26 Since the vertex is 2 units to the right of the focus the directrix must be

26 since the vertex is 2 units to the right of the

This preview shows page 60 - 64 out of 72 pages.

26. Since the vertex is 2 units to the right of the focus, the directrix must be 4 units to the right of the focus and perpendicular to the x -axis. Therefore, r = 4 1 + cos θ . 27. Since the vertex is 1 4 units below the focus, the directrix must be 1 2 units below the focus and parallel to the x - axis. Therefore, r = 1 / 2 1 - sin θ . 28. Since the vertex is 3 2 units above the focus, the directrix must be 3 units above the focus and parallel to the x -axis. Therefore, r = 3 1 + sin θ . 29. This is the parabola r = 4 1 + cos θ rotated counterclockwise by π/ 4 . The original parabola had its focus at the origin and its directrix at x = 4 . The original vertex therefore had polar coordinates (2 , 0). After rotation, the vertex is located at (2 , π/ 4) .
Image of page 60
728 CHAPTER 10. CONICS AND POLAR COORDINATES 30. This is the ellipse r = 5 3 + 2 cos θ = 5 / 3 1 + 2 / 3 cos θ rotated counterclockwise by π/ 3. The original ellipse had vertices at θ = 0 and θ = π. The polar coordinates of the vertices were (1 , 0) and (5 , π ). After rotation, the vertices are located at (1 , π/ 3) and (5 , 4 π/ 3) . 31. This is the ellipse r = 10 2 - sin θ = 5 2 - 1 2 sin θ rotated clockwise by π/ 6 . The original ellipse had vertices at θ = π/ 2 and θ = 3 π/ 2 . The polar coorinates of the vertices were (10 , π/ 2) and (10 . 3 , 3 π/ 2). After rotation, the vertices are located at (10 .π/ 3) and (10 / 3 , 4 π/ 3) . 32. This is the hyperbola r = 6 1 + 2 sin θ rotated clockwise by π/ 3 . The original hyperbola had vertices at θ = π/ 2 and θ = 3 π/ 2. The polar coordinates of the vertices were (2 , π/ 2) and ( - 6 , 3 π/ 2). After rotation, the vertices are located at (2 , π/ 6) and ( - 6 , 7 π/ 6) . 33. Identifying r a = 12 , 000 and e = 0 . 2 , we have from (7) in the text 0 . 2 = 12 , 000 - r p 12 , 000 + r p . Solving for r p , we obtain r p = 8 , 000 km. 34. The equation of the orbit is r = 0 . 2 p (1 - 0 . 2 cos θ ) . When θ = 0 , r = 12 , 000 so 12 , 000 = 0 . 2 p (1 - 0 . 2) = p 4 . Thus, p = 48 , 000 and the equation of the orbit is r = 9 , 600 (1 - 0 . 2 cos θ ) . 35. The equation of the orbit is r = ep (1 - e cos θ ) . From (7) in the text, e = 1 . 5 × 10 8 - 1 . 47 × 10 8 1 . 52 × 10 8 + 1 . 47 × 10 8 = 5 299 1 . 67 × 10 - 2 . When θ = 0 , r = r a = 1 . 52 × 10 8 = ep (1 - 1 . 67 × 10 - 2 ) . Thus ep 1 . 52 × 10 8 - 2 . 52 × 10 6 1 . 49 × 10 8 and the equation of the orbit is r = (1 . 49 × 10 8 ) (1 - 1 . 67 × 10 - 2 cos θ ) . 36. (a) The equation of the orbit is r = 0 . 97 p (1 - 0 . 97 cos θ ) . The length of the major axis is the sum of r a = r (0) = 0 . 97 p 0 . 03 and r p = r ( π ) = 0 . 97 p 1 + 0 . 97 = 0 . 97 p 1 . 97 . That is r a + r p = 0 . 97 p 1 0 . 03 + 1 1 . 97 = 3 . 34 × 10 9 . Solving for p we obtain p = 1 . 02 × 10 8 . The equation of the orbit is r = 0 . 97(1 . 02 × 10 8 ) (1 - 0 . 97 cos θ ) . (b) From part (a) r a = r (0) = 9 . 87 × 10 7 1 - 0 . 97 = 9 . 87 × 10 7 0 . 03 3 . 29 × 10 9 miles
Image of page 61
10.7. CONIC SECTIONS IN POLAR COORDINATES 729 and r p = r ( π ) = 9 . 87 × 10 7 1 + 0 . 97 = 9 . 87 × 10 7 1 . 97 5 . 01 × 10 7 miles . 37. polar axis 38. polar axis 39. polar axis 40. polar axis
Image of page 62
730 CHAPTER 10. CONICS AND POLAR COORDINATES 41. polar axis 42. polar axis
Image of page 63
Image of page 64

You've reached the end of your free preview.

Want to read all 72 pages?

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture