CalculusSooTansolution1015 2 - Preface This Complete...

Unformatted text preview: Preface This Complete Solutions Manual contains solutions to all of the exercises in chapters 10–15 of Calculus, First Edition, by Soo T. Tan. These correspond to chapters 9–14 of Calculus, Early Transcendentals, First Edition. A Student Solutions Manual is also available; it contains solutions to only the odd-numbered exercises. I would like to thank Tao Guo, who contributed to this manual by solving the problems independently; Kevin Charlwood, who checked all of the solutions for accuracy; and Andy Bulman-Fleming, who typeset this manual. I also want to thank Jerry Grossman for his help with the accuracy checking of the odd answers for the text. Their efforts played a big role in ensuring the accuracy of this manual, and I sincerely appreciate their contributions. I also wish to thank our development editor Jeannine Lawless and our editor Liz Covello of Cengage Learning for their help and patience in bringing this supplement to completion. Please submit any errors in the solutions manual or suggestions for improvements to me in care of the publisher: Math Editorial, Cengage Learning, 20 Channel Center Street, Boston, MA 02210. Soo T. Tan iii Contents 10 Conic Sections, Plane Curves, and Polar Coordinates 1 10.1/ET 9.1 Conic Sections 10.2/ET 9.2 Plane Curves and Parametric Equations 10.3/ET 9.3 The Calculus of Parametric Equations 10.4/ET 9.4 Polar Coordinates 10.5/ET 9.5 Areas and Arc Length in Polar Coordinates 10.6/ET 9.6 Conic Sections in Polar Coordinates 1 Chapter 10/ET 9 Review Challenge Problems 11 24 38 48 61 67 77 11.1/ET 10.1 Vectors in the Plane 11.2/ET 10.2 Coordinate Systems and Vectors in Three-Space 11.3/ET 10.3 The Dot Product 11.4/ET 10.4 The Cross Product 11.5/ET 10.5 Lines and Planes in Space 11.6/ET 10.6 Surfaces in Space 11.7/ET 10.7 Cylindrical and Spherical Coordinates Chapter 11/ET 10 Review 12 16 Vectors and the Geometry of Space 81 Challenge Problems ET 10 81 88 97 102 111 120 126 131 138 Vector-Valued Functions 141 ET 11 12.1/ET 11.1 Vector-Valued Functions 12.2/ET 11.2 Differentiation and Integration of Vector-Valued Functions 12.3/ET 11.3 Arc Length and Curvature 12.4/ET 11.4 Velocity and Acceleration 12.5/ET 11.5 Tangential and Normal Components of Acceleration Chapter 12/ET 11 Review Challenge Problems ET 9 141 148 156 164 170 181 185 v vi Contents 13 Functions of Several Variables 191 13.1/ET 12.1 Functions of Two or More Variables 13.2/ET 12.2 Limits and Continuity 13.3/ET 12.3 Partial Derivatives 13.4/ET 12.4 Differentials 13.5/ET 12.5 The Chain Rule 13.6/ET 12.6 Directional Derivatives and Gradient Vectors 13.7/ET 12.7 Tangent Planes and Normal Lines 13.8/ET 12.8 Extrema of Functions of Two Variables 13.9/ET 12.9 Lagrange Multipliers Challenge Problems 191 199 204 216 221 Chapter 13/ET 12 Review 14 ET 12 231 238 245 264 280 290 Multiple Integrals 293 ET 13 14.1/ET 13.1 Double Integrals 14.2/ET 13.2 Iterated Integrals 14.3/ET 13.3 Double Integrals in Polar Coordinates 14.4/ET 13.4 Applications of Double Integrals 14.5/ET 13.5 Surface Area 14.6/ET 13.6 Triple Integrals 14.7/ET 13.7 Triple Integrals in Cylindrical and Spherical Coordinates 14.8/ET 13.8 Change of Variables in Multiple Integrals 298 308 316 323 Chapter 14/ET 13 Review Challenge Problems 293 368 327 360 351 341 Contents 15 Vector Analysis 371 ET 14 15.1/ET 14.1 Vector Fields 15.2/ET 14.2 Divergence and Curl 15.3/ET 14.3 Line Integrals 15.4/ET 14.4 Independence of Path and Conservative Vector Fields 15.5/ET 14.5 Green’s Theorem 15.6/ET 14.6 Parametric Surfaces 15.7/ET 14.7 Surface Integrals 15.8/ET 14.8 The Divergence Theorem 15.9/ET 14.9 Stokes’ Theorem 437 Chapter 15/ET 14 Review 446 Challenge Problems 371 374 382 454 400 410 418 431 392 vii 10 Conic Sections, Plane Curves, and Polar Coordinates 10.1 Concept Questions ET 9 ET 9.1 1. a. See page 828 (824 in ET). b. i. See page 829 (825 in ET). ii. See page 830 (826 in ET). 2. a. See page 833 (829 in ET). b. i. See page 834 (830 in ET). ii. See page 834 (830 in ET). 3. a. See page 836 (832 in ET). b. i. See page 838 (834 in ET). ii. See page 838 (834 in ET). 10.1 Conic Sections ET 9.1 1. x 2 = −4y has the form x 2 = 4 py with p = −1, so it represents the parabola with vertex (0, 0), focus (0, −1), and directrix y = 1, labeled h. x2 ⇔ x 2 = 8y has the form x 2 = 4 py with p = 2, so it represents the parabola with vertex (0, 0), focus (0, 2), and 8 directrix y = −2, labeled a. 2. y = 3. y 2 = 8x has the form y 2 = 4 px with p = 2, so it represents the parabola with vertex (0, 0), focus (2, 0), and directrix x = −2, labeled c. 4. x = − 14 y 2 ⇔ y 2 = −4x has the form y 2 = 4 px with p = −1, so it represents the parabola with vertex (0, 0), focus (−1, 0), and directrix x = 1, labeled e. 5. x2 y2 x2 y2 + = 1 has the form 2 + 2 = 1 with a = 3, b = 2, and c = 9 4 a b √ √ vertices (±3, 0) and foci ± 5, 0 , labeled b. Its eccentricity is 35 . x2 y2 + 2 = 1 with a = 2, b = 1, and c = 2 b a √ √ vertices (0, ±2) and foci 0, ± 3 , labeled f. Its eccentricity is 23 . 6. x 2 + 7. y2 = 1 has the form 4 y2 x2 y2 x2 − = 1 has the form 2 − 2 = 1 with a = 4, b = 3, and c = 16 9 a b vertices (±4, 0) and foci (±5, 0), labeled d. Its eccentricity is 54 . x2 x2 y2 8. y 2 − = 1 has the form 2 − 2 = 1 with a = 1, b = 2, and c = 4 a b √ √ vertices (0, ±1) and foci 0, ± 5 , labeled g. Its eccentricity is 5. a 2 − b2 = a 2 − b2 = √ 5, so it represents the ellipse with √ 3, so it represents the ellipse with a 2 + b2 = 5, so it represents the hyperbola with a 2 + b2 = √ 5, so it represents the hyperbola with 1 2 Chapter 10 Conic Sections, Plane Curves, and Polar Coordinates ET Chapter 9 9. y = 2x 2 ⇔ x 2 = 4 py with p = 18 , so the parabola has vertex (0, 0), focus 0, 18 , and directrix y = − 18 . y 10. x 2 = −12y = 4 py with p = −3, so the parabola has vertex (0, 0), focus (0, −3), and directrix y = 3. y _5 _4 _3 _2 _1 1 2 3 4 x 0 40 30 _1 20 10 _4 _2 0 2 4 _2 x 11. x = 2y 2 ⇔ y 2 = 12 x = 4 px with p = 18 , so the parabola 12. y 2 = −8x = 4 px with p = −2, so the parabola has vertex (0, 0), focus (−2, 0), and directrix x = 2. has vertex (0, 0), focus 18 , 0 , and directrix x = − 18 . y 8 y 4 4 2 0 10 _2 20 30 40 x _12 _8 0 _4 _4 _8 _4 3 13. 5y 2 = 12x ⇔ y 2 = 12 5 x = 4 px with p = 5 , so the parabola has vertex (0, 0), focus 35 , 0 , and directrix x = − 35 . x 14. y 2 = −40x = 4 px with p = −10, so the parabola has vertex (0, 0), focus (−10, 0), and directrix x = 10. y 10 y 10 _10 5 _10 0 _5 _10 _8 _6 _4 0 _2 _10 10 20 30 40 50 x _20 x Section 10.1 Conic Sections ET Section 9.1 15. y2 x2 + = 1 has a = 5, b = 2, and 4 25 √ √ c = a 2 − b2 = 21, so the ellipse has foci 0, ± 21 and vertices (0, ±5). 16. 3 √ x 2 y2 + = 1 has a = 4, b = 3, and c = a 2 − b2 = 7, 16 9 √ so the ellipse has foci ± 7, 0 and vertices (±4, 0). y y 2 4 2 _4 _2 0 _2 _2 2 x 0 2 4 x _2 _4 _4 _6 y2 x2 + = 1 has a = 3, b = 2, and 9 4 √ √ a 2 − b2 = 5, so the ellipse has foci ± 5, 0 17. 4x 2 + 9y 2 = 36 ⇔ c= and vertices (±3, 0). 18. 25x 2 + 16y 2 = 400 ⇔ and c = a 2 − b2 = 3, so the ellipse has foci (0, ±3) and vertices (0, ±5). y y 4 2 _4 _2 0 _4 _2 0 x 2 x2 19. x 2 + 4y 2 = 4 ⇔ + y 2 = 1 has a = 2, b = 1, and 4 √ √ c = a 2 − b2 = 3, so the ellipse has foci ± 3, 0 and vertices (±2, 0). y √ x2 y2 + = 1 has a = 2, b = 2, and 2 4 √ c = a 2 − b2 = 2, so the ellipse has foci (0, ±2) and √ vertices 0, ± 2 . 20. 2x 2 + y 2 = 4 ⇔ y 2 1 _1 0 _1 4 x 2 _4 _2 _2 y2 x2 + = 1 has a = 5, b = 4, 16 25 1 2 1 x _2 _1 0 _1 _2 1 x 4 Chapter 10 Conic Sections, Plane Curves, and Polar Coordinates ET Chapter 9 21. y2 x2 − = 1 has a = 5, b = 12, and 25 144 c= 22. a 2 + b2 = 13, so the hyperbola has foci (±13, 0), vertices (±5, 0), and asymptotes y = ± 12 5 x. y x2 y2 − = 1 has a = 4, b = 9, and 16 81 √ c = a 2 + b2 = 97, so the hyperbola has foci √ 0, ± 97 , vertices (0, ±4), and asymptotes y = ± 49 x. y 10 4 0 _10 _5 5 x _12 _8 0 _4 4 x 8 _4 _10 _8 _20 √ x2 = 1 has a = 1, b = 2, and 23. x 2 − y 2 = 1 has a = 1, b = 1, and c = a 2 + b2 = 2, 24. 4y 2 − x 2 = 4 ⇔ y 2 − 4 √ √ so the hyperbola has foci ± 2, 0 , vertices (±1, 0), and c = a 2 + b2 = 5, so the hyperbola has foci √ asymptotes y = ±x. 0, ± 5 , vertices (0, ±1), and asymptotes y = ± 12 x. y y 1 _2 1 0 _1 1 _4 x _2 0 _1 4 x 2 _2 _1 _2 √ √ x2 y2 y2 x2 − = 1 has a = 5, b = 5, and 26. x 2 − 2y 2 = 8 ⇔ − = 1 has a = 2 2, b = 2, and 25. y 2 − 5x 2 = 25 ⇔ 25 5 8 4 √ √ 2 2 2 2 c = a + b = 30, so the hyperbola has foci c = a + b = 2 3, so the hyperbola has foci √ √ √ √ 0, ± 30 , vertices (0, ±5), and asymptotes y = ± 5x. ±2 3, 0 , vertices ±2 2, 0 , and asymptotes 10 √ y y = ± 22 x. 5 _5 0 _5 _10 y 4 5 x 2 _8 _6 _4 _2 0 _2 2 4 6 x _4 _6 27. The parabola with focus (3, 0) and directrix x = −3 has p = 3 and axis the x-axis, so an equation is y 2 = 4 px = 12x. 28. The parabola with focus (0, −2) and directrix y = 2 has p = −2 and axis the y-axis, so an equation is x 2 = 4 py = −8y. 29. The parabola with focus − 52 , 0 and directrix x = 52 has p = − 52 and axis the x-axis, so an equation is y 2 = 4 px = −10x. 30. The parabola with focus 0, 32 and directrix y = − 32 has p = 32 and axis the y-axis, so an equation is x 2 = 4 py = 6y. Section 10.1 Conic Sections ET Section 9.1 31. The ellipse with foci (±1, 0) and vertices (±3, 0) has a = 3, c = 1, and b = √ a 2 − c2 = 2 2, so an equation is 32. The ellipse with foci (0, ±3) and vertices (0, ±5) has a = 5, c = 3, and b = a 2 − c2 = 4, so an equation is y2 x2 + = 1. 9 8 x2 16 + y2 25 = 1. 33. The ellipse with foci (0, ±1) and major axis of length 6 has 2a = 6 ⇔ a = 3, c = 1, and b = equation is x2 y2 + = 1. 8 9 5 √ a 2 − c2 = 2 2, so an y2 4x 2 + = 1. 34. The ellipse with vertices (0, ±5) and minor axis of length 5 has a = 5, 2b = 5 ⇔ b = 52 , so an equation is 25 25 y2 x2 + 2 = 1. Substituting the point 35. The ellipse with vertices (±3, 0) has a = 3, so the ellipse has an equation of the form 9 b √ 2 2 √ 4y 2 12 2 8 x2 1, 2 , we find + + = 1. =1 ⇔ = ⇔ b2 = 94 , so an equation of the ellipse is 2 2 9 9 9 9 b b 36. By plotting the points (1, 5) and (2, 4), we see that the ellipse’s vertices must lie on the y-axis, so we substitute the x2 y2 12 52 22 points into the standard equation 2 + 2 = 1: 2 + 2 = 1 and 2 + b a b a b 2 2 1 5 a 2 + 25b2 = 4a 2 + 16b2 ⇔ a 2 = 3b2 . Substituting, 2 + 2 = 1 ⇔ b2 b 3b y2 3x 2 + = 1. thus 28 28 42 1 25 4 16 =1⇒ 2 + 2 = 2 + 2 ⇔ 2 a b a b a 28 , so a 2 = 28. An equation of the ellipse is = 3 y2 x2 = 1. Substituting the point 37. The ellipse with vertices (0, ±5) has a = 5, so the ellipse has an equation of the form 2 + 25 b √ 4 73x 2 y2 27 400 2, 3 2 3 , we find 2 + = 1 ⇔ b2 = , so an equation is + = 1. 4 · 25 73 400 25 b 38. The ellipse with x-intercepts ±3 and y-intercepts ± 12 has a = 3 and b = 12 with major axis along the x-axis, so an equation x2 + 4y 2 = 1. 9 39. The hyperbola with foci (±5, 0) and vertices (±3, 0) has a = 3, c = 5, and b2 = c2 − a 2 = 16, so an equation is is y2 x2 − = 1. 9 16 40. The hyperbola with foci (0, ±8) and vertices (0, ±4) has a = 4, c = 8, and b2 = c2 − a 2 = 48, so an equation is y2 x2 − = 1. 16 48 41. The hyperbola with foci (0, ±5) and conjugate axis of length 4 has c = 5, 2b = 4 ⇔ b = 2, and a 2 = c2 − b2 = 21, so an equation is y2 x2 − = 1. 21 4 x2 52 y2 92 42. The hyperbola with vertices (±4, 0) has a = 4 and equation 2 − 2 = 1. Substituting 5, 94 gives 2 − 2 2 = 1 ⇔ 4 b 4 4 ·b 2 2 y x − = 1. 25b2 − 81 = 16b2 ⇔ 9b2 = 81 ⇔ b2 = 9, so an equation is 16 9 43. The hyperbola with vertices (±2, 0) and asymptotes y = ± 32 x has a = 2 and ab = 32 , so b = 32 (2) = 3 and an equation is y2 x2 − = 1. 4 9 6 Chapter 10 Conic Sections, Plane Curves, and Polar Coordinates ET Chapter 9 √ 1 x has a = 1 and a = √ 1 , so b = 2 2 and an equation is 44. The hyperbola with y-intercepts ±1 and asymptotes y = ± √ b 2 2 y2 − x2 8 2 2 = 1. 45. Referring to the table on page 840 (836 in ET) and Figure 26, we see that (x + 3)2 = −2 (y − 4) is an equation of the parabola with vertex (−3, 4) opening downward, labeled b. 46. (x − 2)2 (y + 3)2 + = 1 is an equation of an ellipse centered at (2, −3), labeled d. 16 4 47. (y − 3)2 (x + 1)2 − = 1 is an equation of a hyperbola centered at (−1, 3), labeled c. 16 9 48. (y − 1)2 = −4 (x − 2) is an equation of the parabola with vertex (2, 1) opening to the left, labeled a. 49. The parabola with focus (3, 1) and directrix x = 1 has vertex (2, 1) and opens to the right with p = 1, so an equation is (y − 1)2 = 4 (1) (x − 2) or (y − 1)2 = 4 (x − 2). 50. The parabola with focus (−2, 3) and directrix y = 5 has vertex (−2, 4) and opens downward with p = −1, so an equation is (x + 2)2 = 4 (−1) (y − 4) or (x + 2)2 = −4 (y − 4). 51. The parabola with vertex (2, 2) and focus 32 , 2 opens to the left with p = − 12 , so an equation is (y − 2)2 = 4 − 12 (x − 2) or (y − 2)2 = −2 (x − 2). 52. The parabola with vertex (1, −2) and directrix y = 1 opens downward with p = −3, so an equation is (x − 1)2 = 4 (−3) (y + 2) or (x − 1)2 = −12 (y + 2). 53. Referring to the table on page 840 (836 in ET), the parabola with axis parallel to the y-axis has equation ⎧ ⎪ (−3 − h)2 = 4 p (2 − k) ⎪ ⎨ (x − h)2 = 4 p (y − k). Substituting the three known points gives the equations (0 − h)2 = 4 p − 52 − k ⇒ ⎪ ⎪ ⎩ (1 − h)2 = 4 p (−6 − k) ⎧ 2 ⎪ ⎪ ⎨ h + 6h + 9 = 8 p − 4 pk Subtracting the second equation from the first gives 6h + 9 = 18 p ⇔ 2h + 3 = 6 p h 2 = −10 p − 4 pk ⎪ ⎪ ⎩ h 2 − 2h + 1 = −24 p − 4 pk (call this equation α), and subtracting the third from the second gives 2h − 1 = 14 p (equation β). Subtracting β from α gives 4 = −8 p ⇔ p = − 12 . Thus, from α, 2h + 3 = −3 ⇔ h = −3, and finally, from the second original equation (−3)2 = −10 − 12 − 4 − 12 k ⇔ k = 2. An equation of the parabola is thus (x + 3)2 = −2 (y − 2). 54. Referring to the table on page 840 (836 in ET), the parabola with axis parallel to the x-axis has equation ⎧ 2 ⎪ ⎪ ⎨ (6 − k) = 4 p (−6 − h) (y − k)2 = 4 p (x − h). Substituting the three known points gives the equations (0 − k)2 = 4 p (0 − h) ⇒ ⎪ ⎪ ⎩ (2 − k)2 = 4 p (2 − h) ⎧ 2 ⎪ ⎪ ⎨ k − 12k + 36 = −24 p − 4 ph k 2 = −4 ph ⎪ ⎪ ⎩ k 2 − 4k + 4 = 8 p − 4 ph Subtracting the first equation from the second gives 12k − 36 = 24 p ⇔ k = 2 p + 3 (call this equation α), and subtracting the third from the second gives 4k − 4 = −8 p ⇔ k = −2 p + 1 (equation β). Adding α and β gives 2k = 4 ⇔ k = 2. Thus, from α, 2 = 2 p + 3 ⇔ p = − 12 , and finally, from the second original equation 22 = −4 − 12 h ⇔ h = 2. An equation of the parabola is thus (y − 2)2 = −2 (x − 2). Section 10.1 Conic Sections ET Section 9.1 7 55. The ellipse with foci (±1, 3) and vertices (±3, 3) has center (0, 3), a = 3, and c = 1. Thus, b2 = a 2 − c2 = 8, and an equation is (y − 3)2 x2 + = 1. 9 8 56. The ellipse with foci (0, 2) and (4, 2) and vertices (−1, 2) and (5, 2) has center (2, 2), a = 3, and c = 2. Thus, b2 = a 2 − c2 = 5, and an equation is (y − 2)2 (x − 2)2 + = 1. 9 5 57. The ellipse with foci (±1, 2) and major axis of length 2a = 8 ⇔ a = 4 has vertices (±4, 2), center (0, 2), and c = 1. Thus, b2 = a 2 − c2 = 15, and an equation is x2 (y − 2)2 + = 1. 16 15 58. The ellipse with foci (1, ±3) and minor axis of length 2b = 2 ⇔ b = 1 has c = 3, so a 2 = b2 + c2 = 10 and an equation is (x − 1)2 + y2 = 1. 10 59. The ellipse with center (2, 1), one focus at (0, 1), and one vertex at (5, 1) has a = 3 and c = 2, so b2 = a 2 − c2 = 5 and an equation is (x − 2)2 (y − 1)2 + = 1. 9 5 60. The ellipse with foci 2 − √ √ √ 3, −1 and 2 + 3, −1 has center (2, −1) and c = 3 ⇔ c2 = 3, so its equation has the (2 − 2)2 (y + 1)2 (0 + 1)2 (x − 2)2 = 1. Substituting the known point 0), we get = 1 ⇔ 1 = a2 − 3 ⇔ + + (2, a2 a2 − 3 a2 a2 − 3 (x − 2)2 + (y + 1)2 = 1. a 2 = 4, so b2 = 1 and an equation is 4 form 61. The hyperbola with foci (−2, 2) and (8, 2) and vertices (0, 2) and (6, 2) has center (3, 2), so a = 3, c = 5, and b2 = c2 − a 2 = 16. An equation is thus (y − 2)2 (x − 3)2 − = 1. 9 16 62. The hyperbola with foci (−4, 5) and (−4, −15) and vertices (−4, −3) and (−4, −7) has center (−4, −5), so a = 2, c = 10, and b2 = c2 − a 2 = 96. An equation is thus (x + 4)2 (y + 5)2 − = 1. 4 96 63. The hyperbola with foci (6, −3) and (−4, −3) and asymptotes y + 3 = ± 43 (x − 1) has center (1, −3), so c = 5. The slopes of the asymptotes are ± ab = ± 43 and a 2 + b2 = c2 = 25, so by inspection a = 3 and b = 4. An equation is thus (x − 1)2 (y + 3)2 − = 1. 9 16 64. The hyperbola with foci (2, 2) and (2, 6) and asymptotes x = −2 + y and x = 6 − y has center (2, 4), so c = 2. The √ slopes of the asymptotes are ± ab = ±1 and a 2 + b2 = c2 = 4, so by inspection a = b = 2. An equation is thus (y − 4)2 (x − 2)2 − = 1. 2 2 65. The hyperbola with vertices (4, −2) and (4, 4) and asymptotes y − 1 = ± 32 (x − 4) has center (4, 1), so a = 3. The slopes of the asymptotes are ± ab = ± 32 , so b = 2 and an equation is (x − 4)2 (y − 1)2 − = 1. 9 4 66. The hyperbola with vertices (0, −2) and (4, −2) and asymptotes x = −y and x = y + 4 has center (2, −2), so a = 2. The slopes of the asymptotes are ± ab = ±1, so b = 2. An equation is thus (y + 2)2 (x − 2)2 − = 1. 4 4 8 Chapter 10 Conic Sections, Plane Curves, and Polar Coordinates ET Chapter 9 67. We complete the square and put the equation into standard 68. y 2 − 4y − 2x − 4 = 0 ⇔ (y − 2)2 − 4 − 2x − 4 = 0 ⇔ form: y 2 − 2y − 4x + 9 = 0 ⇔ (y − 1)2 − 1 − 4x + 9 = 0 ⇔ (y − 1)2 = 4x − 8 = 4 (x − 2). This equation has p = 1, so it represents a parabola with vertex (2, 1), focus (y − 2)2 = 2 (x + 4) represents a parabola with vertex (−4, 2), focus − 72 , 2 , and directrix x = − 92 . y (3, 1), and directrix x = 1. 6 y 4 6 2 4 2 0 _2 1 2 3 4 5 6 7 8 _4 9 x _2 _2 2 4 6 8 x _4 69. x 2 + 6x − y + 11 = 0 ⇔ (x + 3)2 − 9 − y + 11 = 0 ⇔ (x + 3)2 = y − 2 represents a parabola with vertex (−3, 2), focus −3, 94 , and directrix y = 74 . (x − 2)2 = 12 (y + 3) represents a parabola with vertex 25 (2, −3), focus 2, − 23 8 , and directrix y = − 8 . y y 40 40 30 30 20 20 10 7 y= 4 _10 _8 70. 2x 2 − 8x − y + 5 = 0 ⇔ 2 (x − 2)2 − 8 − y + 5 = 0 ⇔ _6 _4 10 _2 0 2 _2 x 2 0 2 4 y − 12 − 1 − 32x − 31 = 0 ⇔ 2 y − 12 = 8 (x + 1) represents a parabola with vertex 25 _3 _2 _4 y _6 _2 _1 0 _2 2 3 x 5 2 _3 1 − 1 + 9y − 8 = 0 vertex − 13 , 1 , focus − 13 , 34 , and directrix y = 54 . y 4 2 2 x 2 ⇔ x + 13 = − (y − 1) represents a parabola with −1, 12 , focus 1, 12 , and directrix x = −3. 6 6 y=_ 8 72. 9x 2 + 6x + 9y − 8 = 0 ⇔ 9 x + 13 71. 4y 2 − 4y − 32x − 31 = 0 ⇔ _1 0 _2 4 _4 _6 y= 4 1 2 x Section 10.1 Conic Sections ET Section 9.1 y 73. (x − 1)2 + 4 (y + 2)2 = 1 has center (1, −2) and vertices (0, −2) and (2, −2). c= a 2 − b2 = 9 0 √ √ 1 − 14 = 23 , so the foci are 1 ± 23 , −2 . 2 x 1 _1 _2 _3 y 74. We complete the squares and put the equation into standard form: 2 2x 2 + y 2 − 20x + 2y + 43 = 0 ⇔ 2 (x − 5)2 − 50 + (y + 1)2 − 1 + 43 = 0 ⇔ 2 2 + (y+1) = 1 has center (5, −1) and 2 (x − 5)2 + (y + 1)2 = 8 ⇔ (x−5) 4 8 √ vertices 5, −1 ± 2 2 . c = a 2 − b2 = 2, so the foci are (5, −3) and (5, 1). 0 2 4 x 6 _2 _4 y 75. We complete the squares and put the equation into standard form: 0 _1 x 2 + 4y 2 − 2x + 16y + 13 = 0 ⇔ (x − 1)2 − 1 + 4 (y + 2)2 − 16 + 13 = 0 ⇔ 1 3 x 2 _1 2 + (y + 2)2 = 1 has center (1, −2) and (x − 1)2 + 4 (y + 2)2 = 4 ⇔ (x−1) 4 √ vertices (−1, −2) and (3, −2). c = a 2 − b2 = 3, so the foci are √ 1 ± 3, −2 . _2 _3 y 76. 2x 2 + y 2 + 12x − 6y + 25 = 0 ⇔ 2 (x + 3)2 − 18 + (y − 3)2 − 9 + 25 = 0 ⇔ √ 2 = 1 has center (−3, 3) and vertices −3, 3 ± 2 . (x + 3)2 + (y−3) 2 c= 4 3 a 2 − b2 = 1, so the foci are (−3, 2) and (−3, 4). 2 1 _5 _4 _3 _2 _1 x _1 77. 4x 2 + 9y 2 − 18x − 27 = 0 ⇔ 4 x − 94 16 x− 94 189 c= 2 2 − 81 4 + 9y − 27 = 0 ⇔ 2 y 2 √ 2 3 21 9 9 + 4y 21 = 1 has center 4 , 0 and vertices 4 ± 4 , 0 . a 2 − b2 = 189 − 21 = 16 4 √ 105 , so the foci are 4 9 ± 4 √ 105 , 0 . 4 1 _1 0 1 2 3 4 5 x _1 _2 78. 9x 2 + 36y 2 − 36x + 48y + 43 = 0 ⇔ x 2 2 9 (x − 2)2 − 36 + 36 y + 23 − 16 + 43 = 0 ⇔ (x − 2)2 + 4 y + 23 = 1 has 0 √ center 2, − 23 and vertices 1, − 23 and 3, − 23 . c = a 2 − b2 = 23 , so the √ _1 foci are 2 ± 23 , − 23 . 1 2 3 x 10 Chapter 10 Conic Sections, Plane Cu...
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