Analytic Geometry in Two and Three Dimensions

Transcription

1 5144_Demana_Ch08pp /13/06 6:49 AM Page 631 CHAPTER 8 Analtic Geometr in Two and Three Dimensions 8.1 Conic Sections and Parabolas 8. Ellipses 8.3 Hperbolas 8.4 Translation and Rotation of Aes 8.5 Polar Equations of Conics 8.6 Three-Dimensional Cartesian Coordinate Sstem The oval-shaped lawn behind the White House in Washington, D.C. is called the Ellipse. It has views of the Washington Monument, the Jefferson Memorial, the Department of Commerce, and the Old Post Office Building. The Ellipse is 616 ft long, 58 ft wide, and is in the shape of a conic section. Its shape can be modeled using the methods of this chapter. See page

2 5144_Demana_Ch08pp /13/06 6:49 AM Page CHAPTER 8 Analtic Geometr in Two and Three Dimensions HISTORY OF CONIC SECTIONS Parabolas, ellipses, and hperbolas had been studied for man ears when Apollonius (c B.C.) wrote his eight-volume Conic Sections. Apollonius, born in northwestern Asia Minor, was the first to unif these three curves as cross sections of a cone and to view the hperbola as having two branches. Interest in conic sections was renewed in the 17th centur when Galileo proved that projectiles follow parabolic paths and Johannes Kepler ( ) discovered that planets travel in elliptical orbits. Chapter 8 Overview Analtic geometr combines number and form. It is the marriage of algebra and geometr that grew from the works of Frenchmen René Descartes ( ) and Pierre de Fermat ( ). Their achievements allowed geometr problems to be solved algebraicall and algebra problems to be solved geometricall two major themes of this book. Analtic geometr opened the door for Newton and Leibniz to develop calculus. In Sections , we will learn that parabolas, ellipses, and hperbolas are all conic sections and can all be epressed as second-degree equations. We will investigate their uses, including the reflective properties of parabolas and ellipses and how hperbolas are used in long-range navigation. In Section 8.5, we will see how parabolas, ellipses, and hperbolas are unified in the polar-coordinate setting. In Section 8.6, we will move from the two-dimensional plane to revisit the concepts of point, line, midpoint, distance, and vector in three-dimensional space. 8.1 Conic Sections and Parabolas What ou ll learn about Conic Sections Geometr of a Parabola Translations of Parabolas Reflective Propert of a Parabola... and wh Conic sections are the paths of nature: An free-moving object in a gravitational field follows the path of a conic section. Conic Sections Imagine two nonperpendicular lines intersecting at a point V. If we fi one of the lines as an ais and rotate the other line (the generator) around the ais, then the generator sweeps out a right circular cone with verte V, as illustrated in Figure 8.1. Notice that V divides the cone into two parts called nappes, with each nappe of the cone resembling a pointed ice-cream cone. Ais Generator Upper nappe V Lower nappe FIGURE 8.1 A right circular cone (of two nappes). A conic section (or conic) is a cross section of a cone, in other words, the intersection of a plane with a right circular cone. The three basic conic sections are the parabola, the ellipse, and the hperbola (Figure 8.a). Some atpical conics, known as degenerate conic sections, are shown in Figure 8.b. Because it is atpical and lacks some of the features usuall associated with an ellipse,

3 5144_Demana_Ch08pp /13/06 6:49 AM Page 633 SECTION 8.1 Conic Sections and Parabolas 633 OBJECTIVE Students will be able to find the equation, focus, and directri of a parabola. MOTIVATE Ask students if all parabolas are similar (in the geometric sense). (Yes) LESSON GUIDE Da 1: Conic Sections; Geometr of a Parabola Da : Translations of Parabolas; Reflective Propert of a Parabola a circle is considered to be a degenerate ellipse. Other degenerate conic sections can be obtained from cross sections of a degenerate cone; such cones occur when the generator and ais of the cone are parallel or perpendicular. (See Eercise 73.) BIBLIOGRAPHY For students: Practical Conic Sections, J. W. Downs. Dale Semour Publications, Comet, Carl Sagan and Ann Druan. Ballentine Books, For teachers: Astronom: From the Earth to the Universe (5th ed.), J. M. Pasachoff. Saunders College Publishing, Multicultural and Gender Equit in the Mathematics Classroom: The Gift of Diversit (1997 Yearbook), Janet Trentacosta (Ed.) National Council of Teachers of Mathematics, Posters: Conic Sections. Dale Semour Publications Locating Satellites in Elliptical Orbits, National Council of Teachers of Mathematics Ellipse Point: plane through cone's verte onl Parabola (a) Single line: plane tangent to cone (b) Hberbola Intersecting lines TEACHING NOTES You ma wish to construct the nappe of a cone b rolling up a piece of paper in order to illustrate various conic sections as cross sections of a cone. Note that a circle is a special case of an ellipse and thus considered a degenerate conic. FIGURE 8. (a) The three standard tpes of conic sections and (b) three degenerate conic sections. The conic sections can be defined algebraicall as the graphs of second-degree (quadratic) equations in two variables, that is, equations of the form A B C D E F 0, where A, B, and C are not all zero.

4 5144_Demana_Ch08pp /13/06 6:49 AM Page CHAPTER 8 Analtic Geometr in Two and Three Dimensions A DEGENERATE PARABOLA If the focus F lies on the directri l, the parabola degenerates to the line through F perpendicular to l. Henceforth, we will assume F does not lie on l. LOCUS OF A POINT Before the word set was used in mathematics, the Latin word locus, meaning place, was often used in geometric definitions. The locus of a point was the set of possible places a point could be and still fit the conditions of the definition. Sometimes, conics are still defined in terms of loci. Geometr of a Parabola In Section.1 we learned that the graph of a quadratic function is an upward or downward opening parabola. We have seen the role of the parabola in free-fall and projectile motion. We now investigate the geometric properties of parabolas. DEFINITION Parabola A parabola is the set of all points in a plane equidistant from a particular line (the directri) and a particular point (the focus ) in the plane. (See Figure 8.3.) The line passing through the focus and perpendicular to the directri is the (focal) ais of the parabola. The ais is the line of smmetr for the parabola. The point where the parabola intersects its ais is the verte of the parabola. The verte is located midwa between the focus and the directri and is the point of the parabola that is closest to both the focus and the directri. See Figure 8.3. TEACHING NOTE Point on the parabola Eercise 71 eplains how Figure 8.3 was created, and this eercise can be used instead of (or in addition to) Eploration 1. Dist. to directri Dist. to focus Ais EXPLORATION EXTENSIONS Find the -coordinate for a point on the parabola that is a distance of 5 units from the focus. Ans. 4 Verte Focus Directri (0, 1) FIGURE 8.3 Structure of a Parabola. The distance from each point on the parabola to both the focus and the directri is the same. EXPLORATION 1 Understanding the Definition of Parabola 1. Prove that the verte of the parabola with focus 0, 1 and directri 1 is 0, 0. (See Figure 8.4.) FIGURE 8.4 The geometr of a parabola.. Find an equation for the parabola shown in Figure 8.4. = 4 3. Find the coordinates of the points of the parabola that are highlighted in Figure 8.4.

5 5144_Demana_Ch08pp /13/06 6:49 AM Page 635 SECTION 8.1 Conic Sections and Parabolas 635 We can generalize the situation in Eploration 1 to show that an equation for the parabola with focus 0, p and directri p is 4p. (See Figure 8.5.) ALERT Note that the standard form of the equation of an upward or downward opening parabola is not the same as the standard form of a quadratic function. = 4p Directri: = p Focus F(0, p) p p D(, p) P(, ) The verte lies halfwa between directri and focus. (a) l Directri: = p Verte at origin = 4p (b) Focus F(0, p) FIGURE 8.5 Graphs of 4p with (a) p 0 and (b) p 0. We must show first that a point P(, ) that is equidistant from F(0, p) and the line p satisfies the equation 4p, and then that a point satisfing the equation 4p is equidistant from F(0, p) and the line p: Let P(, ) be equidistant from F(0, p) and the line p. Notice that 0 p distance from P, to F 0, p, and p distance from P, to p. NAME GAME Additional features of a parabola are defined in Eercises The focal width is the length of the latus rectum. Equating these distances and squaring ields: 0 p p p 0 p p p p p 4p Simplif. Epand. Combine like terms. B reversing the above steps, we see that a solution, of 4p is equidistant from F 0, p and the line p. The equation 4p is the standard form of the equation of an upward or downward opening parabola with verte at the origin. If p 0, the parabola opens upward; if p 0, it opens downward. An alternative algebraic form for such a parabola is a, where a 1 4p. So the graph of 4p is also the graph of the quadratic function f a. When the equation of an upward or downward opening parabola is written as 4p, the value p is interpreted as the focal length of the parabola the directed distance from the verte to the focus of the parabola. A line segment with endpoints on a parabola is a chord of the parabola. The value 4p is the focal width of the parabola the length of the chord through the focus and perpendicular to the ais.

6 5144_Demana_Ch08pp /13/06 6:49 AM Page CHAPTER 8 Analtic Geometr in Two and Three Dimensions Directri: = p = 4p Parabolas that open to the right or to the left are inverse relations of upward or downward opening parabolas. So equations of parabolas with verte 0, 0 that open to the right or to the left have the standard form 4p. If p 0, the parabola opens to the right, and if p 0, to the left. (See Figure 8.6.) Verte O Focus F(p, 0) Parabolas with Verte (0, 0) Standard equation 4p 4p Opens Upward or To the right or downward to the left Focus 0, p p, 0 (a) Directri p p = 4p Directri: = p Ais -ais -ais Focal length p p Focal width 4p 4p See Figures 8.5 and 8.6. Focus F(p, 0) O Verte EXAMPLE 1 Finding the Focus, Directri, and Focal Width Find the focus, the directri, and the focal width of the parabola 1. (b) FIGURE 8.6 Graph of 4p with (a) p 0 and (b) p 0. SOLUTION Multipling both sides of the equation b ields the standard form. The coefficient of is 4p, and p 1. So the focus is 0, p 0, 1. Because p 1 1, the directri is the line 1. The focal width is 4p. Now tr Eercise 1. NOTES ON EXAMPLES When solving a problem such as Eample, students should be strongl encouraged to draw a sketch of the given information. You ma wish to draw a sketch for the students to model this behavior. EXAMPLE Finding an Equation of a Parabola Find an equation in standard form for the parabola whose directri is the line and whose focus is the point, 0. SOLUTION Because the directri is and the focus is, 0, the focal length is p and the parabola opens to the left. The equation of the parabola in standard form is 4p, or more specificall, 8. Now tr Eercise 15. Translations of Parabolas When a parabola with the equation 4p or 4p is translated horizontall b h units and verticall b k units, the verte of the parabola moves from 0, 0 to h, k. (See Figure 8.7.) Such a translation does not change the focal length, the focal width, or the direction the parabola opens.

7 5144_Demana_Ch08pp /13/06 6:49 AM Page 637 SECTION 8.1 Conic Sections and Parabolas 637 (h, k + p) TEACHING NOTE Students should be able to relate the transformations of parabolas (and other conic sections) to the transformations that were studied beginning in Section 1.5. (a) (h, k) (h, k) (h + p, k) (b) FIGURE 8.7 Parabolas with verte h, k and focus on (a) h and (b) k. Parabolas with Verte (h, k) Standard equation h 4p k k 4p h Opens Upward or To the right or downward to the left Focus h, k p h p, k Directri k p h p Ais h k Focal length p p Focal width 4p 4p See Figure 8.7. EXAMPLE 3 Finding an Equation of a Parabola Find the standard form of the equation for the parabola with verte (3, 4) and focus (5, 4). SOLUTION The ais of the parabola is the line passing through the verte 3, 4 and the focus 5, 4. This is the line 4. So the equation has the form k 4p h. Because the verte h, k 3, 4, h 3 and k 4. The directed distance from the verte 3, 4 to the focus 5, 4 is p 5 3, so 4p 8. Thus the equation we seek is Now tr Eercise 1.

8 5144_Demana_Ch08pp /13/06 6:49 AM Page CHAPTER 8 Analtic Geometr in Two and Three Dimensions When solving a problem like Eample 3, it is a good idea to sketch the verte, the focus, and other features of the parabola as we solve the problem. This makes it eas to see whether the ais of the parabola is horizontal or vertical and the relative positions of its features. Eploration walks us through this process. EXPLORATION Building a Parabola Carr out the following steps using a sheet of rectangular graph paper. EXPLORATION EXTENSIONS Check our work b using a grapher to graph the equation ou found in step Let the focus F of a parabola be, and its directri be 4. Draw the - and -aes on the graph paper. Then sketch and label the focus and directri of the parabola.. Locate, sketch, and label the ais of the parabola. What is the equation of the ais? 3. Locate and plot the verte V of the parabola. Label it b name and coordinates. 4. What are the focal length and focal width of the parabola? 5. Use the focal width to locate, plot, and label the endpoints of a chord of the parabola that parallels the directri. 6. Sketch the parabola. 7. Which direction does it open? downward 8. What is its equation in standard form? ( ) 1( 1) Sometimes it is best to sketch a parabola b hand, as in Eploration ; this helps us see the structure and relationships of the parabola and its features. At other times, we ma want or need an accurate, active graph. If we wish to graph a parabola using a function grapher, we need to solve the equation of the parabola for, as illustrated in Eample 4. EXAMPLE 4 Graphing a Parabola Use a function grapher to graph the parabola of Eample 3. SOLUTION Etract square roots Add 4. Let Y and Y 4 8 3, and graph the two equations in a window centered at the verte, as shown in Figure 8.8. Now tr Eercise 45.

9 5144_Demana_Ch08pp /13/06 6:49 AM Page 639 SECTION 8.1 Conic Sections and Parabolas 639 CLOSING THE GAP In Figure 8.8, we centered the graphing window at the verte (3, 4) of the parabola to ensure that this point would be plotted. This avoids the common grapher error of a gap between the two upper and lower parts of the conic section being plotted. [ 1, 7] b [, 10] FIGURE 8.8 The graphs of Y1 4 8 ( 3) and Y 4 8 ( 3) together form the graph of (Eample 4) Filament (light source) at focus Parabolic radio wave reflector (a) Outgoing ras of light SEARCHLIGHT Incoming radio signals concentrate at focus RADIO TELESCOPE (b) FIGURE 8.9 Eamples of parabolic reflectors. EXAMPLE 5 Using Standard Forms with a Parabola Prove that the graph of is a parabola, and find its verte, focus, and directri. SOLUTION Because this equation is quadratic in the variable, we complete the square with respect to to obtain a standard form Isolate the -terms Complete the square This equation is in the standard form k 4p h, where h, k 1, and p It follows that the verte h, k is, 1 ; the focus h p, k is 3.5, 1, or 7, 1 ; the directri h p is 0.5, or 1. Now tr Eercise 49. Reflective Propert of a Parabola The main applications of parabolas involve their use as reflectors of sound, light, radio waves, and other electromagnetic waves. If we rotate a parabola in three-dimensional space about its ais, the parabola sweeps out a paraboloid of revolution. If we place a signal source at the focus of a reflective paraboloid, the signal reflects off the surface in lines parallel to the ais of smmetr, as illustrated in Figure 8.9a. This propert is used b flashlights, headlights, searchlights, microwave relas, and satellite up-links. The principle works for signals traveling in the reverse direction as well; signals arriving parallel to a parabolic reflector s ais are directed toward the reflector s focus. This propert is used to intensif signals picked up b radio telescopes and television satellite dishes, to focus arriving light in reflecting telescopes, to concentrate heat in solar ovens, and to magnif sound for sideline microphones at football games. See Figure 8.9b.

10 5144_Demana_Ch08pp /13/06 6:49 AM Page CHAPTER 8 Analtic Geometr in Two and Three Dimensions ( 1.5, 1) (1.5, 1) F(0, p) V(0, 0) FIGURE 8.10 Cross section of parabolic reflector in Eample 6. FOLLOW-UP Ask students to eplain wh parabolic mirrors are used in flashlights. ASSIGNMENT GUIDE Da 1: E. 1, 5 19 odds, 31 3, 37, 40, 57 Da : E. 3, 1 54 multiples of 3, 59, 61 COOPERATIVE LEARNING Group Activit: E NOTES ON EXERCISES E are applications of parabolas. E provide practice for standardized tests. E are quite challenging and could be used for a group project. ONGOING ASSESSMENT Self-Assessment: E. 1, 15, 1, 45, 49, 59 Embedded Assessment: E EXAMPLE 6 Studing a Parabolic Microphone On the sidelines of each of its televised football games, the FBTV network uses a parabolic reflector with a microphone at the reflector s focus to capture the conversations among plaers on the field. If the parabolic reflector is 3 ft across and 1 ft deep, where should the microphone be placed? SOLUTION We draw a cross section of the reflector as an upward opening parabola in the Cartesian plane, placing its verte V at the origin (see Figure 8.10). We let the focus F have coordinates 0, p to ield the equation 4p. Because the reflector is 3 ft across and 1 ft deep, the points 1.5, 1 must lie on the parabola. The microphone should be placed at the focus, so we need to find the value of p. We do this b substituting the values we found into the equation: 4p ( 1.5) 4p(1).5 4p p Because p ft, or 6.75 inches, the microphone should be placed inside the reflector along its ais and 6.75 inches from its verte. Now tr Eercise 59. QUICK REVIEW 8.1 (For help, go to Sections P., P.5, and.1.) In Eercises 1 and, find the distance between the given points. 1. 1, 3 and, , 3 and a, b In Eercises 3 and 4, solve for in terms of In Eercises 5 and 6, complete the square to rewrite the equation in verte form In Eercises 7 and 8, find the verte and ais of the graph of f. Describe how the graph of f can be obtained from the graph of g(), and graph f. 7. f f 1 1 In Eercises 9 and 10, write an equation for the quadratic function whose graph contains the given verte and point. 9. Verte 1, 3, point 0, 1 f() ( 1) Verte, 5, point 5, 13 f() ( ) 5

11 5144_Demana_Ch08pp /13/06 6:49 AM Page 641 SECTION 8.1 Conic Sections and Parabolas 641 SECTION 8.1 EXERCISES In Eercises 1 6, find the verte, focus, directri, and focal width of the parabola In Eercises 7 10, match the graph with its equation. 6. Verte (3, 5), directri 7 ( 3) 8( 5) 7. Verte (, 1), opens upward, focal width Verte ( 3, 3), opens downward, focal width 0 9. Verte ( 1, 4), opens to the left, focal width Verte (, 3), opens to the right, focal width 5 In Eercises 31 36, sketch the graph of the parabola b hand In Eercises 37 48, graph the parabola using a function grapher (a) (c) 7. 3 (c) 8. 4 (b) 9. 5 (a) (d) In Eercises 11 30, find an equation in standard form for the parabola that satisfies the given conditions. 11. Verte (0, 0), focus ( 3, 0) 1 1. Verte (0, 0), focus (0, ) Verte (0, 0), directri Verte (0, 0), directri Focus (0, 5), directri Focus ( 4, 0), directri Verte (0, 0), opens to the right, focal width Verte (0, 0), opens to the left, focal width Verte (0, 0), opens downward, focal width Verte (0, 0), opens upward, focal width Focus (, 4), verte ( 4, 4) ( 4) 8( 4). Focus ( 5, 3), verte ( 5, 6) ( 5) 1( 6) 3. Focus (3, 4), directri 1 ( 3) 6( 5/) 4. Focus (, 3), directri 5( 3) 6( 7/) 5. Verte (4, 3), directri 6 ( 3) 8( 4) (b) (d) ( 3) 1( ) 46. ( 1) 4( 5) 47. ( ) 8( 1) 48. ( 6) 16( 4) In Eercises 49 5, prove that the graph of the equation is a parabola, and find its verte, focus, and directri In Eercises 53 56, write an equation for the parabola ( 6, 4) (0, ) (0, ) (, 1) ( 1, 3) 57. Writing to Learn Eplain wh the derivation of 4p is valid regardless of whether p 0 or p Writing to Learn Prove that an equation for the parabola with focus (p, 0) and directri p is 4p. (5.5, 0) (1, 3) (3, 5)

12 5144_Demana_Ch08pp /13/06 6:49 AM Page CHAPTER 8 Analtic Geometr in Two and Three Dimensions 59. Designing a Flashlight Mirror The mirror of a flashlight is a paraboloid of revolution. Its diameter is 6 cm and its depth is cm. How far from the verte should the filament of the light bulb be placed for the flashlight to have its beam run parallel to the ais of its mirror? 64. Group Activit Designing a Bridge Arch Parabolic arches are known to have greater strength than other arches. A bridge with a supporting parabolic arch spans 60 ft with a 30-ft wide road passing underneath the bridge. In order to have a minimum clearance of 16 ft, what is the maimum clearance? 6 cm cm 60. Designing a Satellite Dish The reflector of a television satellite dish is a paraboloid of revolution with diameter 5 ft and a depth of ft. How far from the verte should the receiving antenna be placed? 5 ft 60 ft 30 ft 61. Parabolic Microphones The Sports Channel uses a parabolic microphone to capture all the sounds from golf tournaments throughout a season. If one of its microphones has a parabolic surface generated b the parabola 10, locate the focus (the electronic receiver) of the parabola. 6. Parabolic Headlights Stein Glass, Inc., makes parabolic headlights for a variet of automobiles. If one of its headlights has a parabolic surface generated b the parabola 1, where should its light bulb be placed? 63. Group Activit Designing a Suspension Bridge The main cables of a suspension bridge uniforml distribute the weight of the bridge when in the form of a parabola. The main cables of a particular bridge are attached to towers that are 600 ft apart. The cables are attached to the towers at a height of 110 ft above the roadwa and are 10 ft above the roadwa at their lowest points. If vertical support cables are at 50-ft intervals along the level roadwa, what are the lengths of these vertical cables? 50 ft 600 ft 110 ft ft The maimum clearance is about 1.33 feet. Standardized Test Questions 65. True or False Ever point on a parabola is the same distance from its focus and its ais. Justif our answer. 66. True or False The directri of a parabola is parallel to the parabola s ais. Justif our answer. In Eercises 67 70, solve the problem without using a calculator. 67. Multiple Choice Which of the following curves is not a conic section? D (A) circle (B) ellipse (C) hperbola (D) oval (E) parabola 68. Multiple Choice Which point do all conics of the form 4p have in common? D (A) (1, 1) (B) (1, 0) (C) (0, 1) (D) (0, 0) (E) ( 1, 1) 69. Multiple Choice The focus of 1 is B (A) (3, 3). (B) (3, 0). (C) (0, 3). (D) (0, 0). (E) ( 3, 3).

13 5144_Demana_Ch08pp /13/06 6:49 AM Page 643 SECTION 8.1 Conic Sections and Parabolas Multiple Choice The verte of ( 3) 8( ) is D (A) (3, ). (B) ( 3, ). (C) ( 3, ). (D) (, 3). (E) (, 3). Eplorations 71. Dnamicall Constructing a Parabola Use a geometr software package, such as Cabri Geometr II, The Geometer s Sketchpad, or similar application on a handheld device to construct a parabola geometricall from its definition. (See Figure 8.3.) (a) Start b placing a line l (directri) and a point F (focus) not on the line in the construction window. (b) Construct a point A on the directri, and then the segment AF. (c) Construct a point P where the perpendicular bisector of AF meets the line perpendicular to l through A. (d) What curve does P trace out as A moves? parabola (e) Prove our answer to part (d) is correct. 7. Constructing Points of a Parabola Use a geometr software package, such as Cabri Geometr II, The Geometer s Sketchpad, or similar application on a handheld device, to construct Figure 8.4, associated with Eploration 1. (a) Start b placing the coordinate aes in the construction window. (b) Construct the line 1 as the directri and the point 0, 1 as the focus. (c) Construct the horizontal lines and concentric circles shown in Figure 8.4. (d) Construct the points where these horizontal lines and concentric circles meet. (e) Prove these points lie on the parabola with directri 1 and focus 0, Degenerate Cones and Degenerate Conics Degenerate cones occur when the generator and ais of the cone are parallel or perpendicular. (a) Draw a sketch and describe the cone obtained when the generator and ais of the cone are parallel. (b) Draw sketches and name the tpes of degenerate conics obtained b intersecting the degenerate cone in part (a) with a plane. (c) Draw a sketch and describe the cone obtained when the generator and ais of the cone are perpendicular. (d) Draw sketches and name the tpes of degenerate conics obtained b intersecting the degenerate cone in part (c) with a plane. Etending the Ideas 74. Tangent Lines A tangent line of a parabola is a line that intersects but does not cross the parabola. Prove that a line tangent to the parabola 4p at the point a, b crosses the -ais at 0, b. 75. Focal Chords A focal chord of a parabola is a chord of the parabola that passes through the focus. (a) Prove that the -coordinates of the endpoints of a focal chord of 4p are p m m 1, where m is the slope of the focal chord. (b) Using part (a), prove the minimum length of a focal chord is the focal width 4p. 76. Latus Rectum The focal chord of a parabola perpendicular to the ais of the parabola is the latus rectum, which is Latin for right chord. Using the results from Eercises 74 and 75, prove: (a) For a parabola the two endpoints of the latus rectum and the point of intersection of the ais and directri are the vertices of an isosceles right triangle. (b) The legs of this isosceles right triangle are tangent to the parabola.

14 5144_Demana_Ch08pp /13/06 6:49 AM Page CHAPTER 8 Analtic Geometr in Two and Three Dimensions 8. Ellipses What ou ll learn about Geometr of an Ellipse Translations of Ellipses Orbits and Eccentricit Reflective Propert of an Ellipse... and wh Ellipses are the paths of planets and comets around the Sun, or of moons around planets. OBJECTIVE Students will be able to find the equation, vertices, and foci of an ellipse. Verte Focus Center Focal ais FIGURE 8.11 Ke points on the focal ais of an ellipse. d 1 Focus P d F 1 F Verte Geometr of an Ellipse When a plane intersects one nappe of a right circular clinder and forms a simple closed curve, the curve is an ellipse. DEFINITION Ellipse An ellipse is the set of all points in a plane whose distances from two fied points in the plane have a constant sum. The fied points are the foci (plural of focus) of the ellipse. The line through the foci is the focal ais. The point on the focal ais midwa between the foci is the center. The points where the ellipse intersects its ais are the vertices of the ellipse. (See Figure 8.11.) Figure 8.1 shows a point P, of an ellipse. The fied points F 1 and F are the foci of the ellipse, and the distances whose sum is constant are d 1 and d. We can construct an ellipse using a pencil, a loop of string, and two pushpins. Put the loop around the two pins placed at F 1 and F, pull the string taut with a pencil point P, and move the pencil around to trace out the ellipse (Figure 8.13). We now use the definition to derive an equation for an ellipse. For some constants a and c with a c 0, let F 1 c, 0 and F c, 0 be the foci (Figure 8.14). Then an ellipse is defined b the set of points P, such that PF 1 PF a. Using the distance formula, the equation becomes c 0 c 0 a. c a c c c 4a 4a c c c a c a c a c c a 4 a c c a c a a a c Simplif. Square. Simplif. Square. d 1 + d = constant FIGURE 8.1 Structure of an Ellipse. The sum of the distances from the foci to each point on the ellipse is a constant. Letting b a c, we have which is usuall written as b a a b, a b 1.

15 5144_Demana_Ch08pp /13/06 6:49 AM Page 645 SECTION 8. Ellipses 645 P(, ) F 1 F FIGURE 8.13 How to draw an ellipse. FIGURE 8.14 The ellipse defined b PF 1 PF a is the graph of the equation a b 1, where b a c. b P(, ) Focus Focus F 1 ( c, 0) O Center F (c, 0) a Because these steps can be reversed, a point P, satisfies this last equation if and onl if the point lies on the ellipse defined b PF 1 PF a, provided that a c 0and b a c. The Pthagorean relation b a c can be written man was, including c a b and a b c. The equation a b 1 is the standard form of the equation of an ellipse centered at the origin with the -ais as its focal ais. An ellipse centered at the origin with the -ais as its focal ais is the inverse of a b 1, and thus has an equation of the form a b 1. As with circles and parabolas, a line segment with endpoints on an ellipse is a chord of the ellipse. The chord ling on the focal ais is the major ais of the ellipse. The chord through the center perpendicular to the focal ais is the minor ais of the ellipse. The length of the major ais is a, and of the minor ais is b. The number a is the semimajor ais, and b is the semiminor ais. Ellipses with Center (0, 0) Standard equation a b 1 a b 1 Focal ais -ais -ais Foci c, 0 0, c Vertices a, 0 0, a Semimajor ais a a Semiminor ais b b Pthagorean relation a b c a b c See Figure AXIS ALERT For an ellipse, the word ais is used in several was. The focal ais is a line. The major and minor aes are line segments. The semimajor and semiminor aes are numbers. MOTIVATE Ask students how the graph of (/) 1 is related to the graph of 1. ( a, 0) ( c, 0) (0, b) a b c (c, 0) (a, 0) ( b, 0) (0, a) (0, c) c (0, c) b a (b, 0) LESSON GUIDE Da 1: Geometr of an Ellipse; Translations of Ellipses Da : Orbits and Eccentricit; Reflective Propert of an Ellipse (a) (0, b) (0, a) FIGURE 8.15 Ellipses centered at the origin with foci on (a) the -ais and (b) the -ais. In each case, a right triangle illustrating the Pthagorean relation is shown. (b)

16 5144_Demana_Ch08pp /13/06 6:49 AM Page CHAPTER 8 Analtic Geometr in Two and Three Dimensions EXAMPLE 1 Finding the Vertices and Foci of an Ellipse Find the vertices and the foci of the ellipse SOLUTION Dividing both sides of the equation b 36 ields the standard form Because the larger number is the denominator of, the focal ais is the -ais. So a 9, b 4, and c a b Thus the vertices are 3, 0, and the foci are 5, 0. Now tr Eercise 1. An ellipse centered at the origin with its focal ais on a coordinate ais is smmetric with respect to the origin and both coordinate aes. Such an ellipse can be sketched b first drawing a guiding rectangle centered at the origin with sides parallel to the coordinate aes and then sketching the ellipse inside the rectangle, as shown in the Drawing Lesson. Drawing Lesson How to Sketch the Ellipse /a /b 1 1. Sketch line segments at a and b and complete the rectangle the determine. b b. Inscribe an ellipse that is tangent to the rectangle at ( a, 0) and (0, b). a b a a b a If we wish to graph an ellipse using a function grapher, we need to solve the equation of the ellipse for, as illustrated in Eample. NOTES ON EXAMPLES Notice that we chose square viewing windows in Figure A nonsquare window would give a distorted view of an ellipse. EXAMPLE Finding an Equation and Graphing an Ellipse Find an equation of the ellipse with foci (0, 3) and (0, 3) whose minor ais has length 4. Sketch the ellipse and support our sketch with a grapher. SOLUTION The center is 0, 0. The foci are on the -ais with c 3. The semiminor ais is b 4. Using a b c, we have a So the standard form of the equation for the ellipse is Using a and b, we can sketch a guiding rectangle and then the ellipse itself, as eplained in the Drawing Lesson. (Tr doing this.) To graph the ellipse using a function grapher, we solve for in terms of continued

17 5144_Demana_Ch08pp /13/06 6:49 AM Page 647 SECTION 8. Ellipses 647 Figure 8.16 shows three views of the graphs of Y and Y We must select the viewing window carefull to avoid grapher failure. Now tr Eercise 17. [ 6, 6] b [ 4, 4] (a) [ 4.7, 4.7] b [ 3.1, 3.1] (b) [ 9.4, 9.4] b [ 6., 6.] (c) FIGURE 8.16 Three views of the ellipse All of the windows are square or approimatel square viewing windows so we can see the true shape. Notice that the gaps between the upper and lower function branches do not show when the grapher window includes columns of piels whose -coordinates are as in (b) and (c). (Eample ) (h c, k) (h a, k) (h, k) (h + c, k) (h + a, k) Translations of Ellipses When an ellipse with center 0, 0 is translated horizontall b h units and verticall b k units, the center of the ellipse moves from 0, 0 to h, k, as shown in Figure Such a translation does not change the length of the major or minor ais or the Pthagorean relation. (a) Ellipses with Center (h, k) (h, k + a) (h, k) (h, k + c) Standard equation a h b k 1 a k b h 1 Focal ais k h Foci h c, k h, k c Vertices h a, k h, k a Semimajor a a ais (b) (h, k a) (h, k c) FIGURE 8.17 Ellipses with center h, k and foci on (a) k and (b) h. Semiminor b b ais Pthagorean a b c a b c relation See Figure 8.17.

18 5144_Demana_Ch08pp /16/06 11:5 AM Page CHAPTER 8 Analtic Geometr in Two and Three Dimensions NOTES ON EXAMPLES When solving a problem such as Eample 3, students should be strongl encouraged to draw a sketch of the given information. 6 (, 1) (8, 1) 8 FIGURE 8.18 Given information for Eample EXAMPLE 3 Finding an Equation of an Ellipse Find the standard form of the equation for the ellipse whose major ais has endpoints, 1 and 8, 1, and whose minor ais has length 8. SOLUTION Figure 8.18 shows the major-ais endpoints, the minor ais, and the center of the ellipse. The standard equation of this ellipse has the form ( h) a ( k) b 1, where the center h, k is the midpoint 3, 1 of the major ais. The semimajor ais and semiminor ais are a 8 5 and b 8 4. So the equation we seek is ( 3) 5 ( ( 1)) 4 1, Now tr Eercise 31. EXAMPLE 4 Locating Ke Points of an Ellipse Find the center, vertices, and foci of the ellipse SOLUTION The standard equation of this ellipse has the form The center h, k is, 5. Because the semimajor ais a 4 9 7, the vertices h, k a are Because h, k a, 5 7, 1 and h, k a, 5 7,. c a b the foci h, k c are, 5 4 0, or approimatel, 11.3 and, 1.3. Now tr Eercise 37. With the information found about the ellipse in Eample 4 and knowing that its semiminor ais b 9 3, we could easil sketch the ellipse. Obtaining an accurate graph of the ellipse using a function grapher is another matter. Generall, the best wa to graph an ellipse using a graphing utilit is to use parametric equations.

19 5144_Demana_Ch08pp /13/06 6:49 AM Page 649 SECTION 8. Ellipses 649 EXPLORATION EXTENSIONS Give an equation in standard form for the ellipse defined b the parametric equations 3 cos t, 4 5 sin t, 0 t. EXPLORATION 1 Graphing an Ellipse Using Its Parametric Equations 1. Use the Pthagorean trigonometr identit cos t sin t 1 to prove that the parameterization 3 cos t, 5 7 sin t, 0 t will produce a graph of the ellipse Graph 3 cos t, 5 7 sin t, 0 t in a square viewing window to support part 1 graphicall. 3. Create parameterizations for the ellipses in Eamples 1,, and Graph each of our parameterizations in part 3 and check the features of the obtained graph to see whether the match the epected geometric features of the ellipse. Revise our parameterization and regraph until all features match. 5. Prove that each of our parameterizations is valid. Orbits and Eccentricit Kepler s first law of planetar motion, published in 1609, states that the path of a planet s orbit is an ellipse with the Sun at one of the foci. Asteroids, comets, and other bodies that orbit the Sun follow elliptical paths. The closest point to the Sun in such an orbit is the perihelion, and the farthest point is the aphelion (Figure 8.19). The shape of an ellipse is related to its eccentricit. Aphelion a c a + c Sun at focus Center Semimajor ais a Perihelion Orbiting object FIGURE 8.19 Man celestial objects have elliptical orbits around the Sun.

20 5144_Demana_Ch08pp /13/06 6:49 AM Page CHAPTER 8 Analtic Geometr in Two and Three Dimensions A NEW e Tr not to confuse the eccentricit e with the natural base e used in eponential and logarithmic functions. The contet should clarif which meaning is intended. DEFINITION Eccentricit of an Ellipse The eccentricit of an ellipse is c a e b, a a where a is the semimajor ais, b is the semiminor ais, and c is the distance from the center of the ellipse to either focus. FOLLOW-UP Ask our students if the agree or disagree with the following statement: The sound ou hear in a whispering galler should be mudd, because the sound waves will take varing lengths of time to travel from one focus to the other, depending on where the reflect off the wall. (Disagree) ASSIGNMENT GUIDE Da 1: E. 3 4, multiples of 3 Da : E. 19, 31, 37, 41, 47, 5, 53, 55, 58 COOPERATIVE LEARNING Group Activit: E NOTES ON EXERCISES E and require students to find equations of ellipses. E are application problems involving astronom or lithotripters. E provide practice for standardized tests. ONGOING ASSESSMENT Self-Assessment: E. 1, 17, 31, 37, 53, 59 Embedded Assessment: E. 51, 5, 60 The noun eccentricit comes from the adjective eccentric, which means off-center. Mathematicall, the eccentricit is the ratio of c to a. The larger c is, compared to a, the more off-center the foci are. In an ellipse, a c 0. Dividing this inequalit b a shows that 0 e 1. So the eccentricit of an ellipse is between 0 and 1. Ellipses with highl off-center foci are elongated and have eccentricities close to 1; for eample, the orbit of Halle s comet has eccentricit e Ellipses with foci near the center are almost circular and have eccentricities close to 0; for instance, Venus s orbit has an eccentricit of What happens when the eccentricit e 0? In an ellipse, because a is positive, e c a 0 implies that c 0 and thus a b. In this case, the ellipse degenerates into a circle. Because the ellipse is a circle when a b, it is customar to denote this common value as r and call it the radius of the circle. Surprising things happen when an ellipse is nearl but not quite a circle, as in the orbit of our planet, Earth. EXAMPLE 5 Analzing Earth s Orbit Earth s orbit has a semimajor ais a Gm (gigameters) and an eccentricit of e Calculate and interpret b and c. SOLUTION Because e c a, c ea and b a c The semiminor ais b Gm is onl 0.014% shorter than the semimajor ais a Gm. The aphelion distance of Earth from the Sun is a c Gm, and the perihelion distance is a c Gm. Thus Earth s orbit is nearl a perfect circle, but the distance between the center of the Sun at one focus and the center of Earth s orbit is c.498 Gm, more than orders of magnitude greater than a b. The eccentricit as a percentage is 1.67%; this measures how far off-center the Sun is. Now tr Eercise 53.

21 5144_Demana_Ch08pp /13/06 6:49 AM Page 651 SECTION 8. Ellipses 651 EXPLORATION EXTENSIONS Determine how far apart the pushpins would need to be for the ellipse to have a b/a ratio of 1/. Draw this ellipse. NOTE ON EXPLORATION Have students work in groups on Eploration. To save time and trouble, make a 0-cm closed loop of string for each group in advance. Results will var from group to group. EXPLORATION Constructing Ellipses to Understand Eccentricit Each group will need a pencil, a centimeter ruler, scissors, some string, several sheets of unlined paper, two pushpins, and a foam board or other appropriate backing material. 1. Make a closed loop of string that is 0 cm in circumference.. Place a sheet of unlined paper on the backing material, and carefull place the two pushpins cm apart near the center of the paper. Construct an ellipse using the loop of string and a pencil as shown in Figure Measure and record the resulting values of a, b, and c for the ellipse, and compute the ratios e c a and b a for the ellipse. 3. On separate sheets of paper repeat step three more times, placing the pushpins 4, 6, and 8 cm apart. Record the values of a, b, c and the ratios e and b a for each ellipse. 4. Write our observations about the ratio b a as the eccentricit ratio e increases. Which of these two ratios measures the shape of the ellipse? Which measures how off-center the foci are? 5. Plot the ordered pairs (e, b a), determine a formula for the ratio b a as a function of the eccentricit e, and overla this function s graph on the scatter plot. WHISPERING GALLERIES In architecture, ceilings in the shape of an ellipsoid are used to create whispering galleries. A person whispering at one focus can be heard across the room b a person at the other focus. An ellipsoid is part of the design of the Teas state capitol; a hand clap made in the center of the main vestibule (at one focus of the ellipsoid) bounces off the inner elliptical dome, passes through the other focus, bounces off the dome a second time, and returns to the person as a distinct echo. Reflective Propert of an Ellipse Because of their shape, ellipses are used to make reflectors of sound, light, and other waves. If we rotate an ellipse in three-dimensional space about its focal ais, the ellipse sweeps out an ellipsoid of revolution. If we place a signal source at one focus of a reflective ellipsoid, the signal reflects off the elliptical surface to the other focus, as illustrated in Figure 8.0. This propert is used to make mirrors for optical equipment and to stud aircraft noise in wind tunnels. F 1 F FIGURE 8.0 The reflective propert of an ellipse.

22 5144_Demana_Ch08pp /13/06 6:49 AM Page CHAPTER 8 Analtic Geometr in Two and Three Dimensions Ellipsoids are used in health care to avoid surger in the treatment of kidne stones. An elliptical lithotripter emits underwater ultrahigh-frequenc (UHF) shock waves from one focus, with the patient s kidne carefull positioned at the other focus (Figure 8.1). Patient s kidne stone Source of UHF shock waves F Kidne Stone (0, 0) F 1 Source FIGURE 8.1 How a lithotripter breaks up kidne stones. EXAMPLE 6 Focusing a Lithotripter The ellipse used to generate the ellipsoid of a lithotripter has a major ais of 1 ft and a minor ais of 5 ft. How far from the center are the foci? SOLUTION From the given information, we know a 1 6 and b 5.5. So c a b So the foci are about 5 ft 5.5 inches from the center of the lithotripter. Now tr Eercise 59. White House CHAPTER OPENER PROBLEM (from page 631) The Ellipse 616 ft 58 ft PROBLEM: If the Ellipse at the White House is 616 ft long and 58 ft wide, what is its equation? SOLUTION: For simplicit s sake, we model the Ellipse as centered at 0, 0 with the -ais as its focal ais. Because the Ellipse is 616 ft long, a , and because the Ellipse is 58 ft wide, b Using a b 1, we obtain , 1. 94,864 69,696 Other models are possible.

23 5144_Demana_Ch08pp /13/06 6:49 AM Page 653 SECTION 8. Ellipses 653 QUICK REVIEW 8. (For help, go to Sections P. and P.5.) In Eercises 1 and, find the distance between the given points. 1. 3, and, , 4 and a, b (a ) 3 (b 4) In Eercises 3 and 4, solve for in terms of In Eercises 5 8, solve for algebraicall , , In Eercises 9 and 10, find eact solutions b completing the square SECTION 8. EXERCISES In Eercises 1 6, find the vertices and foci of the ellipse = In Eercises 7 10, match the graph with its equation, given that the ticks on all aes are 1 unit apart. (a) (c) 7. 1 (d) (c) (b) (d) In Eercises 11 16, sketch the graph of the ellipse b hand In Eercises 17 0, graph the ellipse using a function grapher In Eercises 1 36, find an equation in standard form for the ellipse that satisfies the given conditions. 1. Major ais length 6 on -ais, minor ais length 4. Major ais length 14 on -ais, minor ais length Foci, 0, major ais length 10 /5 / Foci 0, 3, major ais length 10 /5 / Endpoints of aes are 4, 0 and 0, 5 6. Endpoints of aes are 7, 0 and 0, 4 7. Major ais endpoints 0, 6, minor ais length 8 8. Major ais endpoints 5, 0, minor ais length 4 9. Minor ais endpoints 0, 4, major ais length Minor ais endpoints 1, 0, major ais length Major ais endpoints 1, 4 and 1, 8, minor ais length 8

24 5144_Demana_Ch08pp /13/06 6:49 AM Page CHAPTER 8 Analtic Geometr in Two and Three Dimensions 3. Major ais endpoints are, 3 and, 7, minor ais length 4 ( ) /5 ( ) / The foci are 1, 4 and 5, 4 ; the major ais endpoints are 0, 4 and 6, 4. ( 3) /9 ( 4) / The foci are, 1 and, 5 ; the major ais endpoints are, 1 and, 7. ( 3) /16 ( ) / The major ais endpoints are 3, 7 and 3, 3 ; the minor ais length is 6. ( ) /5 ( 3) / The major ais endpoints are 5, and 3, ; the minor ais length is 6. ( 1) /16 ( ) /9 1 In Eercises 37 40, find the center, vertices, and foci of the ellipse. ( 1) ( ) ( 3) ( 5) ( 3) ( 7) ( 1) 5 ( ) 1 16 In Eercises 41 44, graph the ellipse using a parametric grapher In Eercises 45 48, prove that the graph of the equation is an ellipse, and find its vertices, foci, and eccentricit In Eercises 49 and 50, write an equation for the ellipse (, 6) (, 3) (6, 3) ( 4, 5) ( 4, ) 51. Writing to Learn Prove that an equation for the ellipse with center 0, 0, foci 0, c, and semimajor ais a c 0 is a b 1, where b a c. [Hint: Refer to derivation at the beginning of the section.] 5. Writing to Learn Dancing Planets Using the data in Table 8.1, prove that the planet with the most eccentric orbit sometimes is closer to the Sun than the planet with the least eccentric orbit. (0, ) 53. The Moon s Orbit The Moon s apogee (farthest distance from Earth) is 5,710 miles, and perigee (closest distance to Earth) is 1,463 miles. Assuming the Moon s orbit of Earth is elliptical with Earth at one focus, calculate and interpret a, b, c, and e. 54. Hot Mercur Given that the diameter of the Sun is about 1.39 Gm, how close does Mercur get to the Sun s surface? 45.3 Gm 55. Saturn Find the perihelion and aphelion distances of Saturn Gm, 1507 Gm 56. Venus and Mars Write equations for the orbits of Venus and Mars in the form a b Sungrazers One comet group, known as the sungrazers, passes within a Sun s diameter (1.39 Gm) of the solar surface. What can ou conclude about a c for orbits of the sungrazers? a c 1.5(1.39) Halle s Comet The orbit of Halle s comet is AU long and 9.1 AU wide. What is its eccentricit? Lithotripter For an ellipse that generates the ellipsoid of a lithotripter, the major ais has endpoints ( 8, 0) and (8, 0). One endpoint of the minor ais is (0, 3.5). Find the coordinates of the foci. ( 51.75, 0) ( 7.19, 0) 60. Lithotripter (Refer to Figure 8.1.) A lithotripter s shape is formed b rotating the portion of an ellipse below its minor ais about its major ais. If the length of the major ais is 6 in. and the length of the minor ais is 10 in., where should the shockwave source and the patient be placed for maimum effect? Group Activities In Eercises 61 and 6, solve the sstem of equations algebraicall and support our answer graphicall. Table 8.1 Semimajor Aes and Eccentricities of the Planets Planet Semimajor Ais (Gm) Eccentricit Mercur Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto Source: Shupe, et al., National Geographic Atlas of the World (rev. 6th ed.). Washington, DC: National Geographic Societ, 199, plate 116, and other sources (, 0), (, 0) ( 3, 0), (0, 1)

25 5144_Demana_Ch08pp /13/06 6:49 AM Page 655 SECTION 8. Ellipses Group Activit Consider the sstem of equations (a) Solve the sstem graphicall. (b) If ou have access to a grapher that also does smbolic algebra, use it to find the eact solutions to the sstem. 64. Writing to Learn Look up the adjective eccentric in a dictionar and read its various definitions. Notice that the word is derived from e-centric, meaning out-of-center or off-center. Eplain how this is related to the word s everda meanings as well as its mathematical meaning for ellipses. Standardized Test Questions 65. True or False The distance from a focus of an ellipse to the closer verte is a(1 e), where a is the semimajor ais and e is the eccentricit. Justif our answer. 66. True or False The distance from a focus of an ellipse to either endpoint of the minor ais is half the length of the major ais. Justif our answer. In Eercises 67 70, ou ma use a graphing calculator to solve the problem. 67. Multiple Choice One focus of 4 4 is C (A) (4, 0). (B) (, 0). (C) ( 3, 0). (D) (, 0). (E) (1, 0). ( ) ( 3) 68. Multiple Choice The focal ais of is C (A) 1. (B). (C) 3. (D) 4. (E) Multiple Choice The center of is B (A) (4, ). (B) (4, 3). (C) (4, 4). (D) (4, 5). (E) (4, 6). 70. Multiple Choice The perimeter of a triangle with one verte on the ellipse a b 1 and the other two vertices at the foci of the ellipse would be C (A) a b. (B) a b. (C) a c. (D) b c. (E) a b c. Eplorations 71. Area and Perimeter The area of an ellipse is A ab, but the perimeter cannot be epressed so simpl: 3 a b a 3 b a b P a b ( 3 ) (a) Prove that, when a b r, these become the familiar formulas for the area and perimeter (circumference) of a circle. (b) Find a pair of ellipses such that the one with greater area has smaller perimeter. Answers will var. 7. Writing to Learn Kepler s Laws We have encountered Kepler s First and Third Laws (p. 193). Using a librar or the Internet, (a) Read about Kepler s life, and write in our own words how he came to discover his three laws of planetar motion. Answers will var. (b) What is Kepler s Second Law? Eplain it with both pictures and words. Eplanations and drawings will var. 73. Pendulum Velocit vs. Position As a pendulum swings toward and awa from a motion detector, its distance (in meters) from the detector is given b the position function (t) 3 cos(t 5), where t represents time (in seconds). The velocit (in m/sec) of the pendulum is given b (t) sin (t 5), (a) Using parametric mode on our grapher, plot the (, ) relation for velocit versus position for 0 t (b) Write the equation of the resulting conic in standard form, in terms of and, and eliminating the parameter t. 74. Pendulum Velocit vs. Position A pendulum that swings toward and awa from a motion detector has a distance (in feet) from the detector of (t) 5 3 sin( t ) and a velocit (in ft/sec) of (t) 3 cos( t ), where t represents time (in seconds). (a) Prove that the plot of velocit versus position (distance) is an ellipse. (b) Writing to Learn Describe the motion of the pendulum. Etending the Ideas 75. Prove that a nondegenerate graph of the equation A C D E F 0 is an ellipse if AC Writing to Learn The graph of the equation ( a h) ( b k) 0 is considered to be a degenerate ellipse. Describe the graph. How is it like a full-fledged ellipse, and how is it different?

26 5144_Demana_Ch08pp /13/06 6:49 AM Page CHAPTER 8 Analtic Geometr in Two and Three Dimensions 8.3 Hperbolas What ou ll learn about Geometr of a Hperbola Translations of Hperbolas Eccentricit and Orbits Reflective Propert of a Hperbola Long-Range Navigation... and wh The hperbola is the least known conic section, et it is used in astronom, optics, and navigation. Geometr of a Hperbola When a plane intersects both nappes of a right circular clinder, the intersection is a hperbola. The definition, features, and derivation for a hperbola closel resemble those for an ellipse. As ou read on, ou ma find it helpful to compare the nature of the hperbola with the nature of the ellipse. DEFINITION Hperbola A hperbola is the set of all points in a plane whose distances from two fied points in the plane have a constant difference. The fied points are the foci of the hperbola. The line through the foci is the focal ais. The point on the focal ais midwa between the foci is the center. The points where the hperbola intersects its focal ais are the vertices of the hperbola. (See Figure 8..) Focus Verte Verte Center FIGURE 8. Ke points on the focal ais of a hperbola. = a Focal ais Focus = a P(, ) Figure 8.3 shows a hperbola centered at the origin with its focal ais on the -ais. The vertices are at ( a, 0) and (a, 0), where a is some positive constant. The fied points F 1 ( c, 0) and F (c, 0) are the foci of the hperbola, with c a. Notice that the hperbola has two branches. For a point P(, ) on the right-hand branch, PF 1 PF a. On the left-hand branch, PF PF 1 a. Combining these two equations gives us PF 1 PF a. Using the distance formula, the equation becomes c 0 c 0 a. c a c c c 4a 4a ( ) c c c a c ) a c a c c a 4 a c c Simplif. Square. Square. F 1 ( c, 0) O F (c, 0) c a ) a a c a Simplif. FIGURE 8.3 Structure of a Hperbola. The difference of the distances from the foci to each point on the hperbola is a constant. Letting b c a, we have which is usuall written as b a a b, a b 1.

27 5144_Demana_Ch08pp /13/06 6:49 AM Page 657 SECTION 8.3 Hperbolas 657 OBJECTIVE Students will be able to find the equation, vertices, and foci of a hperbola. MOTIVATE Have students use a graphing utilit to graph 1 and describe the graph. LESSON GUIDE Da 1: Geometr of a Hperbola; Translations of Hperbolas Da : Eccentricit and Orbits; Reflective Propert of a Hperbola; Long-Range Navigation NAMING AXES The word transverse comes from the Latin trans vertere: to go across. The transverse ais goes across from one verte to the other. The conjugate ais is the transverse ais for the conjugate hperbola, defined in Eercise 73. Because these steps can be reversed, a point P, satisfies this last equation if and onl if the point lies on the hperbola defined b PF 1 PF a, provided that c a 0 and b c a. The Pthagorean relation b c a can be written man was, including a c b and c a b. The equation a b 1 is the standard form of the equation of a hperbola centered at the origin with the -ais as its focal ais. A hperbola centered at the origin with the -ais as its focal ais is the inverse relation of a b 1, and thus has an equation of the form a b 1. As with other conics, a line segment with endpoints on a hperbola is a chord of the hperbola. The chord ling on the focal ais connecting the vertices is the transverse ais of the hperbola. The length of the transverse ais is a. The line segment of length b that is perpendicular to the focal ais and that has the center of the hperbola as its midpoint is the conjugate ais of the hperbola. The number a is the semitransverse ais, and b is the semiconjugate ais. The hperbola a b 1 has two asmptotes. These asmptotes are slant lines that can be found b replacing the 1 on the right-hand side of the hperbola s equation b a 0: a b 1 a b 0 b a hperbola Replace 1 b 0. asmptotes A hperbola centered at the origin with its focal ais one of the coordinate aes is smmetric with respect to the origin and both coordinate aes. Such a hperbola can be sketched b drawing a rectangle centered at the origin with sides parallel to the coordinate aes, followed b drawing the asmptotes through opposite corners of the rectangle, and finall sketching the hperbola using the central rectangle and asmptotes as guides, as shown in the Drawing Lesson. Drawing Lesson How to Sketch the Hperbola /a /b 1 1. Sketch line segments at a and b, and complete the rectangle the determine.. Sketch the asmptotes b etending the rectangle s diagonals. 3. Use the rectangle and asmptotes to guide our drawing. b a a b b = a b a a b b = a b a a b = 1 a b

28 5144_Demana_Ch08pp /13/06 6:50 AM Page CHAPTER 8 Analtic Geometr in Two and Three Dimensions (a) = ( c, 0) ( a, 0) (a, 0) (c, 0) = 1 a b (0, c) (b) b a b = a (0, a) (0, a) (0, c) = = 1 a b a b a = b FIGURE 8.4 Hperbolas centered at the origin with foci on (a) the -ais and (b) the -ais. Hperbolas with Center (0, 0) Standard equation a b 1 a b 1 Focal ais -ais -ais Foci c, 0 0, c Vertices a, 0 0, a Semitransverse ais a a Semiconjugate ais b b Pthagorean relation c a b c a b Asmptotes b a a b See Figure 8.4. EXAMPLE 1 Finding the Vertices and Foci of a Hperbola Find the vertices and the foci of the hperbola SOLUTION Dividing both sides of the equation b 36 ields the standard form So a 9, b 4, and c a b Thus the vertices are 3, 0, and the foci are 1 3, 0. Now tr Eercise 1. If we wish to graph a hperbola using a function grapher, we need to solve the equation of the hperbola for, as illustrated in Eample. EXAMPLE Finding an Equation and Graphing a Hperbola Find an equation of the hperbola with foci 0, 3 and 0, 3 whose conjugate ais has length 4. Sketch the hperbola and its asmptotes, and support our sketch with a grapher. SOLUTION The center is 0, 0. The foci are on the -ais with c 3. The semiconjugate ais is b 4. Thus a c b 3 5. So the standard form of the equation for the hperbola is Using a 5.4 and b, we can sketch the central rectangle, the asmptotes, and the hperbola itself. Tr doing this. To graph the hperbola using a function grapher, we solve for in terms of. 1 Add [ 9.4, 9.4] b [ 6., 6.] FIGURE 8.5 The hperbola 5 4 1, shown with its asmptotes. (Eample ) Multipl b Etract square roots. Figure 8.5 shows the graphs of and 5 1 4, together with the asmptotes of the hperbola 3 5 and 4 5. Now tr Eercise 17.

29 5144_Demana_Ch08pp /13/06 6:50 AM Page 659 SECTION 8.3 Hperbolas 659 = b ( h) + k a In Eample, because the hperbola had a vertical focal ais, selecting a viewing rectangle was eas. When a hperbola has a horizontal focal ais, we tr to select a viewing window to include the two vertices in the plot and thus avoid gaps in the graph of the hperbola. (h a, k) (h c, k) (h, k) (h + a, k) (h + c, k) = k Translations of Hperbolas When a hperbola with center 0, 0 is translated horizontall b h units and verticall b k units, the center of the hperbola moves from 0, 0 to h, k, as shown in Figure 8.6. Such a translation does not change the length of the transverse or conjugate ais or the Pthagorean relation. = b ( h) + k a (h, k + c) (h, k + a) (h, k a) (h, k c) = h (h, k) (a) (b) ( h) + k = a b = a ( h) + k b FIGURE 8.6 Hperbolas with center h, k and foci on (a) k and (b) h. Hperbolas with Center (h, k) Standard h a k b 1 equation k a h b 1 Focal ais k h Foci h c, k h, k c Vertices h a, k h, k a Semitransverse a a ais Semiconjugate b b ais Pthagorean c a b c a b relation Asmptotes b a h k a h k b See Figure EXAMPLE 3 Finding an Equation of a Hperbola Find the standard form of the equation for the hperbola whose transverse ais has endpoints, 1 and 8, 1, and whose conjugate ais has length 8. SOLUTION Figure 8.7 shows the transverse-ais endpoints, the conjugate ais, and the center of the hperbola. The standard equation of this hperbola has the form h a k b 1, (, 1) (8, 1) 8 10 where the center h, k is the midpoint 3, 1 of the transverse ais. The semitransverse ais and semiconjugate ais are a 8 5 and b 8 4. So the equation we seek is FIGURE 8.7 Given information for Eample , Now tr Eercise

30 5144_Demana_Ch08pp /13/06 6:50 AM Page CHAPTER 8 Analtic Geometr in Two and Three Dimensions EXAMPLE 4 Locating Ke Points of a Hperbola Find the center, vertices, and foci of the hperbola SOLUTION The center h, k is, 5. Because the semitransverse ais a 9 3, the vertices are h a, k 3, 5 1, 5 and (h a, k 3, 5 5, 5. Because c a b , the foci h c, k are 5 8, 5, or approimatel 5.6, 5 and 9.6, 5. Now tr Eercise 39. With the information found about the hperbola in Eample 4 and knowing that its semiconjugate ais b 4 9 7, we could easil sketch the hperbola. Obtaining an accurate graph of the hperbola using a function grapher is another matter. Often, the best wa to graph a hperbola using a graphing utilit is to use parametric equations. EXPLORATION EXTENSIONS Create a parameterization for the general hperbola ( h) a ( k) b 1. FOLLOW-UP Ask wh it would not be convenient to use the hperbola with foci at Q and R in Eample 6. ASSIGNMENT GUIDE Da 1: E. 3 5, multiples of 3 Da : E. 17, 3, 37, 49, 55, 57, 58 COOPERATIVE LEARNING Group Activit: E NOTES ON EXERCISES E and 51 5 require students to write equations of hperbolas. E provide practice for standardized tests. E introduce the topics of conjugate hperbolas and the focal width of a hperbola. ONGOING ASSESSMENT Self-Assessment: E. 1, 17, 31, 39, 55, 57 Embedded Assessment: E. 53, 69, 7 EXPLORATION 1 Graphing a Hperbola Using Its Parametric Equations 1. Use the Pthagorean trigonometr identit sec t tan t 1 to prove that the parameterization 1 3 cos t, 1 tan t 0 t will produce a graph of the hperbola ( Using Dot graphing mode, graph 1 3 cos t, 1 tan t 0 t in a square viewing window to support part 1 graphicall. Switch to Connected graphing mode, and regraph the equation. What do ou observe? Eplain. 3. Create parameterizations for the hperbolas in Eamples 1,, 3, and Graph each of our parameterizations in part 3 and check the features of the obtained graph to see whether the match the epected geometric features of the hperbola. If necessar, revise our parameterization and regraph until all features match. 5. Prove that each of our parameterizations is valid. Eccentricit and Orbits DEFINITION Eccentricit of a Hperbola The eccentricit of a hperbola is c a e b, a a where a is the semitransverse ais, b is the semiconjugate ais, and c is the distance from the center to either focus.

31 5144_Demana_Ch08pp /13/06 6:50 AM Page 661 SECTION 8.3 Hperbolas Gm 90 Gm 81.5 Gm Sun (170, 0) Path of comet FIGURE 8.8 The graph of one branch of /6400 /, (Eample 5) For a hperbola, because c a, the eccentricit e 1. In Section 8. we learned that the eccentricit of an ellipse satisfies the inequalit 0 e 1 and that, for e 0, the ellipse is a circle. In Section 8.5 we will generalize the concept of eccentricit to all tpes of conics and learn that the eccentricit of a parabola is e 1. Kepler s first law of planetar motion sas that a planet s orbit is elliptical with the Sun at one focus. Since 1609, astronomers have generalized Kepler s law; the current theor states: A celestial bod that travels within the gravitational field of a much more massive bod follows a path that closel approimates a conic section that has the more massive bod as a focus. Two bodies that do not differ greatl in mass (such as Earth and the Moon, or Pluto and its moon Charon) actuall revolve around their balance point, or barcenter. In theor, a comet can approach the Sun from interstellar space, make a partial loop about the Sun, and then leave the solar sstem returning to deep space; such a comet follows a path that is one branch of a hperbola. EXAMPLE 5 Analzing a Comet s Orbit A comet following a hperbolic path about the Sun has a perihelion distance of 90 Gm. When the line from the comet to the Sun is perpendicular to the focal ais of the orbit, the comet is 81.5 Gm from the Sun. Calculate a, b, c, and e. What are the coordinates of the center of the Sun if we coordinatize space so that the hperbola is given b a b 1? F P = F H Parabola Hperbola Ellipse F H = F E Primar mirror F E FIGURE 8.9 Cross section of a reflecting telescope. SOLUTION The perihelion distance is c a 90. When c, b a (see Eercise 74). So b a 81.5, or b 81.5a. Because b c a, we have the sstem c a 90 and c a 81.5a, which ields the equation: a 90 a 81.5a a 180a 8100 a 81.5a a a 80 So a 80 Gm, b 150 Gm, c 170 Gm, and e If the comet s path is the branch of the hperbola with positive -coordinates, then the Sun is at the focus c, 0 170, 0. See Figure 8.8. Now tr Eercise 55. Reflective Propert of a Hperbola Like other conics, a hperbola can be used to make a reflector of sound, light, and other waves. If we rotate a hperbola in three-dimensional space about its focal ais, the hperbola sweeps out a hperboloid of revolution. If a signal is directed toward a focus of a reflective hperboloid, the signal reflects off the hperbolic surface to the other focus. In Figure 8.9 light reflects off a primar parabolic mirror toward the mirror s focus F P F H, which is also the focus of a small hperbolic mirror. The light is then reflected off the hperbolic mirror, toward the hperboloid s other focus F H F E, which is also the focus of an elliptical mirror. Finall the light is reflect into the observer s ee, which is at the second focus of the ellipsoid F E.

32 5144_Demana_Ch08pp /13/06 6:50 AM Page CHAPTER 8 Analtic Geometr in Two and Three Dimensions R Reflecting telescopes date back to the 1600s when Isaac Newton used a primar parabolic mirror in combination with a flat secondar mirror, slanted to reflect the light out the side to the eepiece. French optician G. Cassegrain was the first to use a hperbolic secondar mirror, which directed the light through a hole at the verte of the primar mirror (see Eercise 70). Toda, reflecting telescopes such as the Hubble Space Telescope have become quite sophisticated and must have nearl perfect mirrors to focus properl. 100 mi O 80 mi FIGURE 8.30 Strategicall located LORAN transmitters O, Q, and R. (Eample 6) FIGURE 8.31 Graphs for Eample 6. Q [ 00, 400] b [ 00, 400] Long-Range Navigation Hperbolas and radio signals are the basis of the LORAN (long-range navigation) sstem. Eample 6 illustrates this sstem using the definition of hperbola and the fact that radio signals travel 980 ft per microsecond 1 microsecond 1 sec 10 6 sec. EXAMPLE 6 Using the LORAN Sstem Radio signals are sent simultaneousl from transmitters located at points O, Q, and R (Figure 8.30). R is 100 mi due north of O, and Q is 80 mi due east of O. The LORAN receiver on sloop Gloria receives the signal from O 33.7 sec after the signal from R, and sec after the signal from Q. What is the sloop s bearing and distance from O? SOLUTION The Gloria is at a point of intersection between two hperbolas: one with foci O and R, the other with foci O and Q. The hperbola with foci O 0, 0 and R 0, 100 has center 0, 50 and transverse ais a 33.7 sec 980 ft sec 1 mi 580 ft 60 mi. Thus a 30 and b c a , ielding the equation ( The hperbola with foci O 0, 0 and Q 80, 0 has center 40, 0 and transverse ais a sec 980 ft sec 1 mi 580 ft 48 mi. Thus a 4 and b c a , ielding the equation The Gloria is at point P where upper and right-hand branches of the hperbolas meet (see Figure 8.31). Using a grapher we find that P , So the bearing from point O is ) 1( 90 tan and the distance from point O is 44.04, d So the Gloria is about mi east and mi north of point O on a bearing of roughl 44, and the sloop is about 69 mi from point O. Now tr Eercise 57.

33 5144_Demana_Ch08pp /13/06 6:50 AM Page 663 SECTION 8.3 Hperbolas 663 QUICK REVIEW 8.3 (For help, go to Sections P., P.5, and 7.1.) In Eercises 1 and, find the distance between the given points. 1. 4, 3 and 7, a, 3 and b, c (b a ) (c 3) In Eercises 3 and 4, solve for in terms of In Eercises 5 8, solve for algebraicall no solution , , 5.55 In Eercises 9 and 10, solve the sstem of equations. 9. c a and c a 16a 3 a 3, c c a 1 and c a 5a 1 a 1, c 13 SECTION 8.3 EXERCISES In Eercises 1 6, find the vertices and foci of the hperbola In Eercises 17, graph the hperbola using a function grapher In Eercises 7 10, match the graph with its equation (c) 8. 1 (b) (a) (d) In Eercises 11 16, sketch the graph of the hperbola b hand (a) (b) (c) (d) In Eercises 3 38, find an equation in standard form for the hperbola that satisfies the given conditions. 3. Foci ( 3, 0), transverse ais length 4 /4 / Foci 0, 3, transverse ais length 4 /4 / Foci 0, 15, transverse ais length 8 /16 / Foci 5, 0, transverse ais length 3 7. Center at 0, 0, a 5, e, horizontal focal ais 8. Center at 0, 0, a 4, e 3, vertical focal ais 9. Center at 0, 0, b 5, e 13 1, vertical focal ais 30. Center at 0, 0, c 6, e, horizontal focal ais 31. Transverse ais endpoints (, 3) and (, 1), conjugate ais length 6 3. Transverse ais endpoints (5, 3) and ( 7, 3), conjugate ais length Transverse ais endpoints 1, 3 and 5, 3, slope of one asmptote Transverse ais endpoints, and, 7, slope of one asmptote Foci 4, and,, transverse ais endpoints 3, and 1, 36. Foci 3, 11 and 3, 0, transverse ais endpoints 3, 9 and 3, 37. Center at 3, 6, a 5, e, vertical focal ais 38. Center at 1, 4, c 6, e, horizontal focal ais

34 5144_Demana_Ch08pp /13/06 6:50 AM Page CHAPTER 8 Analtic Geometr in Two and Three Dimensions In Eercises 39 4, find the center, vertices, and the foci of the hperbola. ( 1) ( ) 39. = ( 4) ( 6) signals from the three transmitters. The signal from O arrives 33.7 sec after the signal from R, and sec after the signal from Q. Determine the ship s bearing and distance from point O. 41. ( 3) ( ) 1 4. ( 1) ( 5) In Eercises 43 46, graph the hperbola using a parametric grapher in Dot graphing mode In Eercises 47 50, graph the hperbola, and find its vertices, foci, and eccentricit In Eercises 51 and 5, write an equation for the hperbola a0, b R 80 mi O 58. Gun Location Observers are located at positions A, B, and C with A due north of B. A cannon is located somewhere in the first quadrant as illustrated in the figure. A hears the sound of the cannon sec before B, and C hears the sound 4 sec before B. Determine the bearing and distance of the cannon from point B. (Assume that sound travels at 1100 ft sec.) 00 mi Q (3, ) (, 0) (, 0) (, ) A a0, b 53. Writing to Learn Prove that an equation for the hperbola with center 0, 0, foci 0, c, and semitransverse ais a is a b 1, where c a 0 and b c a. [Hint: Refer to derivation at the beginning of the section.] 54. Degenerate Hperbolas Graph the degenerate hperbola. (a) 0 (b) Rogue Comet A comet following a hperbolic path about the Sun has a perihelion of 10 Gm. When the line from the comet to the Sun is perpendicular to the focal ais of the orbit, the comet is 50 Gm from the Sun. Calculate a, b, c, and e. What are the coordinates of the center of the Sun if the center of the hperbolic orbit is 0, 0 and the Sun lies on the positive -ais? 56. Rogue Comet A comet following a hperbolic path about the Sun has a perihelion of 140 Gm. When the line from the comet to the Sun is perpendicular to the focal ais of the orbit, the comet is 405 Gm from the Sun. Calculate a, b, c, and e. What are the coordinates of the center of the Sun if the center of the hperbolic orbit is 0, 0 and the Sun lies on the positive -ais? 57. Long-Range Navigation Three LORAN radio transmitters are positioned as shown in the figure, with R due north of O and Q due east of O. The cruise ship Princess Ann receives simultaneous 4000 ft Group Activities In Eercises 59 and 60, solve the sstem of equations algebraicall and support our answer graphicall. B 7000 ft Group Activit Consider the sstem of equations (a) Solve the sstem graphicall. (b) If ou have access to a grapher that also does smbolic algebra, use it to find the the eact solutions to the sstem. C

35 5144_Demana_Ch08pp /13/06 6:50 AM Page 665 SECTION 8.3 Hperbolas Writing to Learn Escape of the Unbound When NASA launches a space probe, the probe reaches a speed sufficient for it to become unbound from Earth and escape along a hperbolic trajector. Look up escape speed in an astronom tetbook or on the Internet, and write a paragraph in our own words about what ou find. Standardized Test Questions 63. True or False The distance from a focus of a hperbola to the closer verte is a(e 1), where a is the semitransverse ais and e is the eccentricit. Justif our answer. 64. True or False Unlike that of an ellipse, the Pthagorean relation for a hperbola is the usual a b c. Justif our answer. In Eercises 65 68, ou ma use a graphing calculator to solve the problem. 65. Multiple Choice One focus of 4 4 is B (A) (4, 0). (B) ( 5, 0). (C) (, 0). (D) ( 3 0). (E) (1, 0). 66. Multiple Choice The focal ais of ( 5) ( 6) 1 is E 9 16 (A). (B) 3. (C) 4. (D) 5. (E) Multiple Choice The center of is B (A) (, ). (B) (, 3). (C) (, 4). (D) (, 5). (E) (, 6). 68. Multiple Choice The slopes of the asmptotes of the hperbola 1 are C 4 3 (A) 1. (B) 3. (C) 3. (D) 3. (E) 4 3. Eplorations 69. Constructing Points of a Hperbola Use a geometr software package, such as Cabri Geometr II, The Geometer s Sketchpad, or a similar application on a hand-held device, to carr out the following construction. (a) Start b placing the coordinate aes in the construction window. (b) Construct two points on the -ais at 5, 0 as the foci. (c) Construct concentric circles of radii r 1,, 3,, 1 centered at these two foci. (d) Construct the points where these concentric circles meet and have a difference of radii of a 6, and overla the conic that passes through these points if the software has a conic tool. (e) Find the equation whose graph includes all of these points. /9 / Cassegrain Telescope A Cassegrain telescope as described in the section has the dimensions shown in the figure. Find the standard form for the equation of the hperbola centered at the origin with the focal ais the -ais. Eepiece Primar Parabolic Mirror F H Etending the Ideas Secondar Hperbolic Mirror F P F H 80 cm 100 cm 10 cm 71. Prove that a nondegenerate graph of the equation A C D E F 0 is a hperbola if AC Writing to Learn The graph of the equation h k 0 a b is considered to be a degenerate hperbola. Describe the graph. How is it like a full-fledged hperbola, and how is it different? 73. Conjugate Hperbolas The hperbolas h k k h 1 and 1 a b b a obtained b switching the order of subtraction in their standard equations are conjugate hperbolas. Prove that these hperbolas have the same asmptotes and that the conjugate ais of each of these hperbolas is the transverse ais of the other hperbola. 74. Focal Width of a Hperbola Prove that, for the hperbola 1, a b if c, then b a. Wh is it reasonable to define the focal width of such hperbolas to be b a? 75. Writing to Learn Eplain how the standard form equations for the conics are related to A B C D E F 0.

36 5144_Demana_Ch08pp /13/06 6:50 AM Page CHAPTER 8 Analtic Geometr in Two and Three Dimensions 8.4 Translation and Rotation of Aes What ou ll learn about Second-Degree Equations in Two Variables Translating Aes versus Translating Graphs Rotation of Aes Discriminant Test... and wh You will see ellipses, hperbolas, and parabolas as members of the famil of conic sections rather than as separate tpes of curves. OBJECTIVE Students will be able to determine equations for translated and rotated aes for conic sections. MOTIVATE Have students solve 4 for and graph the result using a function grapher. Describe the graph. LESSON GUIDE Da 1: Second-Degree Equations in Two Variables; Translating Aes versus Translating Graphs Da : Rotation of Aes; Discriminant Test Second-Degree Equations in Two Variables In Section 8.1, we began with a unified approach to conic sections, learning that parabolas, ellipses, and hperbolas are all cross sections of a right circular cone. In Sections , we gave separate plane-geometr definitions for parabolas, ellipses, and hperbolas that led to separate kinds of equations for each tpe of curve. In this section and the net, we once again consider parabolas, ellipses, and hperbolas as a unified famil of interrelated curves. In Section 8.1, we claimed that the conic sections can be defined algebraicall in the Cartesian plane as the graphs of second-degree equations in two variables, that is, equations of the form EXAMPLE 1 Graphing a Second-Degree Equation Solve for, and use a function grapher to graph SOLUTION Rearranging terms ields the equation: The quadratic formula gives us A B C D E F 0, where A, B, and C are not all zero. In this section, we investigate equations of this tpe, which are reall just quadratic equations in and. Because the are quadratic equations, we can adapt familiar methods to this unfamiliar setting. That is eactl what we do in Eamples [ 9.4, 9.4] b [ 6., 6.] FIGURE 8.3 The graph of (Eample 1) Let Y and Y , and graph the two equations in the same viewing window, as shown in Figure 8.3. The combined figure appears to be an ellipse. Now tr Eercise 1.

37 5144_Demana_Ch08pp /13/06 6:50 AM Page 667 SECTION 8.4 Translation and Rotation of Aes 667 In the equation in Eample 1, there was no B term. None of the eamples in Sections included such a cross-product term. A cross-product term in the equation causes the graph to tilt relative to the coordinate aes, as illustrated in Eamples and 3. [ 9.4, 9.4] b [ 6., 6.] FIGURE 8.33 The graph of 9 0. (Eample ) [ 3, 3] b [ 5, 5] (a) X=3.75 Y=9.375 [ 3, 3] b [ 5, 5] (b) FIGURE 8.34 The graph of (a) with a gap and (b) with the trace feature activated at the connecting point. (Eample 3) EXAMPLE Graphing a Second-Degree Equation Solve for, and use a function grapher to graph 9 0. SOLUTION This equation can be rewritten as 9 or as 9. The graph of this equation is shown in Figure It appears to be a hperbola with a slant focal ais. Now tr Eercise 5. EXAMPLE 3 Graphing a Second-Degree Equation Solve for, and use a function grapher to graph SOLUTION We rearrange the terms as a quadratic equation in : The quadratic formula gives us (4 90) (4 90) 4(4)( 30 0) 45 (4) Let and, 4 4 and graph the two equations in the same viewing window, as shown in Figure 8.34a. The combined figure appears to be a parabola, with a slight gap due to grapher failure. The combined graph should connect at a point for which the radicand , that is, when Figure 8.34b supports this analsis. Now tr Eercise 9. The graphs obtained in Eamples 1 3 all appear to be conic sections, but how can we be sure? If the are conics, then we probabl have classified Eamples 1 and correctl, but couldn t the graph in Eample 3 (Figure 8.34) be part of an ellipse or one branch of a hperbola? We now set out to answer these questions and to develop methods for simplifing and classifing second-degree equations in two variables. Translating Aes versus Translating Graphs The coordinate aes are often viewed as a permanent fiture of the plane, but this just isn t so. We can shift the position of aes just as we have been shifting the position of graphs since Chapter 1. Such a translation of aes produces a new set of aes parallel to the original aes, as shown in Figure 8.35 on the net page.

38 5144_Demana_Ch08pp /13/06 6:50 AM Page CHAPTER 8 Analtic Geometr in Two and Three Dimensions P(, ) = P(, ) O O (h, k) h k FIGURE 8.35 A translation of Cartesian coordinate aes. Figure 8.35 shows a plane containing a point P that is named in two was: using the coordinates, and the coordinates,. The coordinates, are based on the original - and -aes and the original origin O, while, are based on the translated - and -aes and the corresponding origin O. Translation-of-Aes Formulas The coordinates, and, based on parallel sets of aes are related b either of the following translation formulas : h and k or h and k. We use the second pair of translation formulas in Eample 4. EXAMPLE 4 Revisiting Eample 1 Prove that is the equation of an ellipse. Translate the coordinate aes so that the origin is at the center of this ellipse. SOLUTION We complete the square of both and : 8 1 FIGURE 8.36 The graph of (Eample 4) This is a standard equation of an ellipse. If we let 1 and, then the equation of the ellipse becomes Figure 8.36 shows the graph of this final equation in the new coordinate sstem, with the original -aes overlaid. Compare Figures 8.3 and Now tr Eercise 1.

39 5144_Demana_Ch08pp /13/06 6:50 AM Page 669 SECTION 8.4 Translation and Rotation of Aes 669 α O P(, ) = P(, ) α α Rotation of Aes To show that the equation in Eample or 3 is the equation of a conic section, we need to rotate the coordinate aes so that one ais aligns with the (focal) ais of the conic. In such a rotation of aes, the origin stas fied, and we rotate the - and -aes through an angle to obtain the - and -aes. (See Figure 8.37.) Figure 8.37 shows a plane containing a point P named in two was: as, and as,. The coordinates, are based on the original - and -aes, while (, ) are based on the rotated - and -aes. FIGURE 8.37 A rotation of Cartesian coordinate aes. Rotation-of-Aes Formulas The coordinates, and, based on rotated sets of aes are related b either of the following rotation formulas: cos sin and sin cos, or cos sin and sin cos. where, 0, is the angle of rotation. The first pair of equations was established in Eample 10 of Section 7.. The second pair can be derived directl from the geometr of Figure 8.37 (see Eercise 55) and is used in Eample 5. ' 3 3 FIGURE 8.38 The graph of 9 0. (Eample 5) ' EXAMPLE 5 Revisiting Eample Prove that 9 0 is the equation of a hperbola b rotating the coordinate aes through an angle 4. SOLUTION Because cos 4) sin 4 1, the rotation equations become and. So b rotating the aes, the equation 9 0 becomes To see that this is the equation of a hperbola, we put it in standard form: Figure 8.38 shows the graph of the original equation in the original sstem with the -aes overlaid. Now tr Eercise 37.

40 5144_Demana_Ch08pp /13/06 6:50 AM Page CHAPTER 8 Analtic Geometr in Two and Three Dimensions In Eample 5 we converted a second-degree equation in and into a second-degree equation in and using the rotation formulas. B choosing the angle of rotation appropriatel, there was no cross-product term in the final equation, which allowed us to put it in standard form. We now generalize this process. Coefficients for a Conic in a Rotated Sstem If we appl the rotation formulas to the general second-degree equation in and, we obtain a second-degree equation in and of the form A B C D E F 0, where the coefficients are A A cos B cos sin C sin B B cos C A sin C C cos B cos sin A sin D D cos E sin E E cos D sin F F In order to eliminate the cross-product term and thus align the coordinate aes with the focal ais of the conic, we rotate the coordinate aes through an angle that causes B to equal 0. Setting B B cos C A sin 0 leads to the following useful result. Angle of Rotation to Eliminate the Cross-Product Term If B 0, an angle of rotation such that cot A C and 0 < < B will eliminate the term B from the second-degree equation in the rotated coordinate sstem. EXAMPLE 6 Revisiting Eample 3 Prove that is the equation of a parabola b rotating the coordinate aes through a suitable angle. SOLUTION The angle of rotation must satisf the equation cot A C B 4 4. So cos 3 5, continued

41 5144_Demana_Ch08pp /13/06 6:50 AM Page 671 SECTION 8.4 Translation and Rotation of Aes 671 FOLLOW-UP Ask students to eplain wh it was oka to assume cos 0 and sin 0 in Eample FIGURE 8.39 The graph of (Eample 6) and thus cos 1 c o s sin 1 c o s , 5. 5 Therefore the coefficients of the transformed equation are A B 0 C D E F 450 So the equation becomes After completing the square of the -terms, the equation becomes If we translate using h 1 5 and k , then the equation becomes 6, 5 a standard equation of a parabola. Figure 8.39 shows the graph of the original equation in the original coordinate sstem, with the -aes overlaid. Now tr Eercise 39. Discriminant Test Eample 6 demonstrates that the algebra of rotation can get ugl. Fortunatel, we can determine which tpe of conic a second-degree equation represents b looking at the sign of the discriminant B 4AC. Discriminant Test The second-degree equation A B C D E F 0 graphs as a hperbola if B 4AC 0, a parabola if B 4AC 0, an ellipse if B 4AC 0, ecept for degenerate cases.

42 5144_Demana_Ch08pp /13/06 6:50 AM Page CHAPTER 8 Analtic Geometr in Two and Three Dimensions ASSIGNMENT GUIDE Da 1: E. 3 30, multiples of 3 Da : E , multiples of 3 COOPERATIVE LEARNING Group Activit: E. 65 NOTES ON EXERCISES E ask students to prove the translation formulas. E are basic problems involving rotation of aes. E ask students to prove the rotation formulas. E provide practice for standardized tests. ONGOING ASSESSMENT Self-Assessment: E. 1, 5, 9, 1, 37, 39, 43 Embedded Assessment: E. 31 This test hinges on the fact that the discriminant B 4AC is invariant under rotation ; in other words, even though A, B, and C do change when we rotate the coordinate aes, the combination B 4AC maintains its value. EXAMPLE 7 Revisiting Eamples 5 and 6 (a) In Eample 5, before the rotation B 4AC , and after the rotation B 4A C The positive discriminant tells us the conic is a hperbola. (b) In Eample 6, before the rotation B 4AC , and after the rotation B 4 A C The zero discriminant tells us the conic is a parabola. Now tr Eercise 43. Not onl is the discriminant B 4AC invariant under rotation, but also its sign is invariant under translation and under algebraic manipulations that preserve the equivalence of the equation, such as multipling both sides of the equation b a nonzero constant. The discriminant test can be applied to degenerate conics. Table 8. displas the three basic tpes of conic sections grouped with their associated degenerate conics. Each conic or degenerate conic is shown with a sample equation and the sign of its discriminant. Table 8. Conics and the Equation A B C D E F 0 Sample Sign of Conic Equation A B C D E F Discriminant Hperbola Positive Intersecting lines Positive Parabola 1 Zero Parallel lines Zero One line 0 1 Zero No graph Zero Ellipse Negative Circle Negative Point Negative No graph Negative QUICK REVIEW 8.4 (For help, go to Sections 4.7 and 5.4.) In Eercises 1 10, assume Given that cot 5 1, find cos. cos 5/13. Given that cot 8 15, find cos. cos 8/17 3. Given that cot 1 3, find cos. cos 1/ 4. Given that cot 5, find cos. cos /3 5. Given that cot 0, find. /4 6. Given that cot 3, find. /1 7. Given that cot 3 4, find cos. cos / 5 8. Given that cot 3 7, find cos. cos 14 /4 9. Given that cot 5/ 1 1, find sin. sin 1/ Given that cot 45 8, find sin. sin / 53

43 5144_Demana_Ch08pp /13/06 6:50 AM Page 673 SECTION 8.4 Translation and Rotation of Aes 673 SECTION 8.4 EXERCISES In Eercises 1 1, solve for, and use a function grapher to graph the conic In Eercises 13 16, write an equation in standard form for the conic shown (, 1) 3 In Eercises 17 0, using the point P, and the translation information, find the coordinates of P in the translated coordinate sstem. 17. P,, 3, h, k 4 (, ) (4, 1) 18. P,, 5, h 4, k 7 (, ) (, 1) 19. P, 6, 3, h 1, k 5 (, ) (5, 3 5 ) 0. P, 5, 4, h, k 3 (, ) ( 5, 1) (, 4) 3 4 In Eercises 1 30, identif the tpe of conic, write the equation in standard form, translate the conic to the origin, and sketch it in the translated coordinate sstem Writing to Learn Translation Formulas Use the geometric relationships illustrated in Figure 8.35 to eplain the translation formulas h and k. 3. Translation Formulas Prove that if h and k, then h and k. In Eercises 33 36, using the point P, and the rotation information, find the coordinates of P in the rotated coordinate sstem. 33. P,, 5, 4 (3 /, 7 /) 34. P, 6, 3, P, 5, 4, cot 3 5 ( 5.94,.38) 36. P,, 3, cot 0 (5 /, /) In Eercises 37 40, identif the tpe of conic, and rotate the coordinate aes to eliminate the term. Write and graph the transformed equation In Eercises 41 and 4, identif the tpe of conic, solve for, and graph the conic. Approimate the angle of rotation needed to eliminate the term In Eercises 43 5, use the discriminant B 4AC to decide whether the equation represents a parabola, an ellipse, or a hperbola ; ellipse ; hperbola ; parabola ; hperbola ; ellipse ; hperbola

44 5144_Demana_Ch08pp /13/06 6:50 AM Page CHAPTER 8 Analtic Geometr in Two and Three Dimensions ; hperbola ; ellipse ; ellipse ; ellipse 53. Revisiting Eample 5 Using the results of Eample 5, find the center, vertices, and foci of the hperbola 9 0 in the original coordinate sstem. 54. Revisiting Eamples 3 and 6 Use information from Eamples 3 and 6 (a) to prove that the point P, 3.75, where the graphs of Y and Y meet is not the verte of the parabola, (b) to prove that the point V, 3.6, 8.7 is the verte of the parabola. 55. Rotation Formulas Prove cos sin and sin cos using the geometric relationships illustrated in Figure Rotation Formulas Prove that if cos sin and sin cos, then cos sin and sin cos. Standardized Test Questions 57. True or False The graph of the equation A C D E F 0 (A and C not both zero) has a focal ais aligned with the coordinate aes. Justif our answer. 58. True or False The graph of the equation D E F 0 is a circle or a degenerate circle. Justif our answer. In Eercises 59 6, solve the problem without using a calculator. 59. Multiple Choice Which of the following is not a reason to translate the aes of a conic? B (A) to simplif its equation (B) to eliminate the cross-product term (C) to place its center or verte at the origin (D) to make it easier to identif its tpe (E) to make it easier to sketch b hand 60. Multiple Choice Which of the following is not a reason to rotate the aes of a conic? C (A) to simplif its equation (B) to eliminate the cross-product term (C) to place its center or verte at the origin (D) to make it easier to identif its tpe (E) to make it easier to sketch b hand 61. Multiple Choice The vertices of are A (A) (1 4, ) (B) (1 3, ) (C) (4 1, 3) (D) (4, 3) (E) (1, 3) 6. Multiple Choice The asmptotes of the hperbola 4 are E (A),. (C),. (E) the coordinate aes Eplorations (B),. (D) 4, Aes of Oblique Conics The aes of conics that are not aligned with the coordinate aes are often included in the graphs of conics. (a) Recreate the graph shown in Figure 8.38 using a function grapher including the - and -aes. What are the equations of these rotated aes? (b) Recreate the graph shown in Figure 8.39 using a function grapher including the - and -aes. What are the equations of these rotated and translated aes? 3/, ( 1/) 1/ 64. The Discriminant Determine what happens to the sign of B 4AC within the equation A B C D E F 0 when (a) the aes are translated h units horizontall and k units verticall, no sign change (b) both sides of the equation are multiplied b the same nonzero constant k. no sign change Etending the Ideas 65. Group Activit Working together prove that the formulas for the coefficients A, B, C, D, E, and F in a rotated sstem given on page 670 are correct. 66. Identifing a Conic Develop a wa to decide whether A C D E F 0, with A and C not both 0, represents a parabola, an ellipse, or a hperbola. Write an eample to illustrate each of the three cases. 67. Rotational Invariant Prove that B 4A C B 4AC when the coordinate sstem is rotated through an angle. 68. Other Rotational Invariants Prove each of the following are invariants under rotation: (a) F, (b) A C, (c) D E. 69. Degenerate Conics Graph all of the degenerate conics listed in Table 8.. Recall that degenerate cones occur when the generator and ais of the cone are parallel or perpendicular. (See Figure 8..) Eplain the occurrence of all of the degenerate conics listed on the basis of cross sections of tpical or degenerate right circular cones.

45 5144_Demana_Ch08pp /13/06 6:50 AM Page 675 SECTION 8.5 Polar Equations of Conics Polar Equations of Conics What ou ll learn about Eccentricit Revisited Writing Polar Equations for Conics Analzing Polar Equations of Conics Orbits Revisited Eccentricit Revisited Eccentricit and polar coordinates provide was to see once again that parabolas, ellipses, and hperbolas are a unified famil of interrelated curves. We can define these three curves simultaneousl b generalizing the focus-directri definition of parabola given in Section and wh You will learn the approach to conics used b astronomers. Focus-Directri Definition Conic Section A conic section is the set of all points in a plane whose distances from a particular point (the focus ) and a particular line (the directri ) in the plane have a constant ratio. (We assume that the focus does not lie on the directri.) Conic section F P Focus Verte D Directri Focal ais The line passing through the focus and perpendicular to the directri is the (focal) ais of the conic section. The ais is a line of smmetr for the conic. The point where the conic intersects its ais is a verte of the conic. If P is a point of the conic, F is the focus, and D is the point of the directri closest to P, then the constant ratio PF PD is the eccentricit e of the conic (see Figure 8.40). A parabola has one focus and one directri. Ellipses and hperbolas have two focus-directri pairs, and either focus-directri pair can be used with the eccentricit to generate the entire conic section. FIGURE 8.40 The geometric structure of a conic section. OBJECTIVE Students will understand the general focus-directri definition of a conic section and will be able to write equations of conic sections in polar form. MOTIVATE Have students use a grapher to graph the polar equation 5 r. 1 cos Ask what the observe. (The graph is a parabola.) Focus-Directri-Eccentricit Relationship If P is a point of a conic section, F is the conic s focus, and D is the point of the directri closest to P, then PF e and PF e PD, P D where e is a constant and the eccentricit of the conic. Moreover, the conic is a hperbola if e 1, a parabola if e 1, an ellipse if e 1. In this approach to conic sections, the eccentricit e is a strictl positive constant, and there are no circles or other degenerate conics.

46 5144_Demana_Ch08pp /13/06 6:50 AM Page CHAPTER 8 Analtic Geometr in Two and Three Dimensions REMARKS To be consistent with our work on parabolas, we could use p for the distance from the focus to the directri, but following George B. Thomas, Jr. we use k for this distance. This simplifies our polar equations of conics. Rather than religiousl using polar coordinates and equations, we use a miture of the polar and Cartesian sstems. So, for eample, we use k for the directri rather than r cos k or r k sec. Focus at the pole Conic section F r θ r cos θ P(r, θ) k r cos θ Directri D Writing Polar Equations for Conics Our focus-directri definition of conics works best in combination with polar coordinates. Recall that in polar coordinates the origin is the pole and the -ais is the polar ais. To obtain a polar equation for a conic section, we position the pole at the conic s focus and the polar ais along the focal ais with the directri to the right of the pole (Figure 8.41). If the distance from the focus to the directri is k, the Cartesian equation of the directri is k. From Figure 8.41, we see that So the equation PF e PD becomes which when solved for r is PF r and PD k r cos. r e k r cos, ke r. 1 e cos In Eercise 53, ou are asked to show that this equation is still valid if r 0 or r cos k. This one equation can produce all sizes and shapes of nondegenerate conic sections. Figure 8.4 shows three tpical graphs for this equation. In Eploration 1, ou will investigate how changing the value of e affects the graph of r ke 1 e cos. = k FIGURE 8.41 A conic section in the polar plane. Directri Directri Directri P D P D P D F(0, 0) F(0, 0) F(0, 0) = k = k = k PF e = < 1 PD Ellipse (a) PF e = = 1 PD Parabola (b) PF e = > 1 PD Hperbola (c) FIGURE 8.4 The three tpes of conics possible for r ke 1 e cos.

47 5144_Demana_Ch08pp /13/06 6:50 AM Page 677 SECTION 8.5 Polar Equations of Conics 677 EXPLORATION EXTENSIONS How would the graphs change if ou used the equation ke r? 1 e cos ( /) Simplif this equation. EXPLORATION 1 Graphing Polar Equations of Conics Set our grapher to Polar and Dot graphing modes, and to Radian mode. Using k 3, an window of 1, 4 b 1, 1, min 0, ma, and step 48, graph ke r 1 e cos for e 0.7, 0.8, 1, 1.5, 3. Identif the tpe of conic section obtained for each e value. Overla the five graphs, and eplain how changing the value of e affects the graph of r ke 1 e cos. Eplain how the five graphs are similar and how the are different. NOTES ON EXAMPLES If possible, work with students through all of the eamples in this section, allowing time for eploration. Consider having some students present a few of the eamples in class. Polar Equations for Conics The four standard orientations of a conic in the polar plane are as follows. ke ke (a) r (b) r 1 e cos 1 e cos Focus at pole Focus at pole TEACHING NOTE This section provides a good opportunit to observe students selection of viewing window settings in the polar mode on the grapher. Students should have a good idea of how the graphs of conics should appear and can now be in a position to refine their ideas about viewing windows. Directri = k (a) ke (c) r 1 e sin Directri = k (b) ke (d) r 1 e sin Directri = k Focus at pole Focus at pole Directri = k (c) (d)

48 5144_Demana_Ch08pp /13/06 6:50 AM Page CHAPTER 8 Analtic Geometr in Two and Three Dimensions EXAMPLE 1 Writing and Graphing Polar Equations of Conics Given that the focus is at the pole, write a polar equation for the specified conic, and graph it. (a) Eccentricit e 3 5, directri. (b) Eccentricit e 1, directri. (c) Eccentricit e 3, directri 4. SOLUTION ke (a) Setting e 3 5 and k in r ields 1 e cos 3 5 r cos cos Figure 8.43a shows this ellipse and the given directri. ke (b) Setting e 1 and k in r ields 1 e cos r. 1 cos Figure 8.43b shows this parabola and its directri. ke (c) Setting e 3 and k 4 in r ields 1 e sin 4 3 r 1 3 sin 1. 3 sin Figure 8.43c shows this hperbola and the given directri. Now tr Eercise 1. [ 4.7, 4.7] b [ 3.1, 3.1] (a) FIGURE 8.43 Graphs for Eample 1. [ 4.7, 4.7] b [ 3.1, 3.1] (b) [ 15,15] b [ 5, 15] (c)

49 5144_Demana_Ch08pp /13/06 6:50 AM Page 679 SECTION 8.5 Polar Equations of Conics 679 Analzing Polar Equations of Conics The first step in analzing the polar equations of a conic section is to use the eccentricit to identif which tpe of conic the equation represents. Then we determine the equation of the directri. EXAMPLE Identifing Conics from Their Polar Equations Determine the eccentricit, the tpe of conic, and the directri. 6 6 (a) r (b) r 3 cos 4 3 sin SOLUTION (a) Dividing numerator and denominator b ields r cos. So the eccentricit e 1.5, and thus the conic is a hperbola. The numerator ke 3, so k, and thus the equation of the directri is. (b) Dividing numerator and denominator b 4 ields r sin. So the eccentricit e 0.75, and thus the conic is an ellipse. The numerator ke 1.5, so k, and thus the equation of the directri is. Now tr Eercise 7. All of the geometric properties and features of parabolas, ellipses, and hperbolas developed in Sections still appl in the polar coordinate setting. In Eample 3 we use this prior knowledge. [ 5, 10] b [ 5, 5] FIGURE 8.44 Graph of the ellipse r cos. (Eample 3) EXAMPLE 3 Analzing a Conic Analze the conic section given b the equation r cos. Include in the analsis the values of e, a, b, and c. SOLUTION Dividing numerator and denominator b 5 ields 3. r cos So the eccentricit e 0.6, and thus the conic is an ellipse. Figure 8.44 shows this ellipse. The vertices (endpoints of the major ais) have polar coordinates 8, 0 and,. So a 8 10, and thus a 5. The verte, is units to the left of the pole, and the pole is a focus of the ellipse. So a c, and thus c 3. An alternative wa to find c is to use the fact that the eccentricit of an ellipse is e c a, and thus c ae To find b we use the Pthagorean relation of an ellipse: b a c With all of this information, we can write the Cartesian equation of the ellipse: Now tr Eercise 31.

50 5144_Demana_Ch08pp /13/06 6:50 AM Page CHAPTER 8 Analtic Geometr in Two and Three Dimensions Orbits Revisited Polar equations for conics are used etensivel in celestial mechanics, the branch of astronom based on the work of Kepler and others who have studied the motion of celestial bodies. The polar equations of conic sections are well suited to the two-bod problem of celestial mechanics for several reasons. First, the same equations are used for ellipses, parabolas, and hperbolas the paths of one bod traveling about another. Second, a focus of the conic is alwas at the pole. This arrangement has two immediate advantages: The pole can be thought of as the center of the larger bod, such as the Sun, with the smaller bod, such as Earth, following a conic path about the larger bod. The coordinates given b a polar equation are the distance r between the two bodies and the direction from the larger bod to the smaller bod relative to the ais of the conic path of motion. For these reasons, polar coordinates are preferred over Cartesian coordinates for studing orbital motion. Table 8.3 Semimajor Aes and Eccentricities of the Planets Planet Semimajor Ais (Gm) Eccentricit Mercur Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto Source: Shupe, et al., National Geographic Atlas of the World (rev. 6th ed.). Washington, DC: National Geographic Societ, 199, plate 116, and other sources. Directri = k Focus Verte Center at pole C F P D c a c + k FIGURE 8.45 Geometric relationships within an ellipse. To use the data in Table 8.3 to create polar equations for the elliptical orbits of the planets, we need to epress the equation r ke 1 e cos in terms of a and e. We appl the formula PF e PD to the ellipse shown in Figure 8.45: e PD PF e c k a a c From Figure 8.45 e ae k a a ae Use e c/a. ae ke ae a ae Distribute the e. ae ke a Add ae. ke a ae Subtract ae. ke a 1 e Factor.

51 5144_Demana_Ch08pp /13/06 6:50 AM Page 681 SECTION 8.5 Polar Equations of Conics 681 So the equation r ke (1 e cos ) can be rewritten as follows: FOLLOW-UP Ask students how to determine what tpe of conic is represented b an equation of the form a r. b c cos ASSIGNMENT GUIDE E. 3 4, multiples of 3, 43, 54 COOPERATIVE LEARNING Group Activit: E NOTES ON EXERCISES These eercises require students to combine their knowledge of polar-coordinate graphing and conic sections. E deal with circular orbits. E provide practice for standardized tests. E. 58 and 59 require students to work with the directri of an ellipse and hperbola in rectangular coordinates. ONGOING ASSESSMENT Self-Assessment: E. 1, 7, 31, 41 Embedded Assessment: E Ellipse with Eccentricit e and Semimajor Ais a In this form of the equation, when e 0, the equation reduces to r a, the equation of a circle with radius a. EXAMPLE 4 a 1 e r 1 e cos Analzing a Planetar Orbit Find a polar equation for the orbit of Mercur, and use it to approimate its aphelion (farthest distance from the Sun) and perihelion (closest distance to the Sun). SOLUTION Setting e and a 57.9 in a 1 e r ields r e cos cos Mercur s aphelion is r Gm Mercur s perihelion is r Gm Now tr Eercise 41. QUICK REVIEW 8.5 (For help, go to Section 6.4.) In Eercises 1 and, solve for r. In Eercises 5 and 6, find the focus and directri of the parabola. 1. 3, r, r , r, r In Eercises 7 10, find the foci and vertices of the conic. In Eercises 3 and 4, solve for , 6 1.5,, 4. 3, 4 3 3,, /3, 5 / , The focus is (0, 4), and the directri is The focus is ( 3, 0), and the directri is Foci: ( 5, 0); Vertices: ( 3, 0) 8. Foci: (0, 4); Vertices: (0, 5) 9. Foci: ( 5, 0); Vertices: ( 4, 0) 10. Foci: (0, 4 ); Vertices: (0, 6)

52 5144_Demana_Ch08pp /13/06 6:50 AM Page CHAPTER 8 Analtic Geometr in Two and Three Dimensions SECTION 8.5 EXERCISES In Eercises 1 6, find a polar equation for the conic with a focus at the pole and the given eccentricit and directri. Identif the conic, and graph it. 1. e 1,. e 5 4, 4 3. e 3 5, 4 4. e 1, 5. e 7 3, 1 6. e 3, 5 In Eercises 7 14, determine the eccentricit, tpe of conic, and directri r 8. r 1 cos 1 cos 5 9. r 10. r sin 4 cos r 1. r 6 5 sin 7 sin r 14. r 5 cos 5 sin In Eercises 15 0, match the polar equation with its graph, and identif the viewing window. In Eercises 1 4, find a polar equation for the ellipse with a focus at the pole and the given polar coordinates as the endpoints of its major ais , 0 and 6,. 1.5, 0 and 1, 3. 1, and 3, , and 0.75, In Eercises 5 8, find a polar equation for the hperbola with a focus at the pole and the given polar coordinates as the endpoints of its transverse ais. 5. 3, 0 and 15, 6. 3, 0 and 1.5, 7. (.4, ) ( and 1, 3 ) 8. ( 6, ) ( and, 3 ) In Eercises 9 and 30, find a polar equation for the conic with a focus at the pole (3, π ) (0.75, 0) π b a1, (a) (c) (e) r 16. r 3 4 cos 3 cos r 18. r sin 5 3 sin r 0. r 5 sin 4 4 cos (b) (d) (f) In Eercises 31 36, graph the conic, and find the values of e, a, b, and c r 3. r 5 cos 6 5 sin r 34. r 4 sin 5 3 cos r 36. r 3 5 cos 1 5 sin In Eercises 37 and 38, determine a Cartesian equation for the given polar equation r 38. r sin 1 cos In Eercises 39 and 40, use the fact that k p is twice the focal length and half the focal width, to determine a Cartesian equation of the parabola whose polar equation is given r 40. r cos 3 3 cos 41. Halle s Comet The orbit of Halle s comet has a semimajor ais of AU and an orbital eccentricit of Compute its perihelion and aphelion distances.

53 5144_Demana_Ch08pp /13/06 6:50 AM Page 683 SECTION 8.5 Polar Equations of Conics Uranus The orbit of the planet Uranus has a semimajor ais of AU and an orbital eccentricit of Compute its perihelion and aphelion distances. In Eercises 43 and 44, the velocit of an object traveling in a circular orbit of radius r (distance from center of planet in meters) around a planet is given b v r k m sec, where k is a constant related to the mass of the planet and the orbiting object. 43. Group Activit Lunar Module A lunar ecursion module is in a circular orbit 50 km above the surface of the Moon. Assume that the Moon s radius is 1740 km and that k Find the following. (a) the velocit of the lunar module v 1551 m sec km sec (b) the length of time required for the lunar module to circle the moon once about hr 14 min 44. Group Activit Mars Satellite A satellite is in a circular orbit 1000 mi above Mars. Assume that the radius of Mars is 100 mi and that k Find the velocit of the satellite. 965 m sec.965 km sec mi sec Standardized Test Questions 45. True or False The equation r ke (1 e cos ) ields no true circles. Justif our answer. 46. True or False The equation r a(1 e ) (1 e cos ) ields no true parabolas. Justif our answer. In Eercises 47 50, solve the problem without using a calculator. 47. Multiple Choice Which ratio of distances is constant for a point on a nondegenerate conic? D (A) distance to center : distance to directri (B) distance to focus : distance to verte (C) distance to verte : distance to directri (D) distance to focus : distance to directri (E) distance to center : distance to verte 48. Multiple Choice Which tpe of conic section has an eccentricit greater than one? C (A) an ellipse (B) a parabola (C) a hperbola (D) two parallel lines (E) a circle 49. Multiple Choice For a conic epressed b r ke (1 e sin ), which point is located at the pole? B (A) the center (B) a focus (C) a verte (D) an endpoint of the minor ais (E) an endpoint of the conjugate ais 50. Multiple Choice Which of the following is not a polar equation of a conic? A (A) r 1 cos (B) r 1 (1 sin ) (C) r 3 (D) r 1 ( cos ) (E) r 1 (1 cos ) Eplorations 51. Planetar Orbits Use the polar equation r a(1 e ) (1 e cos ) in completing the following activities. (a) Use the fact that 1 cos 1 to prove that the perihelion distance of an planet is a(1 e) and the aphelion distance is a(1 e). (b) Use e c a to confirm that a(1 e) a c and a(1 e) a c. (c) Use the formulas a(1 e) and a(1 e) to compute the perihelion and aphelion distances of each planet listed in Table 8.4. (d) For which of these planets is the difference between the perihelion and aphelion distance the greatest? Table 8.4 Semimajor Aes and Eccentricities of the Si Innermost Planets Planet Semimajor Ais (AU) Eccentricit Mercur Venus Earth Mars Jupiter Saturn Source: Encrenaz & Bibring. The Solar Sstem (nd ed.). New York: Springer, p Using the Astronomer s Equation for Conics Using Dot mode, a, an window of [ 13, 5] b [ 6, 6], min 0, ma, and step 48, graph r a (1 e ) (1 e cos ) for e 0, 0.3, 0.7, 1.5, 3. Identif the tpe of conic section obtained for each e value. What happens when e 1?

54 5144_Demana_Ch08pp /13/06 6:50 AM Page CHAPTER 8 Analtic Geometr in Two and Three Dimensions Etending the Ideas 53. Revisiting Figure 8.41 In Figure 8.41, if r 0 or r cos k, then we must use PD k r cos and PF r. Prove that, even in these cases, the resulting equation is still r ke 1 e cos. 54. Deriving Other Polar Forms for Conics Using Figure 8.41 as a guide, draw an appropriate diagram for and derive the equation. ke (a) r 1 e cos ke (b) r 1 e sin ke (c) r 1 e sin 55. Revisiting Eample 3 Use the formulas r cos and r 16 to transform the polar equation r 5 3 cos into the Cartesian equation ( 3) Focal Widths Using polar equations, derive formulas for the focal width of an ellipse and the focal width of a hperbola. Begin b defining focal width for these conics in a manner analogous to the definition of the focal width of a parabola given in Section Prove that for a hperbola the formula r ke 1 e cos is equivalent to r a e 1 1 e cos, where a is the semitransverse ais of the hperbola. 58. Connecting Polar to Rectangular Consider the ellipse a 1, b (c) Prove that the distance from F to L is a e ea. ( a, 0) ( c, 0) 59. Connecting Polar to Rectangular Consider the hperbola a b 1, where half the length of the transverse ais is a, and the foci are c, 0 such that c a b. Let L be the vertical line a c. (a) Prove that L is a directri for the hperbola. [Hint: Prove that PF PD is the constant c a, where P is a point on the hperbola, and D is the point on L such that PD is perpendicular to L.] (b) Prove that the eccentricit is e c a. (c) Prove that the distance from F to L is ea a/e P(, ) F(c, 0) L (a, 0) L D = a c where half the length of the major ais is a, and the foci are c, 0 such that c a b. Let L be the vertical line a c. (a) Prove that L is a directri for the ellipse. [Hint: Prove that PF PD is the constant c a, where P is a point on the ellipse, and D is the point on L such that PD is perpendicular to L.] (b) Prove that the eccentricit is e c a. ( c, 0) P ( a, 0) D D P = a c F(c, 0) (a, 0) 55. 5r 3r cos 16 5r So, 5r 5( ) (3 16) Completing the square ields ( 3) 1, the desired result

55 5144_Demana_Ch08pp /13/06 6:50 AM Page 685 SECTION 8.6 Three-Dimensional Cartesian Coordinate Sstem Three-Dimensional Cartesian Coordinate Sstem What ou ll learn about Three-Dimensional Cartesian Coordinates Distance and Midpoint Formulas Equation of a Sphere Planes and Other Surfaces Vectors in Space Lines in Space Three-Dimensional Cartesian Coordinates In Sections P. and P.4, we studied Cartesian coordinates and the associated basic formulas and equations for the two-dimensional plane; we now etend these ideas to three-dimensional space. In the plane, we used two aes and ordered pairs to name points; in space, we use three mutuall perpendicular aes and ordered triples of real numbers to name points. See Figure z... and wh This is the analtic geometr of our phsical world. OBJECTIVE Students will be able to draw threedimensional figures and analze vectors in space. MOTIVATE Ask students wh it might be desirable to create a coordinate sstem representing three-dimensional space. LESSON GUIDE Da 1: Three-Dimensional Cartesian Coordinates; Distance and Midpoint Formulas; Equation of a Sphere Da : Planes and Other Surfaces; Vectors in Space; Lines in Space z 0 0 z (0, 0, 0) Origin 0 FIGURE 8.47 The coordinate planes divide space into eight octants. (, 0, z) (, 0, 0) constant z constant (0, 0, z) P(,, z) constant (,, 0) (0,, z) (0,, 0) FIGURE 8.46 The point P,, z in Cartesian space. Notice that Figure 8.46 ehibits several important features of the three-dimensional Cartesian coordinate sstem: The aes are labeled,, and z, and these three coordinate aes form a right-handed coordinate frame: When ou hold our right hand with fingers curving from the positive -ais toward the positive -ais, our thumb points in the direction of the positive z-ais. A point P in space uniquel corresponds to an ordered triple,, z of real numbers. The numbers,, and z are the Cartesian coordinates of P. Points on the aes have the form, 0,0, 0,, 0, or 0, 0, z, with, 0,0 on the -ais, 0,, 0 on the -ais, and 0, 0, z on the z-ais. In Figure 8.47, the aes are paired to determine the coordinate planes: The coordinate planes are the -plane, the z-plane, and the z-plane, and have equations z 0, 0, and 0, respectivel. Points on the coordinate planes have the form,, 0,, 0,z, or 0,, z, with,, 0 on the -plane,, 0, z on the z-plane, and 0,, z on the z-plane. The coordinate planes meet at the origin, 0, 0, 0. The coordinate planes divide space into eight regions called octants, with the first octant containing all points in space with three positive coordinates.

56 5144_Demana_Ch08pp /13/06 6:50 AM Page CHAPTER 8 Analtic Geometr in Two and Three Dimensions EXAMPLE 1 Locating a Point in Cartesian Space Draw a sketch that shows the point (, 3, 5). SOLUTION To locate the point, 3, 5, we first sketch a right-handed threedimensional coordinate frame. We then draw the planes, 3, and z 5, which parallel the coordinate planes 0, 0, and z 0, respectivel. The point, 3, 5 lies at the intersection of the planes, 3, and z 5, as shown in Figure Now tr Eercise 1. z (0, 0, 5) (, 3, 5) Line 3, z 5 Plane z 5 Line, z 5 Plane Plane 3 (, 0, 0) 0 (0, 3, 0) Line, 3 FIGURE 8.48 The planes, 3, and z 5 determine the point, 3, 5. (Eample 1) Distance and Midpoint Formulas The distance and midpoint formulas for space are natural generalizations of the corresponding formulas for the plane. Distance Formula (Cartesian Space) The distance d P, Q between the points P 1, 1, z 1 and Q,, z in space is d P, Q 1 1 z 1 z.

57 5144_Demana_Ch08pp /13/06 6:50 AM Page 687 SECTION 8.6 Three-Dimensional Cartesian Coordinate Sstem 687 Just as in the plane, the coordinates of the midpoint of a line segment are the averages for the coordinates of the endpoints of the segment. Midpoint Formula (Cartesian Space) The midpoint M of the line segment PQ with endpoints P 1, 1, z 1 and Q,, z is M ( 1, 1, z 1 z ). EXAMPLE Calculating a Distance and Finding a Midpoint Find the distance between the points P, 3, 1 and Q 4, 1, 5, and find the midpoint of line segment PQ. SOLUTION The distance is given b d P, Q The midpoint is M ( 4, 3 1, 1 5 ) 1, 1, 3. Now tr Eercises 5 and 9. TEACHING NOTE Man of the results discussed in this section are fairl intuitive etensions of previousl studied two-dimensional results. You ma wish to present a comparison between the two-dimensional results and their three-dimensional counterparts. Equation of a Sphere A sphere is the three-dimensional analogue of a circle: In space, the set of points that lie a fied distance from a fied point is a sphere. The fied distance is the radius, and the fied point is the center of the sphere. The point P,, z is a point of the sphere with center h, k, l and radius r if and onl if h k z l r. Squaring both sides gives the standard equation shown below. Standard Equation of a Sphere A point P,, z is on the sphere with center h, k, l and radius r if and onl if h k z l r.

58 5144_Demana_Ch08pp /13/06 6:50 AM Page CHAPTER 8 Analtic Geometr in Two and Three Dimensions Drawing Lesson How to Draw Three-Dimensional Objects to Look Three-Dimensional 1. Make the angle between the positive -ais and the positive -ais large enough. z z This Not this. Break lines. When one line passes behind another, break it to show that it doesn t touch and that part of it is hidden. A D A D A D C C C B B B Intersecting CD behind AB AB behind CD 3. Dash or omit hidden portions of lines. Don t let the line touch the boundar of the parallelogram that represents the plane, unless the line lies in the plane. Line below plane Line above plane Line in plane 4. Spheres: Draw the sphere first (outline and equator); draw aes, if an, later. Use line breaks and dashed lines. Hidden part dashed Break z A contact dot sometimes helps Break Sphere first Aes later

59 5144_Demana_Ch08pp /13/06 6:50 AM Page 689 SECTION 8.6 Three-Dimensional Cartesian Coordinate Sstem 689 EXAMPLE 3 Finding the Standard Equation of a Sphere The standard equation of the sphere with center, 0, 3 and radius 7 is z Now tr Eercise 13. Planes and Other Surfaces In Section P.4, we learned that ever line in the Cartesian plane can be written as a firstdegree (linear) equation in two variables; that is, ever line can be written as A B C 0, where A and B are not both zero. Conversel, ever first-degree equation in two variables represents a line in the Cartesian plane. In an analogous wa, ever plane in Cartesian space can be written as a first-degree equation in three variables : Equation for a Plane in Cartesian Space Ever plane can be written as A B Cz D 0, where A, B, and C are not all zero. Conversel, ever first-degree equation in three variables represents a plane in Cartesian space. z (0, 0, 3) (0, 4, 0) EXAMPLE 4 Sketching a Plane in Space Sketch the graph of z 60. SOLUTION Because this is a first-degree equation, its graph is a plane. Three points determine a plane. To find three points, we first divide both sides of z 60 b 60: (5, 0, 0) z = 60 FIGURE 8.49 The intercepts 5, 0, 0, 0, 4, 0, and 0, 0, 3 determine the plane z 60. (Eample 4) z 1. In this form, it is eas to see that the points 5, 0, 0, 0, 4, 0, and 0, 0, 3 satisf the equation. These are the points where the graph crosses the coordinate aes. Figure 8.49 shows the completed sketch. Now tr Eercise 17.

60 5144_Demana_Ch08pp /13/06 6:50 AM Page CHAPTER 8 Analtic Geometr in Two and Three Dimensions Equations in the three variables,, and z generall graph as surfaces in three-dimensional space. Just as in the plane, second-degree equations are of particular interest. Recall that second-degree equations in two variables ield conic sections in the Cartesian plane. In space, second-degree equations in three variables ield quadric surfaces: The paraboloids, ellipsoids, and hperboloids of revolution that have special reflective properties are all quadric surfaces, as are such eotic-sounding surfaces as hperbolic paraboloids and elliptic hperboloids. Other surfaces of interest include graphs of functions of two variables, whose equations have the form z f,. Some eamples are z ln, z sin, and z 1. The last equation graphs as a hemisphere (see Eercise 63). Equations of the form z f, can be graphed using some graphing calculators and most computer algebra software. Quadric surfaces and functions of two variables are studied in most universit-level calculus course sequences. z v 3 v 1, v, v 3 v 0, 0, 1 k 0, 1, 0 v i j 1, 0, 0 v 1 FIGURE 8.50 The vector v v 1, v, v 3. Vectors in Space In space, just as in the plane, the sets of equivalent directed line segments (or arrows) are vectors. The are used to represent forces, displacements, and velocities in three dimensions. In space, we use ordered triples to denote vectors: v v 1, v, v 3. The zero vector is 0 0, 0, 0, and the standard unit vectors are i 1, 0, 0, j 0, 1, 0, and k 0, 0, 1. As shown in Figure 8.50, the vector v can be epressed in terms of these standard unit vectors: v v 1, v, v 3 v 1 i v j v 3 k. The vector v that is represented b the arrow from P a, b, c to Q,, z is v PQ a, b, z c a i b j z c k. A vector v v 1, v, v 3 can be multiplied b a scalar (real number) c as follows: cv c v 1, v, v 3 cv 1, cv, cv 3. Man other properties of vectors generalize in a natural wa when we move from two to three dimensions: Vector Relationships in Space For vectors v v 1, v, v 3 and w w 1, w, w 3, Equalit: v w if and onl if v 1 w 1, v w, and v 3 w 3 Addition: v w v 1 w 1, v w, v 3 w 3 Subtraction: v w v 1 w 1, v w, v 3 w 3 Magnitude: v v 1 v v 3 Dot product: v w v 1 w 1 v w v 3 w 3 Unit vector: u v v, v 0, is the unit vector in the direction of v.

61 5144_Demana_Ch08pp /13/06 6:50 AM Page 691 SECTION 8.6 Three-Dimensional Cartesian Coordinate Sstem 691 EXAMPLE 5 Computing with Vectors (a) 3, 1, 4 3, 3 1, 3 4 6, 3, 1 (b) 0, 6, 7 5, 5, 8 0 5, 6 5, 7 8 5, 11, 1 (c) 1, 3, 4, 4, 5 1, 3 4, 4 5 3, 1, 1 (d), 0, (e) 5, 3, 1 6,, Now tr Eercises 3 6. EXAMPLE 6 Using Vectors in Space A jet airplane just after takeoff is pointed due east. Its air velocit vector makes an angle of 30 with flat ground with an airspeed of 50 mph. If the wind is out of the southeast at 3 mph, calculate a vector that represents the plane s velocit relative to the point of takeoff. SOLUTION Let i point east, j point north, and k point up. The plane s air velocit is a 50 cos 30, 0, 50 sin , 0, 15, and the wind velocit, which is pointing northwest, is w 3 cos 135, 3 sin 135, 0.67,.67, 0. The velocit relative to the ground is v a w, so v , 0, 15.67,.67, ,.63, i.63j 15k Now tr Eercise 33. In Eercise 64, ou will be asked to interpret the meaning of the velocit vector obtained in Eample 6. Lines in Space We have seen that first-degree equations in three variables graph as planes in space. So how do we get lines? There are several was. First notice that to specif the -ais, which is a line, we could use the pair of first-degree equations 0 and z 0. As alternatives to using a pair of Cartesian equations, we can specif an line in space using one vector equation, or a set of three parametric equations.

62 5144_Demana_Ch08pp /13/06 6:50 AM Page CHAPTER 8 Analtic Geometr in Two and Three Dimensions v = a, b, c P(,, z) z P 0 ( 0, 0, z 0 ) Suppose / is a line through the point P 0 0, 0, z 0 and in the direction of a nonzero vector v a, b, c (Figure 8.51). Then for an point P,, z on /, P 0 P tv for some real number t. The vector v is a direction vector for line /. If r OP,, z and r 0 OP 0 0, 0, z 0, then r r 0 tv. So an equation of the line / is r r 0 tv. FIGURE 8.51 The line / is parallel to the direction vector v a, b, c. FOLLOW-UP Ask students to give alternate vector and parametric forms for the line in Eample 7. ASSIGNMENT GUIDE Da 1: E. 1 15, odds Da : E , multiples of 3 COOPERATIVE LEARNING Group Activit: E. 63 NOTES ON EXERCISES E. 55 and 56 are proof eercises. E introduce the cross product of two vectors. ONGOING ASSESSMENT Self-Assessment: E. 1, 5, 9, 13, 17, 3 6, 33, 41 Embedded Assessment: E , 53 Equations for a Line in Space If / is a line through the point P 0 0, 0, z 0 in the direction of a nonzero vector v a, b, c, then a point P,, z) is on / if and onl if Vector form: r r 0 tv, where r,, z and r 0 0, 0, z 0 ; or Parametric form: 0 at, 0 bt, and z z 0 ct, where t is a real number. EXAMPLE 7 Finding Equations for a Line The line through P 0 4, 3, 1 with direction vector v,, 7 can be written in vector form as r 4, 3, 1 t,, 7 ; or in parametric form as 4 t, 3 t, and z 1 7t. Now tr Eercise 35. EXAMPLE 8 Finding Equations for a Line Using the standard unit vectors i, j, and k, write a vector equation for the line containing the points A 3, 0, ) and B 1,, 5, and compare it to the parametric equations for the line. SOLUTION The line is in the direction of v AB 1 3, 0, 5 4,, 3. So using r 0 OA, the vector equation of the line becomes: r r 0 tv,, z 3, 0, t 4,, 3,, z 3 4t, t, 3t i j zk 3 4t i tj 3t k The parametric equations are the three component equations 3 4t, t, and z 3t. Now tr Eercise 41.

63 5144_Demana_Ch08pp /13/06 6:50 AM Page 693 SECTION 8.6 Three-Dimensional Cartesian Coordinate Sstem 693 QUICK REVIEW 8.6 (For help, go to Sections 6.1 and 6.3.) In Eercises 1 3, let P, and Q, 3 be points in the -plane. 1. Compute the distance between P and Q.. Find the midpoint of the line segment PQ. 3. If P is 5 units from Q, describe the position of P. In Eercises 4 6, let v 4, 5 4i 5j be a vector in the -plane. 4. Find the magnitude of v Find a unit vector in the direction of v. 6. Find a vector 7 units long in the direction of v. 7. Give a geometric description of the graph of in the -plane. 8. Give a geometric description of the graph of t, 4 t in the -plane. 9. Find the center and radius of the circle in the -plane. 10. Find a vector from P, 5 to Q 1, 4 in the -plane. SECTION 8.6 EXERCISES In Eercises 1 4, draw a sketch that shows the point. 1. 3, 4,., 3, ,, 4 4., 3, 5 In Eercises 5 8, compute the distance between the points. 5. 1,, 5, 3, 4, , 1, 8, 6, 3, a, b, c, 1, 3, (a 1 ) ( b 3 ) ( c ) 8.,, z, p, q, r ( p ) ( q ) ( z r ) In Eercises 9 1, find the midpoint of the segment PQ. 9. P 1,, 5, Q 3, 4, 6 (1, 1, 11/) 10. P, 1, 8, Q 6, 3, 4 (4,, ) 11. P,, z, Q, 8, 6 ( 1, 4, z 3) 1. P a, b, c, Q 3a, 3b, 3c (a, b, c) In Eercises 13 16, write an equation for the sphere with the given point as its center and the given number as its radius , 1,, , 5, 8, , 3,, a, a p, q, r, 6 In Eercises 17, sketch a graph of the equation. Label all intercepts z z z 3 0. z In Eercises 3 3, evaluate the epression using r 1, 0, 3, v 3, 4, 5, and w 4, 3, r v, 4, 84. r w 3, 3, v w w r v w 0 8. r v r w 0 9. w w 30. i r i v, j v, k v 3. r v w 48, 36, 144 In Eercises 33 and 34, let i point east, j point north, and k point up. 33. Three-Dimensional Velocit An airplane just after takeoff is headed west and is climbing at a 0 angle relative to flat ground with an airspeed of 00 mph. If the wind is out of the northeast at 10 mph, calculate a vector v that represents the plane s velocit relative to the point of takeoff. v i 7.07j 68.40k 34. Three-Dimensional Velocit A rocket soon after takeoff is headed east and is climbing at a 80 angle relative to flat ground with an airspeed of 1,000 mph. If the wind is out of the southwest at 8 mph, calculate a vector v that represents the rocket s velocit relative to the point of takeoff. In Eercises 35 38, write the vector and parametric forms of the line through the point P 0 in the direction of v. 35. P 0 (, 1, 5), v 3,, P 0 ( 3, 8, 1), v 3, 5, 37. P 0 (6, 9, 0), v 1, 0, P 0 (0, 1, 4), v 0, 0, 1

64 5144_Demana_Ch08pp /13/06 6:50 AM Page CHAPTER 8 Analtic Geometr in Two and Three Dimensions In Eercises 39 48, use the points A 1,, 4, B 0, 6, 3, and C, 4, Find the distance from A to the midpoint of BC Find the vector from A to the midpoint of BC., 1, Write a vector equation of the line through A and B. 4. Write a vector equation of the line through A and the midpoint of BC. r 1,, 4 t, 1, Write parametric equations for the line through A and C. 44. Write parametric equations for the line through B and C. 45. Write parametric equations for the line through B and the midpoint of AC. 46. Write parametric equations for the line through C and the midpoint of AB. 47. Is ABC equilateral, isosceles, or scalene? scalene 48. If M is the midpoint of BC, what is the midpoint of AM? In Eercises 49 5, (a) sketch the line defined b the pair of equations, and (b) Writing to Learn give a geometric description of the line, including its direction and its position relative to the coordinate frame , , z 51. 3, , z Write a vector equation for the line through the distinct points P 1, 1, z 1 and Q,, z. 54. Write parametric equations for the line through the distinct points P 1, 1, z 1 and Q,, z. 55. Generalizing the Distance Formula Prove that the distance d(p, Q) between the points P( 1, 1, z 1 ) and Q(,, z ) in space is ( 1 ) ( 1 z ) ( ) 1 z b using the point R(,, z 1 ), the two-dimensional distance formula within the plane z z 1, the one-dimensional distance formula within the line r =,, t, and the Pthagorean theorem. [Hint: A sketch ma help ou visualize the situation.] 56. Generalizing a Propert of the Dot Product Prove u u u where u is a vector in three-dimensional space. Standardized Test Questions 57. True or False 4 1 represents a surface in space. Justif our answer. 58. True or False The parametric equation 1 0t, 0t, z 5 0t represent a line in space. Justif our answer. In Eercises 59 6, solve the problem without using a calculator. 59. Multiple Choice A first-degree equation in three variables graphs as B (A) a line. (B) a plane. (C) a sphere. (D) a paraboloid. (E) an ellipsoid. 60. Multiple Choice Which of the following is not a quadric surface? A (A) a plane (B) a sphere (C) an ellipsoid (D) an elliptic paraboloid (E) a hperbolic paraboloid 61. Multiple Choice If v and w are vectors and c is a scalar, which of these is a scalar? C (A) v w (B) v w (C) v w (D) cv (E) v w 6. Multiple Choice The parametric form of the line r, 3, 0 t 1, 0, 1 is E (A) 3t, 0 1t, z 0 1t. (B) t, 3 0t, z 0 1t. (C) 1 t, 0 3t, z 1 0t. (D) 1 t, 3, z 1t. (E) t, 3, z t. Eplorations 63. Group Activit Writing to Learn The figure shows a graph of the ellipsoid 9 4 z 16 1 drawn in a bo using Mathematica computer software. (a) Describe its cross sections in each of the three coordinate planes, that is, for z 0, 0, and 0. In our description, include the name of each cross section and its position relative to the coordinate frame. (b) Eplain algebraicall wh the graph of z 1 is half of a sphere. What is the equation of the related whole sphere?

65 5144_Demana_Ch08pp /13/06 6:50 AM Page 695 CHAPTER 8 Ke Ideas 695 (c) B hand, sketch the graph of the hemisphere z 1. Check our sketch using a 3D grapher if ou have access to one. (d) Eplain how the graph of an ellipsoid is related to the graph of a sphere and wh a sphere is a degenerate ellipsoid Revisiting Eample 6 Read Eample 6. Then using v i.63j 15k, establish the following: (a) The plane s compass bearing is (b) Its speed downrange (that is, ignoring the vertical component) is 195. mph. (c) The plane is climbing at an angle of (d) The plane s overall speed is 31.8 mph. Etending the Ideas The cross product u v of the vectors u u 1 i u j u 3 k and v v 1 i v j v 3 k is i j k u v u 1 u u 3 v 1 v v 3 u v 3 u 3 v i u 3 v 1 u 1 v 3 j u 1 v u v 1 k. Use this definition in Eercises ,, 3, 1, 1 1, 5, , 1, 1, 3, 4, 6, Prove that i j k. 68. Assuming the theorem about angles between vectors (Section 6.) holds for three-dimensional vectors, prove that u v is perpendicular to both u and v if the are nonzero. CHAPTER 8 Ke Ideas PROPERTIES, THEOREMS, AND FORMULAS Parabolas with Verte (h, k) 637 Ellipses with Center (h, k) 647 Hperbolas with Center (h, k) 659 Translation Formulas 668 Rotation Formulas 669 Discriminant Test 671 Focus Directri Eccentricit Relationship 675 Distance Formula (Cartesian Space) 686 Midpoint Formula (Cartesian Space) 687 Standard Equation of a Sphere 687 Equation for a Plane in Cartesian Space 689 Vector Relationships in Space 690 Equations for a Line in Space 69 PROCEDURES How to Sketch the Ellipse a b How to Sketch the Hperbola a b Translation of Aes Rotation of Aes How to Draw Three-Dimensional Objects to Look Three-Dimensional 688

66 5144_Demana_Ch08pp /13/06 6:51 AM Page CHAPTER 8 Analtic Geometr in Two and Three Dimensions CHAPTER 8 Review Eercises The collection of eercises marked in red could be used as a chapter test. In Eercises 1 4, find the verte, focus, directri, and focal width of the parabola, and sketch the graph In Eercises 5 1, identif the tpe of conic. Find the center, vertices, and foci of the conic, and sketch its graph In Eercises 13 0, match the equation with its graph. (a) (c) (e) (b) (d) (f) (b) (g) (h) (e) (g) (f) 18. (d) (c) 0. 6 (a) In Eercises 1 8, identif the conic. Then complete the square to write the conic in standard form, and sketch the graph Prove that the parabola with focus 0, p and directri p has the equation 4p. 30. Prove that the equation 4p represents a parabola with focus p, 0 and directri p. In Eercises 31 36, identif the conic. Solve the equation for and graph it In Eercises 37 48, find the equation for the conic in standard form. 37. Parabola: verte 0, 0, focus, Parabola: verte 0, 0, opens downward, focal width Parabola: verte 3, 3, directri Parabola: verte 1,, opens to the left, focal length (h)

67 5144_Demana_Ch08pp /13/06 6:51 AM Page 697 CHAPTER 8 Review Eercises Ellipse: center 0, 0, foci 1, 0, vertices 13, 0 4. Ellipse: center 0, 0, foci 0,, vertices 0, Ellipse: center 0,, semimajor ais 3, one focus is,. /9 ( ) / Ellipse: center 3, 4, semimajor ais 4, one focus is 0, 4. ( 3) /16 ( 4) / Hperbola: center 0, 0, foci 0, 6, vertices 0, Hperbola: center 0, 0, vertices, 0, asmptotes /4 / Hperbola: center, 1, vertices 3, 1, one asmptote is Hperbola: center 5, 0, one focus is 5, 3, one verte is 5,. /4 ( 5) /5 1 In Eercises 49 54, find the equation for the conic in standard form cos t, sin t, 0 t /5 / sin t, 6 cos t, 0 t 4 /36 / cos t, 4 sin t, t cos t, 3 3 sin t, t sec t, 5 tan t, 0 t /9 / sec t, 3 tan t, 0 t /16 /9 1 In Eercises 55 6, identif and graph the conic, and rewrite the equation in Cartesian coordinates r 56. r 1 cos 1 sin r 58. r 3 cos 4 sin r 60. r 7 sin 5 cos r 6. r 1 cos 4 4 cos In Eercises 63 74, use the points P 1, 0, 3 and Q 3,, 4 and the vectors v 3, 1, and w 3, 4, Compute the distance from P to Q Find the midpoint of segment PQ. (1, 1, 1/) 65. Compute v w. 0, 3, 66. Compute v w. 6, 5, 67. Compute v w Compute the magnitude of v Write the unit vector in the direction of w. 3/5, 4/5, Compute v w v w. 0, 39, Write an equation for the sphere centered at P with radius Write parametric equations for the line through P and Q. 73. Write a vector equation for the line through P in the direction of v. r 1, 0, 3 t 3, 1, 74. Write parametric equations for the line in the direction of w through the midpoint of PQ. 75. Parabolic Microphones B-Ball Network uses a parabolic microphone to capture all the sounds from the basketball plaers and coaches during each regular season game. If one of its microphones has a parabolic surface generated b the parabola 18, locate the focus (the electronic receiver) of the parabola. (0, 4.5) 76. Parabolic Headlights Specific Electric makes parabolic headlights for a variet of automobiles. If one of its headlights has a parabolic surface generated b the parabola 15 (see figure), where should its lightbulb be placed? (3.75, 0) Focus Light ras 77. Writing to Learn Elliptical Billiard Table Elliptical billiard tables have been constructed with spots marking the foci. Suppose such a table has a major ais of 6 ft and minor ais of 4 ft. (a) Eplain the strateg that a pool shark who knows conic geometr would use to hit a blocked spot on this table. (b) If the table surface is coordinatized so that (0, 0) represents the center of the table and the -ais is along the focal ais of the ellipse, at which point(s) should the ball be aimed? 78. Weather Satellite The Nimbus weather satellite travels in a north-south circular orbit 500 meters above Earth. Find the following. (Assume Earth s radius is 6380 km.) (a) The velocit of the satellite using the formula for velocit v given for Eercises 43 and 44 in Section 8.5 with k km/sec (b) The time required for Nimbus to circle Earth once 1 hr 5 min 79. Elliptical Orbits The velocit of a bod in an elliptical Earth orbit at a distance r (in meters) from the focus (the center of Earth) is v ( 14 r ) 1 a m sec, where a is the semimajor ais of the ellipse. An Earth satellite has a maimum altitude (at apogee) of 18,000 km and has a minimum altitude (at perigee) of 170 km. Assuming Earth s radius is 6380 km, find the velocit of the satellite at its apogee and perigee. 80. Icarus The asteroid Icarus is about 1 mi wide. It revolves around the Sun once ever 409 Earth das and has an orbital eccentricit of Use Kepler s first and third laws to determine Icarus s semimajor ais, perihelion distance, and aphelion distance.