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A. The Equation of a Hyperbola

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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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



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