670 CHAPTER 9 Analytic Geometry

Polynomial equations define parabolas whenever they involve two vari-ables that are quadratic in one variable and linear in the other. To discuss thistype of equation, we first complete the square of the variable that is quadratic.

EXAMPLE 9 Discussing the Equation of a Parabola

Figure 15X2 + 4x - 4y = 0

Axis ofsymmetryx= -2

Discuss the equation: X2 + 4x - 4y = 0

To discuss the equation X2 + 4x - 4Y = 0, we complete the square involv-ing the variable x.

X2 + 4x - 4y = 0X2 + 4x = 4y Isolate the terms involving x on the left side.

X2 + 4x + 4 = 4y + 4 Complete the square on the left side.

(x + 2)2 = 4(y + 1) Factor.

This equation is of the form (x - h? = 4a(y - k), with h = -2, k = -1,and a = 1. The graph is a parabola with vertex at (h, k) = (-2, -1) thatopens up. The focus is at (-2,0), and the directrix is the line y = -2. SeeFigure 15. •

Solution

y

4

4 x

_3D:y=-2

QrnC=::;;;>-- NOW WORK PROBLEM 41.

2..J Parabolas find their way into many applications. For example, as we dis-cussed in Section 3.1,suspension bridges have cables in the shape of a parabo-la.Another property of parabolas that is used in applications is their reflectingproperty.

Reflecting PropertySuppose that a mirror is shaped like a paraboloid of revolution, a surfaceformed by rotating a parabola about its axis of symmetry. If a light (or anyother emitting source) is placed at the focus of the parabola, all the rays em-anating from the light will reflect off the mirror in lines parallel to the axisof symmetry. This principle is used in the design of searchlights, flashlights,certain automobile headlights, and other such devices. See Figure 16.

Conversely, suppose that rays of light (or other signals) emanate from adistant source so that they are essentially parallel. When these rays strikethe surface of a parabolic mirror whose axis of symmetry is parallel to theserays, they are reflected to a single point at the focus. This principle is used inthe design of some solar energy devices, satellite dishes, and the mirrors usedin some types of telescopes. See Figure 17.

Figure 16Searchlight

Figure 17Telescope

~Light at --~~focus I L-~I _---,-

i,> -~-_~

SECTION 9.2 The Parabola 671

EXAMPLE 10 Satellite Dish

A satellite dish is shaped like a paraboloid of revolution. The signals thatemanate from a satellite strike the surface of the dish and are reflected to asingle point, where the receiver is located. If the dish is 8 feet across at itsopening and is 3 feet deep at its center, at what position should the receiverbe placed?

Sol uti 0 n Figure 18(a) shows the satellite dish. We draw the parabola used to form thedish on a rectangular coordinate system so that the vertex of the parabola isat the origin and its focus is on the positive y-axis. See Figure 18(b).

Figure 18

y1--8·'-+----+1

(-4,3) 4 (4,3)32 F=(O,a)

1

2 3 4 x

(a) (b)

The form of the equation of the parabola is

X2 = 4ay

and its focus is at (0, a). Since (4,3) is a point on the graph, we have

42 = 4a(3)4

a=-3

~:y~:~~~.ShOUld be located 1~ feet from the base of the dish, along its a=Cl!l!=====>'- NOW WORK PROBLEM 57.

9.2 Concepts and VocabularyIn Problems 1-3, Jill in the blanks.

1. A(n) is the collection of all points in the plane such that the distance from each point to a fixed point equalsits distance to a fixed line.

2. The surface formed by rotating a parabola about its axis of symmetry is called a _3. The line segment joining the two points on a parabola above and below its focus is called the _

In Problems 4-6, answer True or False to each statement.

4. The vertex of a parabola is a point on the parabola that also is on its axis of symmetry.5. If a light is placed at the focus of a parabola, all the rays reflected off the parabola will be parallel to the axis of

symmetry.6. The graph of a quadratic function is a parabola.

In Problems 31-48, find the vertex, focus, and directrix of each parabola. Graph the equation (a) by hand and (b) by using agraphing utility.

31. x2 = 4y

35. (y - 2)2 = 8(x + 1)

39. (y + 3)2 = 8(x - 2)

43. X2 + 8x = 4Y - 8

47. x2 - 4x = Y + 4

17. Focus at (0, - 3); vertex at (0,0)

19. Focus at (-2,0); directrix the line x = 2

21. Directrix the line y = -~; vertex at (0,0)

'23. Vertex at (2, -3); focus at (2, -5)

25. Vertex at (0,0); axis of symmetry the y-axis;containing the point (2,3)

27. Focus at (- 3,4); directrix the line y = 2

29. Focus at (-3, -2); directrix the line x = 1

32. y2 = 8x

36. (x + 4)2 = 16(y + 2)

40. (x - 2)2 = 4(y - 3)

44. i -2Y = 8x - 1

48. y2 + 12y = -x + 1

In Problems 49-56, write an equation for each parabola.

49. 50.y

2

(0,1)

-2 -2x

-2 -2

53. y 54. Y2 2

(2,2)

(0,1) (0,1)

-2 2 x -2

(1, -1)-2 -2

SECTION 9.2 The Parabola 673

18. Focus at (-4,0); vertex at (0,0)

20. Focus at (0, -1); directrix the line y = 1

22. Directrix the line x = -~; vertex at (0,0)

24. Vertex at (4,-2); focus at (6, -2)

26. Vertex at (0,0); axis of symmetry the x-axis;containing the point (2,3)

28. Focus at (2,4); directrix the line x = -4

30. Focus at ( -4,4); directrix the line y = -2

33. i = -16x

37. (x - 3)2 = -(y + L)

41. i -4y + 4x + 4 = 0

45. i + 2y - x = 0

34. X2 = -4y

38. (y + 1? = -4(x - 2)

42. X2 + 6x - 4Y + 1 = 0

46. X2 - 4x = 2y

51. 52. y

2

x -2 x

-2(0, -1)

-2

55. 56.

2 x -2x x

-2 -2

57. Satellite Dish A satellite dish is shaped like a paraboloidof revolution. The signals that emanate from a satellitestrike the surface of the dish and are reflected to a singlepoint, where the receiver is located. If the dish is 10 feetacross at its opening and is 4 feet deep at its center, atwhat position should the receiver be placed?

58. Constructing a TV Dish A cable TV receiving dish is inthe shape of a paraboloid of revolution. Find the location

of the receiver, which is placed at the focus, if the dish is6 feet across at its opening and 2 feet deep.

59. Constructing a Flashlight The reflector of a flashlight isin the shape of a paraboloid of revolution. Its diameter is4 inches and its depth is 1 inch. How far from the vertexshould the light bulb be placed so that the rays will be re-flected parallel to the axis?

674 CHAPTER 9 Analytic Geometry

60. Constructing a Headlight A sealed-beam headlight is inthe shape of a paraboloid of revolution. The bulb, whichis placed at the focus, is 1 inch from the vertex. If the depthis to be 2 inches, what is the diameter of the headlight atits opening?

61. Suspension Bridge The cables of a suspension bridgeare in the shape of a parabola, as shown in the figure. Thetowers supporting the cable are 600 feet apart and 80 feethigh. If the cables touch the road surface midway betweenthe towers, what is the height of the cable at a point 150feet from the center of the bridge?

62. Suspension Bridge The cables of a suspension bridgeare in the shape of a parabola. The towers supporting thecable are 400 feet apart and 100 feet high. If the cablesare at a height of 10 feet midway between the towers, whatis the height of the cable at a point 50 feet from the cen-ter of the bridge?

63. Searchlight A searchlight is shaped like a paraboloid ofrevolution. If the light source is located 2 feet from thebase along the axis of symmetry and the opening is 5 feetacross, now deep should the searchlight be?

64. Searchlight A searchlight is shaped like a paraboloid ofrevolution. If the light source is located 2 feet from the basealong the axis of symmetry and the depth of the search-light is 4 feet, what should the width of the opening be?

65. Solar Heat A mirror is shaped like a paraboloid of rev-olution and will be used to concentrate the rays of the sunat its focus, creating a heat source. (See the figure.) If themirror is 20 feet across at its opening and is 6 feet deep,where will the heat source be concentrated?

Sun'srays.

66. Reflecting Telescope A reflecting telescope contains amirror shaped like a paraboloid of revolution. If the mir-ror is 4 inches across at its opening and is 3 feet deep,where will the collected light be concentrated?

67. Parabolic Arch Bridge A bridge is built in the shape ofa parabolic arch. The bridge has a span of 120 feet and amaximum height of 25 feet. See the illustration. Choose asuitable rectangular coordinate system and find the heightof the arch at distances of 10, 30, and 50 feet from thecenter.

68. Parabolic Arch Bridge A bridge is to be built in theshape of a parabolic arch and is to have a span of 100 feet.The height of the arch a distance of 40 feet from the cen-ter is to be 10 feet. Find the height of the arch at its center.

69. Show that an equation of the form

AX2 + Ey = 0, A "* 0, E "* 0

is the equation of a parabola with vertex at (0,0) and axisof symmetry the y-axis, Find its focus and directrix.

70. Show that an equation of the form

Cy2+Dx=O, C"*O,D"*O

is the equation of a parabola with vertex at (0,0) and axisof symmetry the x-axis. Find its focus and directrix.

71. Show that the graph of an equation of the form

AX2 + Dx + Ey + F = 0, A"* 0

(a) Is a parabola if E "* O.(b) Is a vertical line if E = 0 and D2 - 4AF = O.(c) Is two vertical lines if E = 0 and D2 - 4AF > O.(d) Contains no points if E = 0 and D2 - 4AF < O.

72. Show that the graph of an equation of the form

Cy2 + Dx + Ey + F = 0, C"* 0

(a) Is a parabola if D "* O.(b) Is a horizontal line if D = 0 and E2 - 4CF = O.(c) Is two horizontallines if D = 0 and E2 - 4CF > O.(d) Contains no points if D = 0 and E2 - 4CF < O.

PREPARING FOR THIS SECTIONBefore getting started, review the following:

./ Distance Formula (Section 1.1, p. 5)

./ Completing the Square (Section A.6, pp. 1033-1035)

./ Intercepts (Section 1.2, pp. 17-18)

./ Square Root Method (Section A.6, 1032-1033)

./ Symmetry (Section 1.3, pp. 20-22)

./ Circles (Section 1.3, pp. 26-30)

./ Graphing Techniques: Transformations(Section 2.5, pp. 135-144)

SECTION 9.3 The Ellipse 683

Applications5 Ellipses are found in many applications in science and engineering. For ex-

ample, the orbits of the planets around the Sun are elliptical, with the Sun'sposition at a focus. See Figure 32.

Figure 32

Stone and concrete bridges are often shaped as semielliptical arches. El-liptical gears are used in machinery when a variable rate of motion is required.

Ellipses also have an interesting reflection property. If a source of light(or sound) is placed at one focus, the waves transmitted by the source will re-flect off the ellipse and concentrate at the other focus.This is the principle be-hind whispering galleries, which are rooms designed with elliptical ceilings.A person standing at one focus of the ellipse can whisper and be heard by aperson standing at the other focus, because all the sound waves that reach theceiling are reflected to the other person.

~XAMPLE 9 Whispering Galleries

Figure 33 shows the specifications for an elliptical ceiling in a hall designedto be a whispering gallery. In a whispering gallery, a person standing at onefocus of the ellipse can whisper and be heard by another person standing atthe other focus, because all the sound waves that reach the ceiling from onefocus are reflected to the other focus.Where are the foci located in the hall?

Sol uti 0n We set up a rectangular coordinate system so that the center of the ellipse isat the origin and the major axis is along the x-axis. See Figure 34.The equa-tion of the ellipse is

where a = 25 and b = 20.

Figure 33 Figure 34

y

30

20

6' "10

T20'

t1-001 o(f----- 50'----~- 1 -30 -20 -10 0 10 20

1 0( 50' -I~ + .i. = 1 a = 25,b = 20a2 b2•

30 x

686 CHAPTER 9 Analytic Geometry

In Problems 55-58, graph each function by hand. Use a graphing utility to verify your results.

[Hint: Notice that each function is half an ellipse.]

56. f(x) = V9 - 9x255. f(x) = V16 - 4X2 58. f(x) = -V4 - 4x257. f(x) = - V64 - 16x2

59. Semielliptical Arch Bridge An arch in the shape of theupper half of an ellipse is used to support a bridge that isto span a river 20 meters wide. The center of the arch is 6meters above the center ofthe river (see the figure). Writean equation for the ellipse in which the x-axis coincideswith the water level and the y-axis passes through the cen-ter of the arch.

60. Semi elliptical Arch Bridge The arch of a bridge is asemiellipse with a horizontal major axis. The span is 30feet, and the top of the arch is 10 feet above the majoraxis.The roadway is horizontal and is 2 feet above the topof the arch. Find the vertical distance from the roadway tothe arch <ilt5-footintervals along the roadway.

61. Whispering Gallery A hall 100 feet in length is to bedesigned asa whispering gallery. If the foci are located 25feet from the center, how high will the ceiling be at thecenter?

62. Whispering Gallery Jim, standing at one focus of a whis-pering gallery, is 6 feet from the nearest wall. His friend isstanding at the other focus, 100 feet away. What is thelength of this whispering gallery? How high is its ellipti-cal ceiling at the center?

63. SemiellipticaI Arch Bridge A bridge is built in the shapeof a semielliptical arch. The bridge has a span of 120 feetand a maximum height of 25 feet. Choose a suitable rec-tangular coordinate system and find the height of the archat distances of 10, 30, and 50 feet from the center.

64. Semielliptical Arch Bridge A bridge is to be built in theshape of a semielliptical arch and is to have a span of 100feet. The height of the arch, at a distance of 40 feet fromthe center, is to be 10 feet. Find the height of the arch atits center.

65. SemiellipticaI Arch An arch in the form of half an ellipseis 40 feet wide and 15 feet high at the center. Find theheight of the arch at intervals of 10 feet along its width.

66. Semielliptical Arch Bridge An arch for a bridge over ahighway is in the form of half an ellipse. The top of thearch is 20 feet above the ground level (the major axis).The highway has four lanes, each 12 feet wide; a centersafety strip 8 feet wide; and two side strips, each 4 feetwide.What should the span of the bridge be (the length ofits major axis) if the height 28 feet from the center is to be13 feet?

In Problems .67-70, use the fact that the orbit of a planet about the Sun is an ellipse, with the Sun at one focus. The aphelion ofa planet is its greatest distance from the Sun, and the perihelion is its shortest distance. The mean distance of a planet from theSun is the length of the semimajor axis of the elliptical orbit. See the illustration.

67. Earth The mean distance of Earth from the Sun is 93million miles. If the aphelion of Earth is 94.5million miles,what is the perihelion? Write an equation for the orbit ofEarth around the Sun.

68. Mars The mean distance of Mars from the Sun is 142million miles. If the perihelion of Mars is 128.5 millionmiles, what is the aphelion? Write an equation for theorbit of Mars about the Sun.

69. Jupiter The aphelion of Jupiter is 507 million miles. Ifthe distance from the Sun to the center of its ellipticalorbit is 23.2 million miles, what is the perihelion? What isthe mean distance? Write an equation for the orbit of

. Jupiter around the Sun.70. Pluto The perihelion of Pluto is 4551 million miles, and

the distance of the Sun from the center of its ellipticalorbit is 897.5 million miles. Find the aphelion of Pluto.What is the mean distance of Pluto from the§tii? Writean equation for the orbit of Pluto about the Sun.

72. Racetrack Design A racetrack is in the shape of an el-lipse 80 feet long and 40 feet wide. What is the width 10

feet from a vertex? ~ 75. The_eccentricitye of an ellipse is defined as the number73. Show that an equation of the form £.., where a and c are the numbers given in equation (2).

AX2 + cy2 + F = 0, A * 0, C * 0, F * ° a-Because a > c, it follows that e < 1. Write a brief para-graph about the general shape of each of the followingellipses. Be sure to justify your conclusions.(a) Eccentricity close to °(b) Eccentricity = 0.5(c) Eccentricity close to 1

71. Racetrack Design Consult the figure. A racetrack is inthe shape of an ellipse, 100 feet long and 50 feet wide.What is the width 10 feet from a vertex?

where A and C are of the same sign and F is of oppositesign,(a) Is the equation of an ellipse with center at (0,0) if

A * C.(b) Is the equation of a circle with center (0,0) ifA = C.

SECTION 9.4 The Hyperbola 687

74. Show that the graph of an equation of the form

AX2 + ci + Dx + Ey + F = 0, A * O,C * °where A and C are of the same sign,

( ) I 11· ·f D2

E2

F· h .a s an e Ipse I 4A + 4C - ISt e same sign as A.

D2 E2

(b) Is a point if 4A + 4C - F = 0.

. D2 E2

(c) ~;:~~~. no points if 4A + 4C - F is of opposite

PREPARING FOR THIS SECTIONBefore getting started, review the following:

.I Distance Formula (Section 1.1, p. 5)

.I Completing the Square (Section A.6, pp. 1033-1035)

.I Symmetry (Section 1.3, pp. 20-22)

.I Asymptotes (Section 3.4, pp. 217-218)

.I Graphing Techniques: Transformations(Section 2.5, pp. 135-144)

.I Square Root Method (Section A.6, pp. 1032-1033)

IEIITHE HYPERBOLA

OBJECTIVES »Find the Equation of a Hyperbola

.v Graph Hyperbolas

2..J Discuss the Equation of a Hyperbola

4) Find the Asymptotes of a Hyperbola

V Work with Hyperbolas with Center at (h, k)6 Solve Applied Problems Involving Hyperbolas

A hyperbola is the collection of all points in the plane the difference ofwhose distances from two fixed points, called the foci, is a constant .

.!.J Figure 35 illustrates a hyperbola with foci FI and F2. The line containingthe foci is called the transverse axis. The midpoint of the line segment join-ing the foci is called the center of the hyperbola. The line through the center

Figure 48(X + 1)2

(y - 2)2 - = 14

Figure 49

Figure 50

d( P, ~) - d( P, F2) = constant

--------- - -------

SECTION 9.4 The Hyperbola 697

This is the equation of a hyperbola with center at (-1,2) and transverseaxis parallel to the y-axis.Also, a2 = 1 and b2 = 4, so c2 = a2 + b2 = 5. Sincethe transverse axis is parallel to the y-axis, the vertices and foci are locateda and c units above and below the center, respectively. The vertices are at(h,k ± a) = (-1,2 ± 1), or (-1,1) and (-1,3). The foci are at

1(h,k ± c) = (-1,2 ± V5). The asymptotes are y - 2 = 2'(x + 1) and

1y - 2 = -2'(x + 1). Figure 48(a) shows the graph drawn by hand. Fig-

ure 48(b) shows the graph obtained using a graphing utility.

Transverseaxis

yY1 = 2 +J (x: 1)2 + 1

6

5 X

V1 = (-1,1)

-7.1 I/~->--':""""'~-+-""':"'~"-"I 5.1

(a) (b) •c..1IJ!! r:r-- NOW WORK PROBLEM 45.

6ApplicationsSee Figure 49. Suppose that a gun is fired from an unknown source S. Anobserver at 01 hears the report (sound of gun shot) 1 second after anotherobserver at O2, Because sound travels at about 1100feet per second, it fol-lows that the point S must be 1100feet closer to O2 than to 01, S lies on onebranch of a hyperbola with foci at 01 and O2, (Do you see why? The differ-ence of the distances from S to 01 and from S to O2 is the constant 1100.)Ifa third observer at 03 hears the same report 2 seconds after 01 hears it, thenS will lie on a branch of a second hyperbola with foci at 01 and 03, The in-tersection of the two hyperbolas will pinpoint the location of S.

LoranIn the LOng RAnge Navigation system (LORAN), a master radio sendingstation and a secondary sending station emit signals that can be received bya ship at sea. See Figure 50. Because a ship monitoring the two signals willusually be nearer to one of the two stations, there will be a difference in thedistance that the two signals travel, which will register as a slight time dif-ference between the signals.As long as the time difference remains constant,the difference of the two distances will also be constant. If the ship followsa path corresponding to the fixed time difference, it will follow the path of ahyperbola whose foci are located at the positions of the two sending stations.So for each time difference a different hyperbolic path results, each bringingthe ship to a different shore location. Navigation charts show the various hy-perbolic paths corresponding to different time differences.

698 CHAPTER 9 Analytic Geometry

EXAMPLE 9

SolutionFigure 51

y200 0.00054 seconds

15()- ::.......

--------.J 00

~(-iO

LORAN

Two LORAN stations are positioned 250 miles apart along a straight shore.

(a) A ship records a time difference of 0.00054 second between the LORANsignals. Set up an appropriate rectangular coordinate system to deter-mine where the ship would reach shore if it were to follow the hyperbolacorresponding to this time difference.

(b) If the ship wants to enter a harbor located between the two stations 25miles from the master station, what time difference should it be lookingfor?

(c) If the ship is 80 miles offshore when the desired time difference is ob-tained, what is the approximate location of the ship?[Note: The speed of each radio signal is 186,000 miles per second.]

(a) We set up a rectangular coordinate system so that the two stations lie onthe x-axis and the origin is midway between them. See Figure 51. Theship lies on a hyperbola whose foci are the locations of the two stations.The reason for this is that the constant time difference of the signalsfrom each station results in a constant difference in the distance of theship from each station. Since the time difference is 0.00054 second andthe speed of the signal is 186,000 miles per second, the difference of thedistances from the ship to each station (foci) is

Distance = Speed X Time = 186,000 X 0.00054 ::::> 100 milesI

The difference of the distances from the ship to each station, 100, equals2a, ~a = 50 and the vertex of the corresponding hyperbola is at (50, 0).Since the focus is at (125,0), following this hyperbola the ship wouldreach shore 75 miles from the master, station.

(b) To reach shore 25 miles from the master station, the ship would followa hyperbola with vertex at (100,0). For this hyperbola, a = 100, so theconstant difference of the distances from the ship to each station is2a = 200. The time difference that the ship should look for is

T" Distance 200ime = Speed = 186,000 ::::> 0.001075 second

(c) To find the approximate location of the ship, we need to find the equa-tion of the hyperbola with vertex at (100,0) and a focus at (125,0). Theform of the equation of this hyperbola is

X2 i---=1a2 b2

where a = 100. Since c = 125, we have

b2 = c2- a2 = 1252 - 1002 = 5625

The equation of the hyperbola isX2 i

1002 - 5625 = 1