Graphics Reference
In-Depth Information
()
()
pt
qt
1
1
x
=
()
()
pt
qt
2
2
y
=
.
Clearing denominators gives
()
-
()
=
xq
t
p
t
0
0
1
1
()
-
()
=
yq
t
p
t
.
(10.49)
2
2
If we could eliminate the variable t, then we would have an implicit equation in x and
y only. For example, consider the parameterization
x=21
y=35
of a line
L
in the plane. If we solve both equations for t, set the results equal, and sim-
plify, we get
3270
x-+=,
which is an equation for
L
. We could also find the equation for a curve
C
defined by
quadratic parameterizations in this way, such as, for example,
2
xt
=-+
=+
31
2 .
t
2
yt
We again simply solve for t and set the results equal. However, as the degree of the
parameterization got higher, solving for t would get more and more complicated. In
fact, if the degree was larger than 4, then this approach would not work at all because
there is a well-known theorem that states that there is no solution by radicals of the
general equation of degree five or higher. Something else is needed if we want to find
implicit definitions for spaces defined parametrically.
Recall resultants. What do they have to do with eliminating variables? Well, con-
sider the equations (10.49) again and rewrite them as
()
=
()
-
()
=
f t
xq
t
p
t
0
0.
1
1
()
=
()
-
()
=
g t
yq
t
p
t
(10.50)
2
2
The polynomials f and g are now thought of as polynomials in t with coefficients in
R
[x,y]. If
()
()
pt
qt
10
10
x
=
0
()
()
,
pt
qt
20
20
y
=
0
