A. The Equation of a Hyperbola
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CHAPTER 8 Analytic Geometry and the Conic Sections
Plotting these points and connecting them
with a smooth curve, while knowing there are no
y-intercepts, produces the graph in the figure.
The point at the origin (in blue) is not a part of
the graph, and is given only to indicate the
“center” of the hyperbola. The points 1Ϫ4, 02
and (4, 0) are called vertices, and the center
of the hyperbola is always the point halfway
between them.
y
Hyperbola
(Ϫ5, 2.25) (5, 2.25)
(Ϫ4, 0)
(4, 0)
(0, 0)
x
(Ϫ5, Ϫ2.25) (5, Ϫ2.25)
Now try Exercises 7 through 22
ᮣ
As with the circle and ellipse, the hyperbola fails the vertical line test and we must
graph the relation on a calculator by writing its equation in two parts, each of which is
a function. For the hyperbola in Example 1, this gives
9x2 Ϫ 16y2 ϭ 144
original equation
2
2
isolate y-term
Ϫ16y ϭ 144 Ϫ 9x
multiply by Ϫ1
16y2 ϭ 9x2 Ϫ 144
9x2 Ϫ 144
y2 ϭ
divide by 16
16
9x2 Ϫ 144
yϭϮ
take square roots
B
16
Ϫ9.4
2
2
9X Ϫ 144
9X Ϫ 144
, Y2 ϭ Ϫ
Y1 ϭ ϩ
B
16
B
16
Figure 8.25
6.2
9.4
The graph is shown in Figure 8.25.
Ϫ6.2
Since the hyperbola in Example 1 crosses a horizontal line of symmetry, it is
referred to as a horizontal hyperbola. If the center is at the origin, we have a central
hyperbola. The line passing through the center and both vertices is called the transverse axis (vertices are always on the transverse axis), and the line passing through the
center and perpendicular to this axis is called the conjugate axis (see Figure 8.26).
In Example 1, the coefficient of x2 was positive and we were subtracting 16y2:
2
9x Ϫ 16y2 ϭ 144. The result was a horizontal hyperbola. If the y2-term is positive
and we subtract the term containing x2, the result is a vertical hyperbola (Figure 8.27).
Figure 8.26
Figure 8.27
y
Conjugate
axis
Center
Vertex
Transverse axis
Transverse
axis
Vertex
Horizontal
hyperbola
x
y
Vertex
Vertex Center
Vertical
hyperbola
Conjugate
axis
x
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Section 8.3 The Hyperbola
EXAMPLE 2
ᮣ
Identifying the Axes, Vertices, and Center of a Hyperbola from Its Graph
For the hyperbola shown, state the location of the
vertices and the equation of the transverse axis.
Then identify the location of the center and the
equation of the conjugate axis.
Solution
ᮣ
By inspection we locate the vertices at (0, 0) and
(0, 4). The equation of the transverse axis is
x ϭ 0. The center is halfway between the vertices
at (0, 2), meaning the equation of the conjugate
axis is y ϭ 2.
y
5
Ϫ5
5
x
Ϫ4
Now try Exercises 23 through 26
ᮣ
Standard Form
As with the ellipse, the polynomial form of the equation is helpful for identifying
hyperbolas, but not very helpful when it comes to graphing a hyperbola (since we still
must go through the laborious process of finding additional points). For graphing, standard form is once again preferred. Consider the hyperbola 9x2 Ϫ 16y2 ϭ 144 from
Example 1. To write the equation in standard form, we divide by 144 and obtain
y2
x2
Ϫ
ϭ 1. By comparing the standard form to the graph, we note a ϭ 4 represents
42
32
the distance from center to vertices, similar to the way we used a previously. But
since the graph has no y-intercepts, what could b ϭ 3 represent? The answer lies in
the fact that branches of a hyperbola are asymptotic, meaning they will approach
and become very close to imaginary lines that can be used to sketch the graph. For
b
a central hyperbola, the slopes of the asymptotic lines are given by the ratios and
a
b
b
b
Ϫ , with the related equations being y ϭ x and y ϭ Ϫ x. The graph from Example 1
a
a
a
is repeated in Figure 8.28, with the asymptotes drawn. For a clearer understanding of
how the equations for the asymptotes were determined, see Exercise 87.
A second method of drawing the asymptotes involves drawing a central rectangle
with dimensions 2a by 2b, as shown in Figure 8.29. The asymptotes will be the extended diagonals of this rectangle. This brings us to the equation of a hyperbola in
standard form.
Figure 8.28
Figure 8.29
y
Slope m ϭ
y
Ϫ 34
rise
bϭ3
(Ϫ4, 0)
Slope m ϭ
3
4
run
aϭ4
(4, 0)
(0, 0)
(Ϫ4, 0)
2b
x
2a
Slope method
Central rectangle method
x
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CHAPTER 8 Analytic Geometry and the Conic Sections
The Equation of a Hyperbola in Standard Form
The equation
1x Ϫ h2 2
a2
Ϫ
1y Ϫ k2 2
b2
The equation
1y Ϫ k2 2
ϭ1
b2
represents a horizontal hyperbola
with center (h, k)
• transverse axis y ϭ k
• conjugate axis x ϭ h
• ͿaͿ gives the distance from
center to vertices.
Ϫ
1x Ϫ h2 2
a2
ϭ1
represents a vertical hyperbola
with center (h, k)
• transverse axis x ϭ h
• conjugate axis y ϭ k
• ͿbͿ gives the distance from
center to vertices.
b
• Asymptotes can be drawn by starting at (h, k) and using slopes m ϭ Ϯ .
a
EXAMPLE 3
ᮣ
Graphing a Hyperbola Using Its Equation in Standard Form
Sketch the graph of 161x Ϫ 22 2 Ϫ 91y Ϫ 12 2 ϭ 144, and label the center, vertices,
and asymptotes.
Solution
ᮣ
Begin by noting a difference of the second-degree terms, with the x2-term
occurring first. This means we’ll be graphing a horizontal hyperbola whose center
is at (2, 1). Continue by writing the equation in standard form.
161x Ϫ 22 2 Ϫ 91y Ϫ 12 2 ϭ 144
91y Ϫ 12 2
161x Ϫ 22 2
144
Ϫ
ϭ
144
144
144
1y Ϫ 12 2
1x Ϫ 22 2
Ϫ
ϭ1
9
16
1x Ϫ 22 2
1y Ϫ 12 2
Ϫ
ϭ1
32
42
given equation
divide by 144
simplify
write denominators in squared form
Since a ϭ 3 the vertices are a horizontal distance of 3 units from the center (2, 1),
giving 12 ϩ 3, 12 S 15, 12 and 12 Ϫ 3, 12 S 1Ϫ1, 12 . After plotting the center and
b
4
vertices, we can begin at the center and count off slopes of m ϭ Ϯ ϭ Ϯ , or
a
3
draw a rectangle centered at (2, 1) with dimensions 2132 ϭ 6 (horizontal dimension)
by 2142 ϭ 8 (vertical dimension) to sketch the asymptotes. The complete graph is
shown here.
Horizontal hyperbola
y
Center at (2, 1)
mϭd
Vertices at (Ϫ1, 1) and (5, 1)
Transverse axis: y ϭ 1
Conjugate axis: x ϭ 2
(2, 1)
(Ϫ1, 1)
(5, 1)
x
m ϭ Ϫd
Width of rectangle
horizontal dimension and
distance between vertices
2a ϭ 2(3) ϭ 6
Length of rectangle
(vertical dimension)
2b ϭ 2(4) ϭ 8
Now try Exercises 27 through 44
ᮣ
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Section 8.3 The Hyperbola
If the hyperbola in Example 3 were a central hyperbola, the equations of the
4
4
asymptotes would be y ϭ x and y ϭ Ϫ x. But the center of this graph has been
3
3
shifted 2 units right and 1 unit up. Using our knowledge of shifts and translations, the
equations for the asymptotes of the shifted hyperbola must be
4
4
5
1. 1y Ϫ 12 ϭ ϩ 1x Ϫ 22 , or y ϭ x Ϫ in simplified form, and
3
3
3
4
11
4
2. 1y Ϫ 12 ϭ Ϫ 1x Ϫ 22 or y ϭ Ϫ x ϩ .
3
3
3
Using Y1 ϭ ϩ
161X Ϫ 22 2 Ϫ 144
161X Ϫ 22 2 Ϫ 144
ϩ 1, and Y2 ϭ Ϫ
ϩ1
B
9
B
9
(obtained by solving for y in the original equation), a calculator generated graph of the
hyperbola and its asymptotes is shown here (Figures 8.30 and 8.31).
Figure 8.31
Figure 8.30
10.3
Ϫ7.4
11.4
Ϫ8.3
Polynomial Form
If the equation is given as a polynomial in expanded form, complete the square in x and
y, then write the equation in standard form.
EXAMPLE 4
ᮣ
Graphing a Hyperbola by Completing the Square
Graph the equation 9y2 Ϫ x2 ϩ 54y ϩ 4x ϩ 68 ϭ 0 by completing the square. Label the center
and vertices and sketch the asymptotes. Then graph the hyperbola on a graphing calculator and
use the TRACE feature with a “friendly” window to locate four additional points whose coordinates
are rational.
Solution
ᮣ
Since the y2-term occurs first, we assume the equation represents a vertical hyperbola, but wait for
the factored form to be sure (see Exercise 91).
9y2 Ϫ x2 ϩ 54y ϩ 4x ϩ 68 ϭ 0
9y2 ϩ 54y Ϫ x2 ϩ 4x ϭ Ϫ68
2
91y ϩ 6y ϩ ___ 2 Ϫ 11x2 Ϫ 4x ϩ ___ 2 ϭ Ϫ68
91y2 ϩ 6y ϩ 92 Ϫ 11x2 Ϫ 4x ϩ 42
ϭ Ϫ68 ϩ 81 ϩ 1Ϫ42
c
c
adds 9 19 2 ϭ 81
c
c
adds Ϫ1 14 2 ϭ Ϫ4
91y ϩ 32 2 Ϫ 11x Ϫ 22 2 ϭ 9
1x Ϫ 22 2
1y ϩ 32 2
Ϫ
ϭ1
1
9
1y ϩ 32 2
1x Ϫ 22 2
Ϫ
ϭ1
12
32
add 81 ϩ 1Ϫ4 2 to right
given
collect like-variable terms; subtract 68
factor out 9 from y-terms and Ϫ1 from x-terms
complete the square
factor S vertical hyperbola
divide by 9 (standard form)
write denominators in squared form
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CHAPTER 8 Analytic Geometry and the Conic Sections
The center of the hyperbola is 12, Ϫ32 with a ϭ 3, b ϭ 1, and a transverse axis of x ϭ 2. The
vertices are at 12, Ϫ3 ϩ 12 and 12, Ϫ3 Ϫ12 S 12, Ϫ22 and 12, Ϫ42 . After plotting the center and
vertices, we draw a rectangle centered at 12, Ϫ32 with a horizontal “width” of 2132 ϭ 6 and a
vertical “length” of 2112 ϭ 2 to sketch the asymptotes. The completed graph is given in Figure 8.32.
Figure 8.32
Vertical hyperbola
y
Center at (2, Ϫ3)
Vertices at (2, Ϫ2) and (2, Ϫ4)
m ϭ Ϫ3
1
(2, Ϫ2)
Transverse axis: x ϭ 2
Conjugate axis: y ϭ Ϫ3
x
1
mϭ3
(2, Ϫ3)
center
Width of rectangle
(horizontal dimension)
2a ϭ 2(3) ϭ 6
(2, Ϫ4)
Length of rectangle
vertical dimension and
distance between vertices
2b ϭ 2(1) ϭ 2
To graph the hyperbola on a calculator, we again solve for y.
91y ϩ 32 2 Ϫ 1x Ϫ 22 2 ϭ 9
factored form
91y ϩ 32 ϭ 9 ϩ 1x Ϫ 22
isolate term containing y
2
9 ϩ 1x Ϫ 22
1y ϩ 32 2 ϭ
divide by 9
9
9 ϩ 1x Ϫ 22 2
yϩ3ϭϮ
take square roots
B
9
9 ϩ 1x Ϫ 22 2
yϭϮ
subtract 3
Ϫ3
B
9
9 ϩ 1X Ϫ 22 2
9 ϩ 1X Ϫ 22 2
Y1 ϭ ϩ
Ϫ3
Ϫ 3, Y2 ϭ Ϫ
Figure 8.33
B
B
9
9
2
2
Using the friendly window shown, we can use the arrow
keys to TRACE though x-values and find the points
1Ϫ5.2, Ϫ0.42 and 19.2, Ϫ0.42 on the upper branch, with
1Ϫ5.2, Ϫ5.62 and 19.2, Ϫ5.62 on the lower branch. Note
how these points show that a hyperbola is symmetric to
its center, as well as the horizontal line and vertical line
through its center. The graph is shown in Figure 8.33.
A. You’ve just seen how
we can use the equation of a
hyperbola to graph central and
noncentral hyperbolas
3.2
Ϫ7.4
11.4
Ϫ9.2
Now try Exercises 45 through 48
ᮣ
B. Distinguishing between the Equations
of Circles, Ellipses, and Hyperbolas
So far we’ve explored numerous graphs of circles, ellipses, and hyperbolas. In Example 5 we’ll attempt to identify a given conic section from its equation alone (without
graphing the equation). As you’ve seen, the corresponding equations have unique characteristics that can help distinguish one from the other.
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Section 8.3 The Hyperbola
EXAMPLE 5
ᮣ
Identifying a Conic Section from Its Equation
Identify each equation as that of a circle, ellipse, or hyperbola. Justify your choice
and name the center, but do not draw the graphs.
a. y2 ϭ 36 ϩ 9x2
b. 4x2 ϭ 16 Ϫ 4y2
c. x2 ϭ 225 Ϫ 25y2
d. 25x2 ϭ 100 ϩ 4y2
2
2
e. 31x Ϫ 22 ϩ 41y ϩ 32 ϭ 12
f. 41x ϩ 52 2 ϭ 36 ϩ 91y Ϫ 42 2
Solution
B. You’ve just seen how
we can distinguish between
the equations of circles,
ellipses, and hyperbolas
ᮣ
a. Writing the equation as y2 Ϫ 9x2 ϭ 36 shows h ϭ 0 and k ϭ 0. Since the
equation contains a difference of second-degree terms, it is the equation of a
(vertical) hyperbola (A and B have opposite signs). The center is at (0, 0).
b. Rewriting the equation as 4x2 ϩ 4y2 ϭ 16 and dividing by 4 gives
x2 ϩ y2 ϭ 4. The equation represents a circle of radius 2 1A ϭ B2 , with the
center at (0, 0).
c. Writing the equation as x2 ϩ 25y2 ϭ 225 we note a sum of second-degree
terms with unequal coefficients. The equation is that of an ellipse 1A B2 ,
with the center at (0, 0).
d. Rewriting the equation as 25x2 Ϫ 4y2 ϭ 100 we note the equation contains a
difference of second-degree terms. The equation represents a central
(horizontal) hyperbola (A and B have opposite signs), whose center is at (0, 0).
e. The equation is in factored form and contains a sum of second-degree terms
with unequal coefficients. This is the equation of an ellipse 1A B2 with the
center at 12, Ϫ32 .
f. Rewriting the equation as 41x ϩ 52 2 Ϫ 91y Ϫ 42 2 ϭ 36 we note a difference of
second-degree terms. The equation represents a horizontal hyperbola (A and B
have opposite signs) with center 1Ϫ5, 42.
Now try Exercises 49 through 60
ᮣ
C. The Foci of a Hyperbola
Like the ellipse, the foci of a hyperbola play an important part in their application. A
long distance radio navigation system (called LORAN for short), can be used to determine the location of ships and airplanes and is based on the characteristics of a hyperbola (see Exercises 85 and 86). Hyperbolic mirrors are also used in some telescopes,
and have the property that a beam of light directed at one focus will be reflected to the
second focus. To understand and appreciate these applications, we use the analytic definition of a hyperbola:
Definition of a Hyperbola
Given two fixed points f1 and f2 in a plane, a hyperbola
is the set of all points (x, y) such that the distance d1
from f1 to (x, y) and the distance d2 from f2 to (x, y),
satisfy the equation
Ϳd1 Ϫ d2Ϳ ϭ k.
In other words, the difference of these two distances
is a positive constant.
The fixed points f1 and f2 are called the foci of the
hyperbola, and all such points (x, y) are on the graph
of the hyperbola.
y
d1
(x, y)
d2
f1
f2
|d1 Ϫ d2| ϭ k
k>0
x