Page 101 - Geometric Modeling and Algebraic Geometry

6 Classification of Surfaces 99
:
containing C t 0
Therefore, we deduce implicit equations of plane Π t 0
⎧
⎨ X − t 0 T =0
: Y − t 0 P =0
Π t 0
Z − t 0 =0.
⎩
is just the point M 0 . We denote by G(t 0 ,u 0 ) the
and Π t 0
The intersection of L u 0
. Hence, G(t 0 ,u 0 ) has dimension 3. Implicit
and Π t 0
affine space generated by L u 0
projective equations of G(t 0 ,u 0 ) are:

X − t 0 T − u Z + u t 0 Q =0
2
2
G(t 0 ,u 0 ): 0 0
Y − u 0 Z − t 0 P + u 0 t 0 Q =0.
We denote by E the set of 4-projective spaces containing G(t 0 ,u 0 ); such a hyperplane
is denoted by H(α, β, t 0 ,u 0 ) and have an equation of type:
α(X − t 0 T − u Z + u t 0 Q)+ β(Y − u 0 Z − t 0 P + u 0 t 0 Q)=0
2
2
0 0
where (α, β) ∈ C −{0}.
2
Each hyperplane H(α, β, t 0 ,u 0 ) cuts the scroll along a curve of degree 4 (because
, the intersection must
and L u 0
the scroll has degree 4). As it already contains C t 0
contain another line of the scroll, let us call it L u . We aim to single out the hyper-
≡L u . By replacing the parametric expressions
plane H(α, β, t 0 ,u 0 ) such that L u 0
of X, Y, Z, T, P, Q of F(2, 2) in the equation above, we obtain the equation of the
intersection of F(2, 2) and H(α, β, t 0 ,u 0 ) in the parameter space of the scroll:
(u − u 0 )(t − t 0 )(αu + αu 0 + β)=0
−(αu 0 + β)
Therefore, u = .(α must be different from 0, otherwise α = β =0).

α
−(αu 0 +β)

We get, u = u 0 if and only if: = u 0 ⇒ β =−2αu 0 . We can take α=1,
α
so we have: β = −2u 0 . In this case H(1, −2u 0 ,t 0 ,u 0 ) (denoted by H(t 0 ,u 0 )) cuts
. The equation of H(t 0 ,u 0 ) becomes:
F(2, 2) in C t 0
and twice in L u 0
X − 2u 0 Y + u Z − t 0 T +2u 0 t 0 P − u t 0 Q =0.
2
2
0 0
∗
The coefficients are the coordinates (A 1 : ... : A 6 ) of this hyperplane in (P ) in
5
the following order:

2 2
.
1 −2u 0 u −t 0 2u 0 t 0 −u t 0
0 0
A 6
A 5
A 1 A 2 A 3 A 4
and satisfy the condition:
⎛ ⎞
1 1
⎜ A 1 2 A 2 A 4 2 A 5 ⎟
1 1
rank ⎝ ⎠ =1.
2 A 2 A 3 2 A 5 A 6