Page 108 - Geometric Modeling and Algebraic Geometry

106 T.-H. Lˆ e and A. Galligo

−2csuv + tv(v +(2c − 1)u)=0
⇐⇒
uv[(2c − 1)u +(1 − 2(d + c))uv +2dv ]=0
2
2
We have that, (t = s =1); (u = v =1) is a double root of the system above if
1
and only if c = + d, d =0. Therefore, the parametric equations of the surface is
2
as follows:
⎧
⎪ x = tu 2
⎨ 2
(S): y =(t − 1)(u − 1) d =0
1
⎪
⎩ z =(d + )u + tu + dt
2
2
If b =0, we obtain the parametric equations of the surface:
⎧
⎨ x = tu 2
(S): y =(t − 1)(u − 1) 2
⎩
z = u + t
2
c) The case (5.2.d) and the system {ψ 1 (t, u), ψ 2 (t, u)} have two different roots:
We can write:

ϕ 1 (t, u)= g(t, u)ψ 1 (t, u)= g(t, u)(tA 1 (u)+ A 2 (u))
ϕ 2 (t, u)= g(t, u)ψ 2 (t, u)= g(t, u)(tB 1 (u)+ B 2 (u))
where A 1 (u) ,A 2 (u) ,B 1 (u) ,B 2 (u) are polynomials of degree 1 in u.

We denote by u 0 the root of g(t, u). We call (t 1 ,u 1 ) and (t 2 ,u 2 ) two roots of
ψ 1 (t, u) and ψ 2 (t, u). We have two cases: either u 1 ,u 2 = u 0 or one of them is equal
to u 0 .
Firstly, we consider the case where u 1 ,u 2 = u 0 . By change of parameters, we
assume that u 0 =0 , (t 1 ,u 1 )=(1, 1) , (t 2 ,u 2 )=(0, ∞). Hence, ϕ 1 (t, u) and
ϕ 2 (t, u) become:

ϕ 1 (t, u)= u(tu − 1)
ϕ 2 (t, u)= u(t − 1)
We deduce the equations of Π A :

X 2 − 2X 6 =0
Π A : X 2 + X 5

By the remark (13) we obtain the parametric equations of the surfaces:
⎧
⎪ X = tu 2
⎪
⎨ Y = u 2
(S):
⎪ Z =2tu − 2u +1
⎪
⎩
T = t