Chapter 4: Geometry

(c) If ¼, the equation again represents a degenerate conic: when and
have different signs, we have a pair of lines Ý ¦ Ô Ü, and,
when they have the same sign, we get a point (the origin).
EXAMPLE
We work out an example for clarity. Suppose the original equation is
Ü ¾ · Ý ¾ ÜÝ ·¿Ü Ý ·½¼ (4.6.6)
In step 1 we apply the substitutions Ü ¾Ü · Ý and Ý ¾Ý Ü. This gives
¾Ü ¾ ·½¼Ü Ý ·½ ¼. Next we interchange Ü and Ý (step 2) and get ¾Ý ¾ ·
½
½¼Ý Ü ·½¼. Replacing Ý by Ý in step 3, we get ¾Ý¾ Ü ¼. Finally, in
step 4a we divide the equation by ¾, thus giving it the form of Equation (4.6.3) with
½ . We have reduced the conic to a parabola with vertex at the origin and focus
¾¼
at ´ ½ ¼µ. To locate the features of the original curve, we work our way back along
¾¼
the chain of substitutions (recall the convention about substitutions and transformations
from Section 4.1.2):
½
ÜÝ
ܾܷÝ
Substitution Ý Ý
ÝÜ
Ý¾Ý Ü
½
Vertex ´¼ ¼µ ´¼
µ ´ ½
¼µ
´ ¾
½
µ
Focus ´ ½ ¾¼ ¼µ
´ ½ ½
¾¼ µ ´ ½
½
¾¼ µ ´
¾¼ ¾¼ µ
We conclude that the original curve, Equation (4.6.6), is a parabola with vertex
¾
´ ½
µ and focus ´ µ.
¾¼ ¾¼
An alternative analysis of Equation (4.6.4) consists in forming the quantities
¡
Ã
¬
½
½ ¾ ¾
¬¬¬¬¬¬
¬
½
¾ ¬
½
Â
¾ ¬ ½
¾ ¬¬¬
½
½
½ Á ·
¾
¾ ¾ ¬ ¬ ½
¾ ¬¬¬ ¬
½
·
¬ ½
¾ ¬¬¬
½
¾ ¾
(4.6.7)
and nding the appropriate case in the following table, where an entry in parentheses
indicates that the equation has no solution in real numbers:
¡ Â ¡Á Ã Type of conic
¼ ¼ Hyperbola
¼ 0 Parabola
¼ ¼ ¼ Ellipse
¼ ¼ ¼ (Imaginary ellipse)
0 ¼ Intersecting lines
0 ¼ Point
0 0 ¼ Distinct parallel lines
0 0 ¼ (Imaginary parallel lines)
0 0 0 Coincident lines
For the central conics (the ellipse, the hyperbola, intersecting lines, and the point),
the center ´Ü ¼ Ý ¼ µ is the solution of the system of equations
¾Ü · Ý · ¼
Ü ·¾Ý · ¼
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