Graphics Reference
In-Depth Information
(
)
ba
-
sin
22
q
+
h
cos
2 0
q
=
.
If a = b, then cos 2q=0, which means that q is ±45 degrees. If a π b, then the solution
can be written either as
h
ab
2
tan2
q=
(3.36a)
-
or
2
(
)
2
ba
-±
ba
-
+
4
h
tan
q=
.
(3.36b)
2
h
Note that if q
1
and q
2
are angles satisfying (3.36b) where we use the + and - sign,
respectively, in the formula, then
tan
q
tan
q
=-
1
,
1
2
which shows that the angles differ by 90 degrees. We shall see that this basically
affirms that the “axes” of the conic are perpendicular. In any case, there is an angle q
which will eliminate the x¢y¢ cross-term in equation (3.35). Thus, we end up with an
equation of the form
2
2
ax
¢¢ + ¢¢ + ¢¢+
by
2
fx
2
gy
¢¢+ ¢=
c
0.
(3.37)
The rest of the steps involved in analyzing equation (3.35) are very simple. Equa-
tion (3.37) still has some linear terms that need to be eliminated if the corresponding
quadratic term is present. This is done by completing the square in the standard way.
For example, if a¢π0, then make the substitution
xx
g
a
¢
¢
¢= ¢¢+
.
From this one sees that, in the nondegenerate case, equation (3.38) splits into two
cases. If a¢b¢=0, then (3.35) represents a parabola; otherwise, (3.35) represents an
ellipse or hyperbola depending on whether the sign of a¢b¢ is positive or negative.
In summary, we have shown how some simple manipulations of equation (3.35)
enable us to determine the geometry of the solution set. However, we have glossed
over some degenerate cases. For example, a degenerate case of equation (3.35) occurs
when it factors into two linear terms, that is, that it can be written in the form
(
)
(
)
=
axbycaxbyc
1
++
++
0
.
(3.38)
1
1
2
2
2
In this case the equation represents a pair of lines. To be able to detect the special
cases, and all the other cases for that matter, in a nice simple way, we need to make
use of some more powerful tools. In particular, we need to switch to homogeneous
coordinates and projective space.
