Graphics Reference
In-Depth Information
p
3
p
3
p
4
p
4
p
4
p
2
p
2
L
L
L
p
1
p
1
p
2
p
3
p
1
(a) Valid case
(b) Disallowed case
(c) Disallowed case
Figure 3.23.
Conic design problem 1.
We now describe solutions to five conic design problems in the plane
R
2
. The fact
that the solutions are indeed correct follows easily from the above, in particular by
repeated use of Lemma 3.6.1.4. See [PenP86] for details.
Conic design problem 1:
To find the equation of the conic passing through
four points
p
1
,
p
2
,
p
3
, and
p
4
that has a given line
L
through one of these points
as tangent line. Assume that at least two of the points do not lie on
L
. If three points
lie off
L
, then no two of them are allowed to be collinear with the fourth. See Figure
3.23.
Solution.
Assume that
L
is the tangent line at
p
1
and that
p
2
and
p
3
do not lie on
L
. Let
L
2
be the line through
p
1
and
p
2
, let
L
3
be the line through
p
1
and
p
3
, and let
L
4
be the line through
p
2
and
p
3
. Let [
L
] = [a,b,c] and [
L
i
] = [a
i
,b
i
,c
i
]. Define symmetric
3 ¥ 3 matrices Q
1
and Q
2
by
1
2
1
2
(
)
T
T
(
) (
)
+
(
) (
)
Q
=
a b c
, ,
a
,
b
,
c
a
,
b
,
c
a b c
, ,
and
1
4
4
4
4
4
t
(
)
T
T
(
) (
)
+
(
) (
)
Qa
=
,
b
,
c
a
,
b
,
c
a
,
b
,
c
a
,
b
,
c
.
2
2
2
2
3
3
3
3
3
3
2
2
2
Let C
i
(x,y) = 0 be the quadratic equation associated to Q
i
. Let
p
4
= (x
4
,y
4
). If
(
)
=
(
)
+-
(
)
(
)
Cxy
,
l
Cxy
,
1
l
Cxy
,
,
l
1
2
then there is a unique l so that C
l
(x
4
,y
4
) = 0 and that is the equation of the conic we
want.
3.6.1.8. Example.
To find the conic that passes through the points
p
1
,
p
2
= (2,-2),
p
3
= (2,2),
p
4
= (5,0), and that has tangent line
L
at
p
1
for the case where
p
1
and
L
have the values
()
(
)
()
=
()
a
p
=-
10
,, :
L
x
+=
1
0
b
p
30
,, :
L
x
-=
3
0
1
1
Solution for (a):
See Figure 3.24(a). We have the following equations for the lines
L
i
:
