Graphics Reference
In-Depth Information
eliminate the quadratic term of one of the variables, say Y, then we could set X = t,
and solve the resulting linear equation in Y for Y giving us Y as a function of t also.
Specifically, after eliminating the Y
2
term in the equation for the conic, we will have
an equation of the form
(
)
=
(
)
2
aX
+
b Y
+
cX
+
dX
+
e
0,
(10.86)
which has solutions
Xt
=
(
)
2
-++
ct
dt
e
Y
=
.
(10.87)
at
+
b
We can see from this that we always have one “point at infinity” (t =•), but we may have
a second one for t =-b/a if a π 0. In general, we will get a rational parameterization, but
if a = 0 in equation (10.86), then we actually get a polynomial parameterization.
To get a better picture of what is happening, consider a degree n curve for a
moment and write
(
)
=++
fXY
,
f
f
...
f
n
,
01
where the f
i
are homogeneous of degree i. The Y
n
term that we want to eliminate is
part of f
n
. Note that over the complex numbers
d
'
1
(
)
=
(
)
fXY
,
aY bX
-
,
(10.88)
n
i
i
i
=
by Proposition 10.5.3, so that f
n
= 0 corresponds to n lines through origin. The curve
will have no Y
n
term if and only if some a
i
is zero and one of the lines is X = 0, that
is, a line parallel to the y-axis meets the curve at infinity. We can see things even better
if we switch to homogeneous coordinates. Homogenizing f by replacing X by X/Z and
Y by Y/Z gives
(
)
=
n
(
)
n
-
1
++
(
)
1
(
)
FX Y Z
,
,
f Z
+
f X YZ
,
...
f
X YZ
,
+
f X Y
,
.
(10.89)
0
1
n
-
1
n
Finding the intersection of the (projective) plane curve
(
)
= 0
FXYZ
,,
with the line at infinity (Z = 0) means solving
(
)
= 0
fXY
n
,
.
