6.4 Hyperbolas 381which display information from the paragraph containing (6.281). Even
though the equation of the hyperbola of the figure has already been
derived, we become acquainted with useful ideas by using the intrinsic
string property
(6.45) IF1FI - IF2PI = ±2a
to derive the equation. Letting F1(-ae,0) and Fz(ae,0) be located on the
x axis with theorigin midway between them as in Figure 6.44, we use the
string property to obtain the equation
1/(x + ae)z + y2- (x - ae)2 + y2 = ±2a.
Transposing the second term, squaring, and simplifying give
± (x - ae)2 + y2 = a - ex.
Squaring and simplifying this give
(e2 - 1)x2 - y2 = a2(e2 - 1),
and dividing by the right side gives the standard form
x2 y2
(6.46) a2 b2=^1 ,
where b2 = a2(e2 - 1), or
(6.47) ae= az+.bz,
As was the case for ellipses, it is of interest to see that memorization of
only a few details enables us to find the graph, foci, directrices, and asymp-
totes of (6.46) when a and b are given constants. Putting y = 0 shows
that the points Vi(-a,0) and V2(a,0) lie on the graph. The line through
these points is the principal axis of the hyperbola, and the foci lie on it.
Putting x = 0 shows that the hyper-
bola contains no points on the y axis,
but we mark the points (0,b) and (0, -b)
anyway. The line through these points
is the conjugate axis of the hyperbola.
The next steps are to draw the box
having horizontal and vertical sides
containing the four points and then
draw the diagonals of the box. These
Figure 6.48diagonals are, as we showed in Problem 7 of Section 3.3 and shall quickly
show again, asymptotes of the hyperbola. A reasonable approximation
to the hyperbola can now be sketched very quickly; it is tangent to the
box at the two vertices we have found, and a pencil point which traces a
part of the hyperbola becomes steadily closer to the asymptote as it
recedes from a vertex. For the part of the hyperbola in the first quad-
rant, this result comes from the fact that
(6.481)a2
a a a x x2 - a2