SECTION 9.5 9.5 4. Ellipse, Copyright © 2013, Cengage Learning. All rights reserved. ■ ■ 10. r 2 1 cos 11. r 6 2 sin 12. r 7 2 5 sin 13. r 8 3 3 cos 14. r 10 3 2 sin 15. r 5 2 3 sin 16. r 8 3 cos eccentricity , directrix x 3 eccentricity 12 , directrix y 4 eccentricity 4, directrix r 5 sec eccentricity 0.6, directrix r 2 csc vertex at 5, 2 7. Parabola, 8. Ellipse, 4 1 3 cos 9. r 4 3 directrix y 2 5. Hyperbola, 6. Ellipse, 9–16 ■ (a) Find the eccentricity, (b) identify the conic, (c) give an equation of the directrix, and (d) sketch the conic. eccentricity 23 , directrix x 3 3. Parabola, eccentricity 0.4, vertex at 2, 0 ■ ■ 1 S Click here for solutions. 1– 8 ■ Write a polar equation of a conic with the focus at the origin and the given data. 2. Hyperbola, ■ CONIC SECTIONS IN POLAR COORDINATES A Click here for answers. 1. Ellipse, CONIC SECTIONS IN POLAR COORDINATES ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 2 ■ SECTION 9.5 CONIC SECTIONS IN POLAR COORDINATES 9.5 ANSWERS E Click here for exercises. S Click here for solutions. 6 3 + 2 cos θ 2. r = 3. r = 2 1 + sin θ 4. r = 4 2 − sin θ 5. r = 20 1 + 4 cos θ 6. r = 6 5 + 3 sin θ 7. r = 10 1 + sin θ 8. r = 8 5 + 2 cos θ 1. r = 9. (a) 3 12 3 − 4 cos θ 4 3 (b) Hyperbola (c) x = (b) Parabola (c) x = −2 (b) Ellipse (c) y = 6 (d) 10. (a) 1 13. (a) 1 8 3 (b) Parabola (c) x = (b) Ellipse (c) y = −5 (b) Hyperbola (c) y = − 35 (b) Ellipse (c) x = 8 (d) 14. (a) 23 (d) 15. (a) 32 (d) (d) 11. (a) 12 Copyright © 2013, Cengage Learning. All rights reserved. (d) 12. (a) 52 (d) 16. (a) 31 (d) (b) Hyperbola (c) y = − 75 SECTION 9.5 9.5 CONIC SECTIONS IN POLAR COORDINATES SOLUTIONS E Click here for exercises. 1. r = 2 ·3 ed 6 3 = = 1 + e cos θ 3 + 2 cos θ 1 + 23 cos θ 4 ·3 12 ed 3 = 2. r = = 4 1 − e cos θ 3 − 4 cos θ 1 − 3 cos θ 3. r = ed 1·2 2 = = 1 + e sin θ 1 + sin θ 1 + sin θ 4. r = 1 ·4 4 ed 2 = = 1 − e sin θ 2 − sin θ 1 − 12 sin θ 5. r = 5 sec θ r= (a) e = 1 (b) Parabola (c) ed = 2 ⇒ d = 2 ⇒ directrix x = −2 ⇔ x = r cos θ = 5, so 11. r = ⇔ y = r sin θ = 2, so ·2 6 ed = = r= 3 1 + e sin θ 5 + 3 sin θ 1 + 5 sin θ 7. Focus (0, 0), vertex 5, π 2 3 1+ (a) e = 3 5 r= 2 1 − cos θ (d) Vertex (−1, 0) = (1, π) ed 4·5 20 = = 1 + e cos θ 1 + 4 cos θ 1 + 4 cos θ 6. r = 2 csc θ 10. r = 1 2 sin θ 1 2 (b) Ellipse ⇒ directrix y = 10 ⇒ (c) ed = 3 ⇒ d = 6 ⇒ directrix y = 6 (d) Vertices 2, π2 and 6, 3π ; center 2, 3π 2 2 ed 10 = 1 + e sin θ 1 + sin θ 8. The directrix is x = 4, so r= 9. r = 2 ·4 ed 8 5 = = 2 1 + e cos θ 5 + 2 cos θ 1 + 5 cos θ 4 1 + 3 cos θ 12. r = (a) e = 3 (b) Since e = 3 > 1, the conic is a hyperbola. (c) ed = 4 ⇒ d = 4 3 ⇒ directrix x = 4 3 Copyright © 2013, Cengage Learning. All rights reserved. (d) The vertices are (1, 0) and (−2, π) = (2, 0); the center is 32 , 0 ; the asymptotes are parallel to θ = ± cos−1 − 13 . 7/2 1 − 52 sin θ (a) e = 5 2 (b) Hyperbola ⇒ d = 75 ⇒ directrix y = − 75 (d) Center 53 , 3π ; vertices − 73 , π2 = 73 , 3π and 2 2 3π 1, 2 (c) ed = 7 2 ■ 3 4 ■ SECTION 9.5 13. r = CONIC SECTIONS IN POLAR COORDINATES 8/3 1 + cos θ 16. r = (a) e = (a) e = 1 8 3 (d) Vertex 14. r = 3 8 3 ⇒ directrix x = 8 3 ,0 10/3 1 − 23 sin θ (a) e = 2 3 (b) Ellipse (c) ed = 10 3 ⇒ d = 5 ⇒ directrix y = −5 (d) Vertices 10, π2 and 2, 3π ; center 4, π2 2 15. r = 5/2 1 − 32 sin θ (a) e = 3 2 (b) Hyperbola (c) ed = 5 2 ⇒ d= 5 3 (d) Vertices −5, π2 = 5, 3π foci (0, 0) and 6, 2 Copyright © 2013, Cengage Learning. All rights reserved. 1 3 (c) ed = ⇒ d= 4 8/3 1 cos θ 3 (b) Ellipse (b) Parabola (c) ed = 1+ ⇒ directrix y = − 53 3π 2 and 1, 3π ; center 3, 3π ; 2 2 8 3 ⇒ d = 8 ⇒ directrix x = 8 (d) Vertices (2, 0) and (4, π); center (−1, 0)
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