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

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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) .
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
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
730 CHAPTER 10. CONICS AND POLAR COORDINATES 41. polar axis 42. polar axis

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