Graphics Reference
In-Depth Information
y
L
4
L
3
y
L
4
L
L
L
2
p
3
p
3
p
4
x
x
p
4
p
1
p
1
p
2
p
2
L
3
L
2
(a)
(b)
Figure 3.24.
The conics that solve Example 3.6.1.6.
L
2
:x ++=
2320
L
3
:x -+=
2320
L
4
:x-=
20
Therefore,
(
)
=+
(
)
(
)
+-
(
)
(
)
(
)
Cxy
,
l
x
1 21 232232
x
-
l
x y
++
x y
-+
.
l
Solving C
l
(5,0) = 0 for l gives l=8/7, so that our conic is the ellipse
2
(
)
=-
(
)
2
Cxy
,
429 60
x
+-=
y
.
87
Solution for (b):
See Figure 3.24(b). We have the following equations for the lines
L
i
:
L
2
:x+-=
2
6
0
L
3
:x--=
2
6
0
L
4
:x-=
20
This time
(
)
=-
(
)
(
)
+-
(
)
(
)
(
)
Cxy
,
l
x
3212
x
-
l
xy
+-
62
xy
--
6
.
l
Solving C
l
(5,0) = 0 for l gives l=8/5, so that our conic is the hyperbola
2
(
)
=-
(
)
2
Cxy
,
443 40
x
-
++=
y
.
85
Conic design problem 2:
To find the equation of the conic passing through three
points
p
1
,
p
2
, and
p
3
that has two given lines
L
1
and
L
2
through two of these points
as tangent lines. Assume that the three points are not collinear and that the intersec-
