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C. The Foci of an Ellipse

C. The Foci of an Ellipse

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Section 8.2 The Circle and the Ellipse



Figure 8.17

y

(0, b)

P(x, y)

(a, 0)

x



(Ϫa, 0)

(Ϫc, 0)



(c, 0)



calculating ease we use a central ellipse). Note the vertices have coordinates 1Ϫa, 02 and

(a, 0), and the endpoints of the minor axis have coordinates 10, Ϫb2 and (0, b) as

before. It is customary to assign foci the coordinates f1 S 1Ϫc, 02 and f2 S 1c, 02. We

can calculate the distance between (c, 0) and any point P(x, y) on the ellipse using the

distance formula:

21x Ϫ c2 2 ϩ 1y Ϫ 02 2



Likewise the distance between 1Ϫc, 02 and any point (x, y) is

21x ϩ c2 2 ϩ 1y Ϫ 02 2



(0, Ϫb)



According to the definition, the sum must be constant:

21x Ϫ c2 2 ϩ y2 ϩ 21x ϩ c2 2 ϩ y2 ϭ k

EXAMPLE 6







Finding the Value of k from the Definition of an Ellipse

Use the definition of an ellipse and the diagram given to determine the constant k

used for this ellipse (also see the following Worthy of Note). Note that

a ϭ 5, b ϭ 3, and c ϭ 4.

y

(0, 3)



P(3, 2.4)



(Ϫ5, 0)

(Ϫ4, 0)



(4, 0)



(5, 0)

x



(0, Ϫ3)



Solution







21x Ϫ c2 2 ϩ 1y Ϫ 02 2 ϩ 21x ϩ c2 2 ϩ 1y Ϫ 02 2 ϭ k



213 Ϫ 42 ϩ 12.4 Ϫ 02 ϩ 213 ϩ 42 ϩ 12.4 Ϫ 02 ϭ k

2



WORTHY OF NOTE

Note that if the foci are coincident

(both at the origin) the “ellipse” will

k

actually be a circle with radius ;

2

2x2 ϩ y2 ϩ 2x2 ϩ y2 ϭ k leads to

k2

x2 ϩ y2 ϭ . In Example 6 we

4

10

ϭ 5, and if

found k ϭ 10, giving

2

we used the “string” to draw the

circle, the pencil would be 5 units

from the center, creating a circle of

radius 5.



2



2



2



21Ϫ12 ϩ 2.4 ϩ 27 ϩ 2.4 ϭ k

16.76 ϩ 154.76 ϭ k

2.6 ϩ 7.4 ϭ k

10 ϭ k

The constant value for this ellipse is 10 units.

2



2



2



2



given

substitute

add

simplify radicals

compute square roots

result



Now try Exercises 45 through 48

In Example 6, the sum of the distances

could also be found by moving the point (x, y)

to the location of a vertex (a, 0), then using

the symmetry of the ellipse. The sum is identical to the length of the major axis, since the

overlapping part of the string from (c, 0) to

(a, 0) is the same length as from (Ϫa, 0) to

(Ϫc, 0) (see Figure 8.18). This shows the

constant k is equal to 2a regardless of the distance between foci.

As we noted, the result is







Figure 8.18

y

d1 ϩ d2 ϭ 2a

d1



d2



(Ϫa, 0)

(Ϫc, 0)



21x Ϫ c2 2 ϩ y2 ϩ 21x ϩ c2 2 ϩ y2 ϭ 2a



(c, 0)



(a, 0)

x



These two segments

are equal

substitute 2a for k



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The details for simplifying this expression are given in Appendix V, and the result

is very close to the standard form seen previously:

y2

x2

ϩ

ϭ1

a2

a2 Ϫ c2

y2

y2

x2

x2

ϩ

ϭ

1

ϩ

ϭ 1, we might

with

a2

b2

a2

a2 Ϫ c 2

suspect that b2 ϭ a2 Ϫ c2, and this is indeed the case. Note from Example 6 the relationship yields

By comparing the standard form



b2 ϭ a2 Ϫ c2

3 2 ϭ 52 Ϫ 42

9 ϭ 25 Ϫ 16

Additionally, when we consider that (0, b) is

Figure 8.19

a point on the ellipse, the distance from (0, b) to

y

(c, 0) must be equal to a due to symmetry (the

(0, b)

“constant distance” used to form the ellipse is always 2a). We then see in Figure 8.19, that

a

a

b2 ϩ c2 ϭ a2 (Pythagorean Theorem), yielding

b

(a, 0)

(Ϫa, 0)

2

2

2

b ϭ a Ϫ c as above.

x

(Ϫc, 0)

(c, 0)

With this development, we now have the

ability to locate the foci of any ellipse —an important step toward using the ellipse in practical

(0, Ϫb)

applications. Because we’re often asked to find

the location of the foci, it’s best to rewrite the relationship in terms of c2, using absolute value bars to allow for a major axis that is vertical: c2 ϭ Ϳa2 Ϫ b2Ϳ.

EXAMPLE 7







Completing the Square to Graph an Ellipse and Locate the Foci

For the ellipse defined by 25x2 ϩ 9y2 Ϫ 100x Ϫ 54y Ϫ 44 ϭ 0, find the

coordinates of the center, vertices, foci, and endpoints of the minor axis. Then

sketch the graph.



Solution







25x2 ϩ 9y2 Ϫ 100x Ϫ 54y Ϫ 44 ϭ 0

25x2 Ϫ 100x ϩ 9y2 Ϫ 54y ϭ 44

2

251x Ϫ 4x ϩ __ 2 ϩ 91y2 Ϫ 6y ϩ __ 2 ϭ 44

251x2 Ϫ 4x ϩ 42 ϩ 91y2 Ϫ 6y ϩ 92 ϭ 44 ϩ 100 ϩ 81

c



c

adds 25142 ϭ 100



c

adds 9192 ϭ 81



c



given

group terms; add 44

factor out lead coefficients



add 100 ϩ 81 to right-hand side



251x Ϫ 22 2 ϩ 91y Ϫ 32 2 ϭ 225

91y Ϫ 32 2

251x Ϫ 22 2

225

ϩ

ϭ

225

225

225

1x Ϫ 22 2

1y Ϫ 32 2

ϩ

ϭ1

9

25

2

2

1x Ϫ 22

1y Ϫ 32

ϩ

ϭ1

32

52



factored form

divide by 225



simplify (standard form)

write denominators

in squared form



The result shows a vertical ellipse with a ϭ 3 and b ϭ 5. The center of the ellipse

is at (2, 3). The vertices are a vertical distance of b ϭ 5 units from center at (2, 8)

and (2, Ϫ2). The endpoints of the minor axis are a horizontal distance of a ϭ 3

units from center at (Ϫ1, 3) and (5, 3). To locate the foci, we use the foci formula



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723



for an ellipse: c2 ϭ Ϳa2 Ϫ b2Ϳ, giving c2 ϭ Ϳ32 Ϫ 52Ϳ ϭ 16. This shows the foci “ ”

are located a vertical distance of 4 units from center at (2, 7) and (2, Ϫ1).

y (2, 8)



Vertical ellipse

Center at (2, 3)



(2, 7)



(Ϫ1, 3)



(2, 3)



(2, Ϫ1)

(2, Ϫ2)



Endpoints of major axis (vertices)

(2, 8) and (2, Ϫ2)

(5, 3)



x



Endpoints of minor axis

(Ϫ1, 3) and (5, 3)

Location of foci

(2, 7) and (2, Ϫ1)

Length of major axis: 2b ϭ 2(5) ϭ 10

Length of minor axis: 2a ϭ 2(3) ϭ 6



Now try Exercises 49 through 54







For an ellipse, a focal chord is a line segment perpendicular to the major axis,

through a focus and with endpoints on the ellipse. In the Exercise Set, you are asked

2m2

to verify that the focal chord of an ellipse has length L ϭ

, where m is the length

n

of the semiminor axis and n is the length of the semimajor axis. This means the

m2

distance from the foci to the graph (along a focal chord) is , a fact can often be used

n

to help graph an ellipse. For Example 7,

m ϭ 3 and n ϭ 5, so the horizontal distance

Figure 8.20

from focus to graph (in either direction) is

9.2

9

32

ϭ . From the upper focus (2, 7), we can

5

5

now graph the additional points 12 Ϫ 1.8, 72 ϭ

11.4

10.2, 72 and (2 ϩ 1.8, 7) ϭ 13.8, 72, and Ϫ7.4

from the lower focus 12, Ϫ12 we obtain

10.2, Ϫ12 and 13.8, Ϫ12 without having to

evaluate the original equation. Graphical verϪ3.2

ification is provided in Figure 8.20. Also see

Exercises 83 and 85.

For future reference, remember the foci of an ellipse always occur on the major

axis, with a 7 c and a2 7 c2 for a horizontal ellipse, with b 7 c and b2 7 c2 for a

vertical ellipse. This makes it easier to remember the foci formula for ellipses:

c2 ϭ Ϳa2 Ϫ b2Ϳ. If any two of the values for a, b, and c are known, the relationship

between them can be used to construct the equation of the ellipse.



EXAMPLE 8







Finding the Equation of an Ellipse

Find the equation of the ellipse (in standard form) that has foci at (0, Ϫ2) and (0, 2),

with a minor axis 6 units in length. Then graph the ellipse

a. By hand.

b. On a graphing calculator.

m2

c. Find the distance from foci to graph along a focal chord ausing

b, and use

n

the result to verify that the endpoints of both focal chords are all on the

graph.



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Solution







LOOKING AHEAD



Since the foci are on the y-axis and an equal distance from (0, 0), we know this is a

vertical and central ellipse with c ϭ 2 and c2 ϭ 4. The minor axis has a length of

2a ϭ 6 units, meaning a ϭ 3 and a2 ϭ 9. To find b2, use the foci equation and solve.

Figure 8.21

foci equation (ellipse)

c2 ϭ Ϳa2 Ϫ b2Ϳ

4 ϭ Ϳ9 Ϫ b2Ϳ

Ϫ4 ϭ 9 Ϫ b2

4 ϭ 9 Ϫ b2

b2 ϭ 13

b2 ϭ 5



For the hyperbola, we’ll find that

c 7 a, and the formula for the foci

of a hyperbola will be c2 ϭ a2 ϩ b2.



y



substitute



(0, √13)



solve the absolute value equation

result



2



Since we know b must be greater than a2 (the major

axis is always longer), b2 ϭ 5 can be discarded. The



(0, 2)

(Ϫ3, 0)



2



y

x2

ϩ

ϭ 1.

2

3

1 2132 2

a. The graph is shown in Figure 8.21.

b. For a calculator generated graph, begin by solving for y.



(3, 0)

(0, Ϫ2)



x



standard form is



(0, Ϫ√13)



y2

x2

ϩ

ϭ1

original equation

9

13

13x2 ϩ 9y2 ϭ 117

clear denominators

isolate y-term

9y2 ϭ 117 Ϫ 13x2

2

117 Ϫ 13x

y2 ϭ

divide by 9

9.4

9

117 Ϫ 13x2

yϭϮ

take square roots

B

9

117 Ϫ 13X2

117 Ϫ 13X2

Y1 ϭ ϩ

, Y2 ϭ Ϫ

B

9

B

9

The graph is shown in Figure 8.22.

c. From the discussion prior to Example 8, the horizontal distance from foci to graph

m2

9

must be

. Using the TRACE feature and entering x ϭ 9/ 213 verifies

ϭ

n

213



Figure 8.22

6.2



Ϫ9.4



Ϫ6.2



Figure 8.23

6.2



Ϫ9.4



9.4



that a



9



, 2b is a point on the graph (Figure 8.23), and that aϪ



9



, 2b,

213

213

9

9



, Ϫ2b, and a

, Ϫ2b must also be on the graph due to symmetry.

213

213



Ϫ6.2



C. You’ve just seen how we

can locate the foci of an ellipse

and use the foci and other

features to write the equation



Now try Exercises 55 through 62







D. Applications Involving Foci

Applications involving the foci of a conic section can take various forms. In many

cases, only partial information about the conic section is available and the ideas from

Example 8 must be used to “fill in the gaps.” In other applications, we must rewrite a

known or given equation to find information related to the values of a, b, and c.



EXAMPLE 9







Solving Applications Using the Characteristics of an Ellipse

In Washington, D.C., there is a park called the Ellipse located between the White

House and the Washington Monument. The park is surrounded by a path that forms

an ellipse with the length of the major axis being about 1502 ft and the minor axis

having a length of 1280 ft. Suppose the park manager wants to install water

fountains at the location of the foci. Find the distance between the fountains

rounded to the nearest foot.



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Solution







725



Since the major axis has length 2a ϭ 1502, we know a ϭ 751 and a2 ϭ 564,001.

The minor axis has length 2b ϭ 1280, meaning b ϭ 640 and b2 ϭ 409,600. To find

c, use the foci equation:

c2 ϭ a2 Ϫ b2

ϭ 564,001 Ϫ 409,600

ϭ 154,401

c Ϸ Ϫ393 and c Ϸ 393



since we know a 7 b

substitute

subtract

square root property



The distance between the water fountains would be 213932 ϭ 786 ft.



D. You’ve just seen how

we can solve applications

involving the foci



Now try Exercises 65 through 80







8.2 EXERCISES





CONCEPTS AND VOCABULARY



Fill in the blank with the appropriate word or phrase. Carefully reread the section if needed.



1. For an ellipse, the relationship between a, b, and c

is given by the foci equation

, since

c 6 a or c 6 b.



2. The greatest distance across an ellipse is called the

and the endpoints are called

.



3. For a vertical ellipse, the length of the minor axis is

and the length of the major axis is

.



4. To write the equation 2x2 ϩ y2 Ϫ 6x ϭ 7 in

standard form,

the

in x.



5. Explain/Discuss how the relations a 7 b, a ϭ b

and a 6 b affect the graph of a conic section with



6. Suppose foci are located at (Ϫ3, 2) and (5, 2).

Discuss/Explain the conditions necessary for the

graph to be an ellipse.



equation





1x Ϫ h2 2

a2



ϩ



1y Ϫ k2 2

b2



ϭ 1.



DEVELOPING YOUR SKILLS



Find an equation of the circle satisfying the conditions

given, then graph the result on a graphing calculator

and locate two additional points on the graph.



15. x2 ϩ y2 Ϫ 4x ϩ 10y ϩ 4 ϭ 0

16. x2 ϩ y2 ϩ 4x ϩ 6y Ϫ 3 ϭ 0



7. center (0, 0), radius 7



17. x2 ϩ y2 ϩ 6x Ϫ 5 ϭ 0



8. center (0, 0), radius 9



18. x2 ϩ y2 Ϫ 8y Ϫ 5 ϭ 0



9. center (5, 0), radius 13



Sketch the graph of each ellipse.



19.



1x Ϫ 12 2



20.



1x Ϫ 32 2



21.



1x Ϫ 22 2



22.



1x ϩ 52 2



10. center (0, 4), radius 15

11. diameter has endpoints (4, 9) and (Ϫ2, 1)

12. diameter has endpoints (Ϫ2, Ϫ32 , and (3, 9)

Write each equation in standard form to identify the

center and radius of the circle, then sketch its graph.



13. x2 ϩ y2 Ϫ 12x Ϫ 10y ϩ 52 ϭ 0

14. x2 ϩ y2 ϩ 8x Ϫ 6y Ϫ 11 ϭ 0



9

4



25

1



ϩ



1y Ϫ 22 2



ϩ



1y Ϫ 12 2



ϩ



1y ϩ 32 2



ϩ



1y Ϫ 22 2



16

25

4



16



ϭ1

ϭ1

ϭ1

ϭ1



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23.



1x ϩ 12 2



24.



1x ϩ 12



16



ϩ



1y ϩ 22 2



ϩ



1y ϩ 32



2



36



47.



ϭ1



9



48.



(0, b) y

(0, 8)



(0, b) y



(4.8, 6)

(0, 28)



2



ϭ1



9



(6, 0) x



(Ϫ6, 0)



(96, 0) x



(Ϫ96, 0)

(0, Ϫ28 )



For each exercise, (a) write the equation in standard form,

then identify the center and the values of a and b, (b) state

the coordinates of the vertices and the coordinates of the

endpoints of the minor axis, (c) sketch the graph, and

(d) for 25–28 (only) graph the relations on a graphing

calculator and identify four additional points on the graph

whose coordinates are rational.



(0, Ϫ8)



(76.8, Ϫ60)



(0, Ϫb)



(0, Ϫb)



Find the coordinates of the (a) center, (b) vertices,

(c) foci, and (d) endpoints of the minor axis. Then

(e) sketch the graph.



49. 4x2 ϩ 25y2 Ϫ 16x Ϫ 50y Ϫ 59 ϭ 0



25. x2 ϩ 4y2 ϭ 16



26. 9x2 ϩ y2 ϭ 36



50. 9x2 ϩ 16y2 Ϫ 54x Ϫ 64y ϩ 1 ϭ 0



27. 16x2 ϩ 9y2 ϭ 144



28. 25x2 ϩ 9y2 ϭ 225



51. 25x2 ϩ 16y2 Ϫ 200x ϩ 96y ϩ 144 ϭ 0



29. 2x2 ϩ 5y2 ϭ 10



30. 3x2 ϩ 7y2 ϭ 21



52. 49x2 ϩ 4y2 ϩ 196x Ϫ 40y ϩ 100 ϭ 0

53. 6x2 ϩ 24x ϩ 9y2 ϩ 36y ϩ 6 ϭ 0



Identify each equation as that of an ellipse or circle,

then sketch its graph.



54. 5x2 Ϫ 50x ϩ 2y2 Ϫ 12y ϩ 93 ϭ 0



31. 1x ϩ 12 ϩ 41y Ϫ 22 ϭ 16

2



2



32. 91x Ϫ 22 2 ϩ 1y ϩ 32 2 ϭ 36



Find the equation of the ellipse (in standard form) that

satisfies the following conditions. Then (a) graph the

ellipse by hand, (b) confirm your graph by graphing the

ellipse on a graphing calculator, and (c) find the length

of the focal chords and verify the endpoints of the

chords are on the graph.



33. 21x Ϫ 22 2 ϩ 21y ϩ 42 2 ϭ 18

34. 1x Ϫ 62 2 ϩ y2 ϭ 49



35. 41x Ϫ 12 2 ϩ 91y Ϫ 42 2 ϭ 36

36. 251x Ϫ 32 2 ϩ 41y ϩ 22 2 ϭ 100

Complete the square in both x and y to write each

equation in standard form. Then draw a complete graph

of the relation and identify all important features,

including the domain and range.



37. 4x2 ϩ y2 ϩ 6y ϩ 5 ϭ 0



55. vertices at (Ϫ6, 0) and (6, 0);

foci at (Ϫ4, 0) and (4, 0)

56. vertices at (Ϫ8, 0) and (8, 0);

foci at (Ϫ5, 0) and (5, 0)

57. foci at (3, Ϫ6) and (3, 2);

length of minor axis: 6 units

58. foci at (Ϫ4, Ϫ3) and (8, Ϫ3);

length of minor axis: 8 units



38. x2 ϩ 3y2 ϩ 8x ϩ 7 ϭ 0

39. x2 ϩ 4y2 Ϫ 8y ϩ 4x Ϫ 8 ϭ 0



Use the characteristics of an ellipse and the graph given to

write the related equation and find the location of the foci.



40. 3x2 ϩ y2 Ϫ 8y ϩ 12x Ϫ 8 ϭ 0

41. 5x2 ϩ 2y2 ϩ 20y Ϫ 30x ϩ 75 ϭ 0



59.



60.



y



y



42. 4x ϩ 9y Ϫ 16x ϩ 18y Ϫ 11 ϭ 0

2



2



43. 2x2 ϩ 5y2 Ϫ 12x ϩ 20y Ϫ 12 ϭ 0

44. 6x2 ϩ 3y2 Ϫ 24x ϩ 18y Ϫ 3 ϭ 0



x



x



Use the definition of an ellipse to find the constant k for

each ellipse (figures are not drawn to scale).



45.



46.



y

(0, 8)

(6, 6.4)

(Ϫa, 0)

(Ϫ6, 0)



(0, Ϫ8)



61.



y



62.



y



y



(0, 12)



(6, 0)



(Ϫ9, 9.6)



(a, 0) (Ϫa, 0)

x



(Ϫ9, 0)



(0, Ϫ12)



(9, 0)



(a, 0)

x



x



x



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WORKING WITH FORMULAS



63. Area of an Ellipse: A ‫␲ ؍‬ab

The area of an ellipse is given by the formula

shown, where a is the distance from the center to

the graph in the horizontal direction and b is the

distance from center to graph in the vertical

direction. Find the area of the ellipse defined by

16x2 ϩ 9y2 ϭ 144.





727



a2 ؉ b2

B 2

The perimeter of an ellipse can be approximated by

the formula shown, where a represents the length

of the semimajor axis and b represents the length

of the semiminor axis. Find the perimeter of the

y2

x2

ellipse defined by the equation

ϩ

ϭ 1.

49

4



64. The Perimeter of an Ellipse: P ‫ ؍‬2␲



APPLICATIONS



65. Decorative fireplaces: A bricklayer intends to

build an elliptical fireplace 3 ft high and 8 ft wide,

with two glass doors that open at the middle. The

hinges to these doors are to be screwed onto a spine

that is perpendicular to the hearth and goes through

the foci of the ellipse. How far from center will the

spines be located? How tall will each spine be?

8 ft



68. Medical procedures: The medical procedure called

lithotripsy is a noninvasive medical procedure that

is used to break up kidney and bladder stones in the

body. A machine called a lithotripter uses its

three-dimensional semielliptical shape and the foci

properties of an ellipse to concentrate shock waves

generated at one focus, on a kidney stone located at

the other focus (see diagram—not drawn to scale). If

the lithotripter has a length (semimajor axis) of 16 cm

and a radius (semiminor axis) of 10 cm, how far

from the vertex should a kidney stone be located for

the best result? Round to the nearest hundredth.



3 ft



Exercise 68

Vertex

Spines



Focus

Lithotripter



66. Decorative gardens: A retired math teacher

decides to present her husband with a beautiful

elliptical garden to help celebrate their 50th

anniversary. The ellipse is to be 8 m long and 5 m

across, with decorative fountains located at the

foci. How far from the center of the ellipse should

the fountains be located (round to the nearest 100th

of a meter)? How far apart are the fountains?

67. Attracting attention to art: As part of an art

show, a gallery owner asks a student from the local

university to design a unique exhibit that will

highlight one of the more significant pieces in the

collection, an ancient sculpture. The student

decides to create an elliptical showroom with

reflective walls, with a rotating laser light on a

stand at one focus, and the sculpture placed at the

other focus on a stand of equal height. The laser

light then points continually at the sculpture as it

rotates. If the elliptical room is 24 ft long and 16 ft

wide, how far from the center of the ellipse should

the stands be located (round to the nearest 10th of a

foot)? How far apart are the stands?



Exercise 69

69. Elliptical arches: In some

situations, bridges are built

using uniform elliptical

8 ft

archways as shown in the

60 ft

figure given. Find the

equation of the ellipse forming each arch if it has a

total width of 30 ft and a maximum center height

(above level ground) of 8 ft. What is the height of a

point 9 ft to the right of the center of each arch?



70. Elliptical arches: An elliptical arch bridge is built

across a one-lane highway. The arch is 20 ft across

and has a maximum center height of 12 ft. Will a

farm truck hauling a load 10 ft wide with a clearance

height of 11 ft be able to go under the bridge

without damage? (Hint: See Exercise 69.)



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CHAPTER 8 Analytic Geometry and the Conic Sections



8–22



71. Plumbing: By allowing the free

flow of air, a properly vented

home enables water to run freely

throughout its plumbing system,

while helping to prevent sewage

gases from entering the home.

Find the equation of the elliptical hole cut in a roof

in order to allow a 3-in. vent pipe to exit, if the roof

4

has a slope of .

3

72. Light projection:

Standing a short

distance from a wall,

Kymani’s flashlight

projects a circle of

radius 30 cm. When

holding the flashlight at

an angle, a vertical

ellipse 50 cm long is

formed, with the focus

10 cm from the vertex

(see Worthy of Note,

page 717). Find the

equation of the circle

and ellipse, and the area of the wall that each

illuminates.



75. Planetary orbits: Except for small variations, a

planet’s orbit around the Sun is elliptical with the

Sun at one focus. The aphelion (maximum distance

from the Sun) of the planet Mars is approximately

156 million miles, while the perihelion (minimum

distance from the Sun) of Mars is about 128 million

miles. Use this information to find the lengths of the

semimajor and semiminor axes, rounded to the

nearest million. If Mars has an orbital velocity of

54,000 miles per hour (1.296 million miles per day),

how many days does it take Mars to orbit the Sun?

(Hint: Use the formula from Exercise 64.)



As a planet orbits around the Sun, it traces out an

ellipse. If the center of the ellipse were placed at (0, 0)

on a coordinate grid, the Sun would be actually offcentered (located at the focus of the ellipse). Use this

information and the graphs provided to complete

Exercises 73 through 78.



79. Area of a race track: Suppose the Toronado 500 is a

car race that is run on an elliptical track. The track is

bounded by two ellipses with equations of

4x2 ϩ 9y2 ϭ 900 and 9x2 ϩ 25y2 ϭ 900, where x

and y are in hundreds of yards. Use the formula given

in Exercise 63 to find the area of the race track.



y



Sun

x



70.5 million miles



Mercury



72 million miles



Exercise 74

y

Pluto



Sun

x



3650 million miles



3540 million miles



74. Orbit of Pluto: The

approximate orbit of the

Kuiper object formerly

known as Pluto is shown in

the figure given. Find an

equation that models this

orbit.



77. Orbital velocity of Earth: The planet Earth has a

perihelion (minimum distance from the Sun) of about

91 million mi, an aphelion (maximum distance from

the Sun) of close to 95 million mi, and completes one

orbit in about 365 days. Use this information and the

formula from Exercise 64 to find Earth’s orbital

speed around the Sun in miles per hour.

78. Orbital velocity of Jupiter: The planet Jupiter has

a perihelion of 460 million mi, an aphelion of

508 million mi, and completes one orbit in about

4329 days. Use this information and the formula

from Exercise 64 to find Jupiter’s orbital speed

around the Sun in miles per hour.



Exercise 73



73. Orbit of Mercury: The

approximate orbit of the

planet Mercury is shown

in the figure given. Find

an equation that models

this orbit.



76. Planetary orbits: The aphelion (maximum distance

from the Sun) of the planet Saturn is approximately

940 million miles, while the perihelion (minimum

distance from the Sun) of Saturn is about 840 million

miles. Use this information to find the lengths of the

semimajor and semiminor axes, rounded to the

nearest million. If Saturn has an orbital velocity of

21,650 miles per hour (about 0.52 million miles per

day), how many days does it take Saturn to orbit the

Sun? How many years?



Exercise 80

80. Area of a border: The

tablecloth for a large oval table

is elliptical in shape. It is

designed with two concentric

ellipses (one within the other)

as shown in the figure. The

equation of the outer ellipse is 9x2 ϩ 25y2 ϭ 225,

and the equation of the inner ellipse is

4x2 ϩ 16y2 ϭ 64 with x and y in feet. Use the

formula given in Exercise 63 to find the area of the

border of the tablecloth.



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8–23



Mid-Chapter Check



729



81. Whispering galleries: Due to their unique properties, ellipses are

used in the construction of whispering galleries like those in

St. Paul’s Cathedral (London) and Statuary Hall in the U.S.

Capitol. Regarding the latter, it is known that John Quincy Adams

(1767–1848), while a member of the House of Representatives,

situated his desk at a focal point of the elliptical ceiling, easily

eavesdropping on the private conversations of other House

members located near the other focal point. Suppose a whispering gallery was built using the equation

y2

x2

ϩ

ϭ 1, with the dimensions in feet. (a) How tall is the ceiling at its highest point? (b) How wide is

2809

2025

the gallery vertex to vertex? (c) How far from the base of the doors at either end, should a young couple stand so

that one can clearly hear the other whispering, “I love you.”?

82. While an elliptical billiard table has little practical value, it offers an excellent illustration of elliptical properties.

A ball placed at one focus and hit with the cue stick from any angle, will hit the cushion and immediately

rebound to the other focus and continue through each focus until coming to rest. Suppose one such table was

y2

x2

ϭ 1 as a model, with the dimensions in feet. (a) How far apart are the

constructed using the equation ϩ

9

4

vertices? (b) How far apart are the foci? As a side note, Lewis Carroll (1832–1898) did invent a game of circular

billiards, complete with rules.





EXTENDING THE CONCEPT



83. For 6x2 ϩ 36x ϩ 3y2 Ϫ 24y ϩ 74 ϭ Ϫ28, does

the equation appear to be that of a circle, ellipse, or

parabola? Write the equation in factored form.

What do you notice? What can you say about the

graph of this equation?





84. Algebraically verify that for the ellipse

y2

x2

ϩ

ϭ 1 with b 7 a, the length of the focal

a2

b2

2a2

chord is still

.

b



MAINTAINING YOUR SKILLS



85. (5.4) Evaluate the expression using the change-ofbase formula: log320.

z1

86. (3.1) Compute the product z1z2 and quotient of:

z2

z1 ϭ 213 ϩ 2i13; z2 ϭ 5 13 Ϫ 5i

87. (2.3) Solve the absolute value inequality

(a) graphically and (b) analytically:

Ϫ2Ϳx Ϫ 3Ϳ ϩ 10 7 4.



88. (2.6) The resistance R to current flow in an

electrical wire varies directly as the length L of the

wire and inversely as the square of its diameter d.

(a) Write the equation of variation; (b) find the

constant of variation if a wire 2 m long with

diameter d ϭ 0.005 m has a resistance of 240 ohms

(⍀); and (c) find the resistance in a similar wire

3 m long and 0.006 m in diameter.



MID-CHAPTER CHECK

Sketch the graph of each conic section.



1. 1x Ϫ 42 2 ϩ 1y ϩ 32 2 ϭ 9



2. x2 ϩ y2 Ϫ 10x ϩ 4y ϩ 4 ϭ 0

3.



1x Ϫ 22 2

16



ϩ



1y ϩ 32 2

1



ϭ1



4. 9x2 ϩ 4y2 ϩ 18x Ϫ 24y ϩ 9 ϭ 0



5.



1x ϩ 32 2

9



ϩ



1y Ϫ 42 2

4



ϭ1



6. 9x2 ϩ 16y2 Ϫ 36x ϩ 96y ϩ 36 ϭ 0

7. Find the equation for all points located an equal

distance from the point (0, 3) and the line

y ϭ Ϫ3.



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8–24



CHAPTER 8 Analytic Geometry and the Conic Sections



9. Find the equation of the ellipse having foci at

(0, 13) and 10, Ϫ132 , with a minor axis of length

10 units.



8. Find the equation of each relation and state its

domain and range.

a.



(Ϫ3, 5)



(Ϫ5, 1)



b.



y



y

10

8

6

4

(Ϫ1, 2) 2



5

4

3

2

(Ϫ1, 1) 1



Ϫ5Ϫ4Ϫ3Ϫ2Ϫ1

Ϫ1

Ϫ2

Ϫ3

(Ϫ3, Ϫ3)Ϫ4

Ϫ5



Ϫ10Ϫ8Ϫ6Ϫ4Ϫ2

Ϫ2

Ϫ4

Ϫ6

Ϫ8

Ϫ10



1 2 3 4 5 x



(3, 6)

(7, 2)

2 4 6 8 10 x



10. Find the equation of the ellipse (in standard form)

if the vertices are (Ϫ4, 0) and (4, 0) and the

distance between the foci is 4 13 units.



(3, Ϫ2)



REINFORCING BASIC CONCEPTS

More on Completing the Square

From our work so far in Chapter 8, we realize the process of completing the square has much greater use than simply as a tool

for working with quadratic equations. It is a valuable tool in the application of the conic sections, as well as other areas. The

purpose of this Reinforcing Basic Concepts is to strengthen the ability and confidence needed to apply the process correctly.

This is important because in some cases the values of a and b are rational or irrational numbers. No matter what the context,

1. The process begins with a coefficient of 1. For 20x2 ϩ 120x ϩ 27y2 Ϫ 54y ϩ 192 ϭ 0, we recognize the

equation of an ellipse, since the coefficients of the squared terms are positive and unequal. To study or graph this

ellipse, we’ll use the standard form to identify the values of a, b, and c. Grouping the like-variable terms gives

120x2 ϩ 120x



2 ϩ 127y2 Ϫ 54y



2 ϩ 192 ϭ 0



and to complete the square, we factor out the lead coefficient of each group (to get a coefficient of 1):

201x2 ϩ 6x



2 ϩ 271y2 Ϫ 2y



2 ϩ 192 ϭ 0



Subtracting 192 from both sides brings us to the fundamental step for completing the square.

2

1

2. The quantity a # linear cofficientb will complete a trinomial square. For this example we obtain

2

2

2

1

1

a # 6b ϭ 9 for x, and a # Ϫ2b ϭ 1 for y, with these numbers inserted in the appropriate group:

2

2

201x2 ϩ 6x ϩ 92 ϩ 271y2 Ϫ 2y ϩ 12 ϭ Ϫ192



complete the square



Due to the distributive property, we have in effect added 20 # 9 ϭ 180 and 27 # 1 ϭ 27 (for a total of 207) to the left

side of the equation:

201x2 ϩ 6x ϩ 92 ϩ 271y2 Ϫ 2y ϩ 12 ϭ Ϫ192

adds 20 # 9 ϭ 180

adds 27 # 1 ϭ 27

to left side



to left side



This brings us to the final step.

3. Keep the equation in balance. Since the left side was increased by 207, we also increase the right side by 207.

201x2 ϩ 6x ϩ 92 ϩ 271y2 Ϫ 2y ϩ 12 ϭ Ϫ192 ϩ 207

adds 20 # 9 ϭ 180

adds 27 # 1 ϭ 27

add 180 ϩ 27 ϭ 207

to left side



to left side



to right side



The quantities in parentheses factor, giving 201x ϩ 32 ϩ 271y Ϫ 12 ϭ 15. We then divide by 15 and simplify, obtaining

2



41x ϩ 32 2



91y Ϫ 12 2



ϭ 1. Note the coefficient of each binomial square is not 1, even after setting

3

5

the equation equal to 1. In the Strengthening Core Skills feature of this chapter, we’ll look at how to write equations of

this type in standard form to obtain the values of a and b. For now, practice completing the square using these exercises.

the standard form



ϩ



2



Exercise 1: 100x2 Ϫ 400x ϩ 18y2 Ϫ 108y ϩ 554 ϭ 0

Exercise 2: 28x2 Ϫ 56x ϩ 48y2 ϩ 192y ϩ 195 ϭ 0



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8.3



The Hyperbola



LEARNING OBJECTIVES

In Section 8.3 you will see

how we can:



A. Use the equation of a

hyperbola to graph

central and noncentral

hyperbolas

B. Distinguish between the

equations of circles,

ellipses, and hyperbolas

C. Locate the foci of a

hyperbola and use the

foci and other features to

write its equation

D. Solve applications

involving foci



EXAMPLE 1







As seen in Section 8.1 (see Figure 8.24), a hyperbola

is a conic section formed by a plane that cuts both

nappes of a right circular cone. A hyperbola has two Axis

symmetric parts called branches, which open in

opposite directions. Although the branches appear

to resemble parabolas, we will soon discover they

are actually a very different curve.



Figure 8.24



Hyperbola



A. The Equation of a Hyperbola

In Section 8.2, we noted that for the equation Ax2 ϩ By2 ϭ F,

if A ϭ B, the equation is that of a circle, if A B, the equation represents an ellipse. Both cases contain a sum of second-degree terms. Perhaps

driven by curiosity, we might wonder what happens if the equation has a difference of

second-degree terms. Consider the equation 9x2 Ϫ 16y2 ϭ 144. It appears the graph

will be centered at (0, 0) since no shifts are applied (h and k are both zero). Using the intercept method to graph this equation reveals an entirely new curve, called a hyperbola.

Graphing a Central Hyperbola

Graph the equation 9x2 Ϫ 16y2 ϭ 144 using intercepts and additional points

as needed.



Solution



9x2 Ϫ 16y2 ϭ 144

9102 2 Ϫ 16y2 ϭ 144

Ϫ16y2 ϭ 144

y2 ϭ Ϫ9







given

substitute 0 for x

simplify

divide by Ϫ16



Since y2 can never be negative, we conclude that the graph has no y-intercepts.

Substituting y ϭ 0 to find the x-intercepts gives

9x2 Ϫ 16y2 ϭ 144

9x2 Ϫ 16102 2 ϭ 144

9x2 ϭ 144

x2 ϭ 16

x ϭ 116 and x ϭ Ϫ 116

x ϭ 4 and x ϭ Ϫ4

(4, 0) and 1Ϫ4, 02



given

substitute 0 for y

simplify

divide by 9

square root property

simplify



x-intercepts



Knowing the graph has no y-intercepts, we select inputs greater than 4 and less

than Ϫ4 to help sketch the graph. Using x ϭ 5 and x ϭ Ϫ5 yields

9x2 Ϫ 16y2 ϭ 144

9152 2 Ϫ 16y2 ϭ 144

91252 Ϫ 16y2 ϭ 144

225 Ϫ 16y2 ϭ 144

Ϫ16y2 ϭ Ϫ81

81

y2 ϭ

16

9

9



yϭϪ

4

4

y ϭ 2.25 y ϭ Ϫ2.25

15, 2.252 15, Ϫ2.252

8–25



given

substitute for x

5 ϭ 1Ϫ52 2 ϭ 25

2



simplify

subtract 225



9x2 Ϫ 16y2

91Ϫ52 2 Ϫ 16y2

91252 Ϫ 16y2

225 Ϫ 16y2

Ϫ16y2



divide by Ϫ16



square root property

decimal form

ordered pairs



9

4

y ϭ 2.25

1Ϫ5, 2.252





ϭ 144

ϭ 144

ϭ 144

ϭ 144

ϭ Ϫ81

81

y2 ϭ

16

9

yϭϪ

4

y ϭ Ϫ2.25

1Ϫ5, Ϫ2.252

731



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