Calculus: Analytic Geometry and Calculus, with Vectors

366 Cones and conics

The formula (6.273) shows that the x and y axes are axes of symmetry of

the hyperbola and that the origin is a center (center of symmetry) of the

hyperbola. This proves that the two branches of the hyperbola are

congruent.

There are two reasons for putting the above equations in the simpler

standard forms

(6.28)

Xz^2 z z

y2 x- y

a2-f-b2 az bz

where a and b are positive constants. In the first place the denominators

in (6.272) and (6.273) are clumsy things to write, and in the second place

equations of the form (6.28) often arise in problems where a and b are not

children of eccentricities. Comparing (6.272) and (6.273) with (6.28)

shows that a and b are determined in terms of e and p by the formulas

(6.281) a=11-e21' b=Vlep 1

1-e 21

On the other hand, the formulas

z

(6.282) 1x11 =

1 1

e pe21 = ae = distance from center to focus

(6.283) 1x1 - pl =

11

p

(^921) e = distance from center to directrix
(6.284) ae = a2 --b2 for ellipse, ae = 1/a2 + b2 for hyperbola
serve to determine other quantities in terms of a and b. The first two of
these formulas are obtained very quickly by comparing the formulas for
x1 and x1 - p in (6.27) with the formula for a in (6.281). To obtain
(6.284), we can square the members of (6.281) and combine the results to
obtain b2/a2 = 11 - eel and then treat separately the cases in which

0 < e < 1 and e>1. Observe that 0 <b <a when 0 < e < 1 and

also I < e < but that b > a when e > V2-. Graphs of ellipses and

hyperbolas, and schemes for remembering essential parts of the above

formulas, will appear in later sections.

Problems 6.29

1 Copy Figure 6.25 and the equation (6.251) of the conic K and look at

them. Then show that the equation of K can be put in the form

12,42 l 2e2AB e2B2 l

(1) (1- i2 +

B2/ x

z _

112 + B2 (1 11z } B=/ y

=

t2 C e2BC l e2C2

Yi)