2c^{2}de^{2} + 4bc^{2}ef − 6abcf^{2} + 8bc^{3}d (3.7)
U_{1}(a, b, c, d, e, f) = S_{1}(b, a, d, c, e, f)
G_{2}(a, b, c, d, e, f) = f(4cd − ef)
S_{2}(a, b, c, d, e, f) = 2(3af^{2} − 2cef − 4c^{2}d)  (3.8)
U_{2}(a, b, c, d, e, f) = S_{2}(b, a, d, c, e, f)

The proof uses the following Lemma 3.4.

1) If the cubic F is irreducible and S_{1} = 0 or U_{1} = 0, then G_{1} = 0.

2) If G_{1} = 0 and either S_{2} = 0 or U_{2} = 0, then F is reducible.

The consequence of this lemma is that the formulas given in (3.2), (3.3), (3.4) and (3.5) cover all the possible situations. The proof is deferred until after the proof of Theorem 3.3.

Proof of Theorem 3.3. First assume that S_{1} and U_{1} are not zero.Then we must show that the point (\overline{s}, 1, \overline{u}) = (\frac{G_{1}}{S_{1}} , 1, \frac{G_{1}}{U_{1}} ) is the singularity of the cubic. A calculation shows F_{s}(\overline{s}, 1, \overline{u}) = ΔΩ_{1}, F_{t}(\overline{s}, 1, \overline{u}) = ΔΩ_{2}, F_{u}(\overline{s}, 1, \overline{u}) = ΔΩ_{3}, where the Ω_{i} are rational functions of the coefficients a, b, c, d, e and f whose denominators do not vanish because they are multiples of S_{1} and U_{1}. So since Δ = 0, it follows that (\overline{s}, 1, \overline{u}) is the singular point.

In the case S_{1} = 0 or U_{1} = 0 we will use Lemma 3.4 to show that the homogeneous coordinates of the singularity are given by (3.4) and (3.5).

Consider the case S_{1} = 0. By Lemma 3.4, G_{1} = 0, and since we have assumed F is irreducible, S_{2} ≠ 0 and U_{2} ≠ 0.

Suppose c and d are not zero. Solving for b in equation (3.7), we get

b = \frac{ce^{3}f − ae^{2}f^{2} + 2c^{2}de^{2} − 12acdef + 12a^{2}df^{2} − 8ac^{2}d^{2}}{cS_{2}} .  (3.9)

Substituting (3.9) in (3.1), Δ can be factored (2a(af − ce)^{2}Δ_{1}^{2})/(c^{2}S_{2}^{3}), where

\begin{matrix}Δ_{1}(a, c, d, e, f) = & 108a^{2}df^{4} − 9ae^{2}f^{4} + 8ce^{3}f^{3} − 144acdef^{3} \\ − 144ac^{2}d^{2}f^{2} + 48c^{2}de^{2}f^{2} + 96c^{3}d^{2}ef + 64c^{4}d^{3}. \end{matrix}

We know that Δ = 0, but we need to know that for F in this case, Δ_{1} = 0. Either af − ce = 0 or Δ_{1} = 0. If af − ce = 0, we find from (3.9) that bc = ad. But then using Lemma 3.2 the cubic is reducible. Hence Δ_{1} = 0.

To show that formulas (3.4) and (3.5) determine the singular point, we check that the derivatives of F (s, t, u) vanish at the point (\frac{G_{2}}{S_{2}} , 1, \frac{G_{2}}{U_{2}} ). Be-

cause this is difficult to check directly, we replace \frac{G_{2}}{U_{2}} by \frac{G_{3}}{H}, where

G_{3}(a, b, c, d, e, f) = −f(8cef − 9af^{2} + 4c^{2}d)

and

H(a, b, c, d, e, f) = 2d(4c^{2}d + 5cef − 6af^{2})

This substitution is allowable (after we check that H ≠ 0) because

G_{1}|(G_{3}U_{2} − G_{2}H).  (3.10)

Evaluating the partial derivatives of F at (\overline{s}, 1, \overline{u}), with \overline{s} = \frac{G_{2}}{S_{2}} and \overline{u} = \frac{G_{3}}{H}, we obtain F_{s}(\overline{s}, 1, \overline{u}) = Δ_{1}Ψ_{1}, F_{t}(\overline{s}, 1, \overline{u}) = Δ_{1}Ψ_{2} and F_{u}(\overline{s}, 1, \overline{u}) = Δ_{1}Ψ_{3}, where Ψ_{1}, Ψ_{2} and Ψ_{3} are rational functions whose denominators are products of H and S_{2}. Because Δ_{1} = 0, it remains only to show that H ≠ 0. From (3.10) it follows that if H = 0, then G_{3} is zero. Since c and d are different from zero, 8cef − 9af^{2} + 4c^{2}d = 0 and 4c^{2}d + 5cef − 6af^{2} = 0, from which it is again possible to deduce that the cubic is reducible, for a contradiction.

If c = 0 or d = 0 the formulas (3.4) and (3.5) can be checked directly. The demonstration when U_{1} = 0 follows the same steps as when S_{1} = 0,

but using that

G_{1}|(G_{3}(a, b, c, d, e, f)S_{2}(b, a, d, c, e, f) − H(a, b, c, d, e, f)G_{2}(b, a, d, c, e, f))

For the proof of (1) of Lemma 3.4, we assume that S_{1} = 0. If c ≠ 0 and S_{2} ≠ 0, again express b as in (3.9). After substituting this expression in (3.6), G_{1} can be factored ((af − ce)Δ_{1})/cS_{2}. Since Δ_{1} = 0, necessarily G_{1} = 0.

If c ≠ 0 and S_{2} = 0, we can conclude first that f ≠ 0. For S_{2} = 0 and f = 0 together imply d = 0, which makes F reducible with factor s. Hence S_{2} = 0 can be solved for a:

a = \frac{2c(ef + 2cd)}{3f^{2}}.  (3.11)

Substituting for a in S_{1}, we obtain c(ef + 2cd)(4cd − ef)^{2} = 0. Since a ≠ 0, this means 4cd − ef = 0. Substituting for a in G_{1}, we obtain

G_{1} = \frac{c(4cd − ef)(−4cd^{2} − 5def + 6bf^{2})}{3f}  (3.12)

so G_{1} = 0.

If c = 0, then

S_{1} = f^{2}a(12ad − e^{2})  (3.13)
G_{1} = f^{2}a(9bf − 8de)  (3.14)

From (3.13) we obtain

d = \frac{e^{2}}{12a}.  (3.15)

Using (3.15) and c = 0, Δ can be factored f^{2}(27abf − 2e^{3})^{2}/27. Substituting (3.15) in (3.14), we obtain G_{1} = f^{2}(27abf − 2e^{3})/3. Since Δ = 0, G_{1} also is zero.

The proof in the case U_{1} = 0 is analogous.

For (2), we know that G_{1} = 0 and S_{2} = 0. We may assume that c and f are different from zero, for otherwise using S_{2} = 0 we see that F is reducible and we are done. Furthermore d ≠ 0. For if d = 0, then G_{1} can be factored bf^{2}(9af − 8ce) and S_{2} reduces to f(3af − 2ce), hence af = 0, which contradicts a ≠ 0 and f ≠ 0. So we assume a, b, c, d and f are all nonzero.

Solving S_{2} = 0 for a, we again obtain (3.11), and when we substitute for a in G_{1} we obtain (3.12). One of the factors in the numerator of (3.12) must be zero.

If ef −4cd = 0, we can deduce from S_{2} = 0 that af −ce = 0. Substituting a = ce/f into (3.1) and simplifying using ef −4cd = 0, we obtain −16c^{3}f(de− bf)^{3} = 0. Now af − ce = 0 and de − bf = 0 imply that F is reducible by Lemma 3.2.

If the other factor of (3.12) is zero,

−4cd^{2} − 5def + 6bf^{2} = 0,  (3.16)

and

c = \frac{(6bf − 5de)f}{4d^{2}} .  (3.17)

Substituting (3.11) and (3.17) into the expression (3.1) for Δ, we obtain Δ = (27(6bf − 5de)f^{2}(2bf − de)^{2}(bf − de)^{3})/4d^{6}. We know Δ = 0. If the first factor 6bf − 5de = 0, then by (3.17) c = 0, a contradiction.

Combining the expression (3.8) for S_{2} and (3.16), we obtain

cde + adf − 2bcf = 0.  (3.18)

If the second factor 2bf − de = 0, then from (3.18) we obtain adf = 0, which is a contradiction.

Consequently, it can only happen that the third factor bf −de = 0. Using this and (3.18) we obtain also ad − bc = 0, so by Lemma 3.2, the cubic F is reducible.

§4 An Application

In this section we give the expressions for the control points and weights of the Bézier form of a singular cubic in terms of its double point (\overline{s}, \overline{t}, \overline{u}) and three of its coefficients a, b and f. Using these formulas together with Theorem 3.3, one can obtain the control points and weights in closed form entirely in terms of the coefficients.

Because we are working in a triangle, the control points can be assigned barycentric coordinates b_{0}(1, 0, 0), b_{1}(X, 1−X, 0), b_{2}(0, 1−Y, Y) and b_{3}(0, 0, 1). We refer to X and Y as the control points ratios of b_{1} and b_{2} respectively.

Theorem 4.1. The control point ratios and weights of the Bézier cubic are

X = \frac{a\overline{s}^{2}\overline{u} + f\overline{t}^{3}}{a\overline{s}^{2}\overline{u} + a\overline{s}\overline{t}\overline{u} + f\overline{t}^{3}}

Y = \frac{b\overline{s}\overline{u}^{2} + f\overline{t}^{3}}{b\overline{s}\overline{u}^{2} + b\overline{s}\overline{t}\overline{u} + f\overline{t}^{3}}

w_{0} = −a\overline{s}\overline{u}^{2}

w_{1} = \frac{1}{3}(a\overline{s}^{2}\overline{u} + a\overline{s}\overline{t}\overline{u} + f\overline{t}^{3})

w_{2} = \frac{1}{3}(b\overline{s}\overline{u}^{2} + b\overline{s}\overline{t}\overline{u} + f\overline{t}^{3})

w_{3} = −b\overline{s}^{2}\overline{u}.

Proof: With the singular point (\overline{s}, \overline{t}, \overline{u}), three of the coefficients of F, given by (2.1), are

c = \frac{−(b\overline{s}\overline{u}^{2} + f\overline{t}^{3})}{\overline{s}\overline{t}^{2}}

d = \frac{−(a\overline{s}^{2}\overline{u} + f\overline{t}^{3})}{\overline{t}^{2}\overline{u}}

e = \frac{−(2a\overline{s}^{2}\overline{u} + 2b\overline{s}\overline{u}^{2} + f\overline{t}^{3})}{\overline{s}\overline{t}\overline{u}}.

After these substitutions are made in F, a parametrization of F = 0 is obtained by the standard method of intersecting the curve with the family of lines through the double point. This parametrization is then converted to Bézier form.

References

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10. Walker, R., Algebraic Curves, Dover, NY, 1950.

M. Paluszny and F. Tovar Laboratorio de Computación Gráfica y Geometría Aplicada Facultad de Ciencias Universidad Central de Venezuela Apartado 47809, Los Chaguaramos Caracas 1041-A. Venezuela

ftovar@euler.ciens.ucv.ve
marco@jade.ciens.ucv.ve

R. Patterson Dept. of Mathematical Sciences Indiana University-Purdue University Indianapolis 402 North Blackford Street Indianapolis, Indiana, 46202-3216

rpatters@math.iupui.edu