Section 7.1 Solutions
1. An ellipse is the set of all points in the plane the sum of whose distances from two fixed points, called the foci, is a constant.
3. This special case would be a circle.
5. It is symmetric about the x-axis, y-axis, and the origin.
7. yes; x232+y222=1x232+y222=1
9. yes; x2(12)2+y2(13)2=1
11. x222+y272=1; Endpoints of major axis (0,7) and (0,−7). Endpoints of minor axis (2,0) and (−2,0). Foci at (0,3√5),(0,−3√5).
13. x2(1)2+y2(13)2=1; Endpoints of major axis (1,0) and (−1,0). Endpoints of minor axis (0,13),(0,−13). Foci at (2√23,0),(−2√23,0).
15. (x−2)272+(y−4)252=1; Endpoints of major axis (9,4),(−5,4). Endpoints of minor axis (2,9),(2,−1). Foci at (2+2√6,4),(2−2√6,4).
17. (x+5)222+(y−7)232=1; Endpoints of major axis (−5,10),(−5,4). Endpoints of minor axis (−3,7),(−7,7). Foci at (−5,7+√5),(−5,7−√5).
19. (x−1)232+(y−4)222=1; Endpoints of major axis (4,4),(−2,4). Endpoints of minor axis (1,6),(1,2). Foci at (1+√5,4),(1−√5,4).
21. (x−3)2(3√2)2+(y−5)2(√2)2=1; Endpoints of major axis (3+3√2,5),(3−3√2,5). Endpoints of minor axis (3,5+√2),(3,5−√2). Foci at (7,5),(−1,5).
23. (x+5)2(5)2+(y−2)2(2)2=1; Endpoints of major axis (0,2),(−10,2). Endpoints of minor axis (−5,4),(−5,0). Foci at (−5+√21,2),(−5−√21,2).
25. (x+3)2(5)2+(y+4)2(2)2=1; Endpoints of major axis (2,−4),(−8,−4). Endpoints of minor axis (−3,−2),(−3,−6). Foci at (−3+√21,−4),(−3−√21,−4).
27. Foci (−3,−1+√11),(−3,−1−√11)
29. Focus (0,0)
31. Foci (−10,30),(−10,−30)
33. Center (0,0), Vertices (4,0),(−4,0),(0,3),(0,−3), Foci (√7,0),(−√7,0)
35. Center (0,0), Vertices (19,0),(−19,0),(0,17),(0,−17), Foci (0,4√263),(0,−4√263)
37. Center (−3,3), Vertices (0,3),(−6,3),(−3,0),(−3,6), Focus (−3,3)
Note that this ellipse is a circle. The circle has only one focus, which coincides with the center.
39. Center (1,1), Vertices (5,1),(−3,1),(1,3),(1,−1), Foci (1,1+4√3),(1,1−4√3)
41. Center (−4,5), Vertices (−2,5),(−6,4),(−4,6),(−4,4), Foci (−4+√3,5),(−4−√3,5)
43. Center (−2,1), Vertices (0,1),(−4,1),(−2,5),(−2,−3), Foci (−2,1+2√3),(−2,1−2√3)
45. Center (−2,−2), Vertices (0,−2),(−4,−2),(−2,0),(−2,−4), Focus (−2,−2)
47. x225+y229=1
49. (x−4)225+(y−2)21=1
51. (x+3)216+(y−4)24=1
53. x281+y29=1
55. (x+2)24+(y−2)29=1
57. Area=12π square units
59. Area=2√5π square units
61. Area 9π square units
63. x24h2+y214h2=1
65. x2400+y2144=1. Distance = 17.32 feet
67. Approximately 51.96 feet
Section 7.2 Solutions
1. A hyperbola is the set of points in a plane the difference of whose distances from two fixed points (foci) is a positive constant.
3. The foci must lie on the transverse axis and be in the interior of the hyperbola.
5. The center must be the midpoint of the line segment joining the foci.
7. yes x262−y232=1
9. yes x242−y252=1
11. x252−y262=1; vertices: (5,0),(−5,0); foci: (√61,0),(−√61,0); asymptotes: y=65x,y=−65x
13. y222−x292=1; vertices: (0,2),(0,−2); foci: (0,√85),(0,−√85); asymptotes: y=29x,y=−29x
15. (x−1)232−(y−2)242=1; vertices: (4,2),(−2,2); foci: (6,2),(−4,2); asymptotes: y=43(x−1)+2,y=−43(x−1)+2
17. (x−2)272−(y+7)272=1; vertices: (9,−7),(−5,−7); foci: (2+7√2,−7),(2−7√2,−7); asymptotes: y=x−9,y=−x−5
19. (x+3)232−(y−3)232=1; vertices: (0,3),(−6,3); foci: (−3+3√2,1),(−3−3√2,1); asymptotes: y=x+6,y=−x
21. (y−4)222−(x−3)242=1; vertices: (3,6),(3,2); foci: (3,4+2√5),(3,4−2√5); asymptotes: y=12(x−3)+4,y=−12(x−3)+4
23. (y+5)272−(x+1)2702=1; vertices: (−1,2),(−1,−12); foci: (−1,−5+7√101),(−1,−5−7√101); asymptotes: y=110(x+1)−5,y=−110(x+1)−5
25. (x+3)252−(y−4)222=1; vertices: (2,4),(−8,4); foci: (−3+√29,4),(−3−√29,4); asymptotes: y=25(x+3)+4,y=−25(x+3)+4
27. y=25(x−3)−4,y=−25(x−3)−4
29. y=34(x−1)+1,y=−34(x−1)+1
31.
33.
35.
37.
39.
41.
43.
45. x29−y216=1
47. (x−6)225−(y−1)211=1
49. (x−4)225−(y−2)21=1
51. y216−x225=1
53. y29−(x+1)29=1
55. (x+3)225−(y+3)225=1
57. y(x)=3√x2+1,y(x)=−3√x2+1
59. y(x)=1+2√x2+4x+5,y(x)=1−2√x2+4x+5
61. x225−y225=1
63. x2100−y225=1
65. x2400−y2225=1
67. (x−1)20.25−y20.75=1
69. (x−3)24−y25=1
Section 7.3 Solutions
1. A parabola is the set of points in the plane that lie equidistant from a fixed point, the focus, and a fixed line, the directrix.
3. The graph will open down.
5. The distance between the focus and directrix will increase.
7. yes y=4(1)x2
9. yes (y−3)2=4(2)(x−2)
11. y2=18x,V:(0,0);F:(132,0);d:x=−132
13. x2=−14y,V:(0,0);F:(0,−116);d:y=116
15. y2=136x,V:(0,0);F:(1144,0);d:x=−1144
17. (x−1)2=4(y−1),V:(1,1);F:(1,2);d:y=0
19. (y−4)2=2(x+3),V:(−3,4);F:(−52,4);d:x=−72
21. (x+4)2=24(y+1),V:(−4,−1);F:(−4,5);d:y=−7
23. (y−3)2=−12(x+1),V:(−1,3);F:(−4,3);d:x=2
25. (x−5)2=45(y+3),V:(5,−3);F:(5,−145);d:y=−165
27. (x−2)2=−2(y−5),V:(2,5);F:(2,92);d:y=112
29. (y−1)2=43(x−5),V:(5,1);F:(163,1);d:x=143
31.
33.
35.
37.
39.
41.
43.
45. x2=−16y
47. (y−2)2=4√2(x−2)
49. (y+√3)2=−4√2(x−√2)
51. x2=y
53. (y−2)2=14(x+2)
55. (y−√3)2=4√5(x+√2)
57. y2=−8x
59. (y+1)2=12(x+3)
61. (0,1)
63. At the point 2.25 feet above the vertex.
65. 0.5625 feet
67. x2=−125(y−20), height is 7.2 feet
69. 2304 feet
Section 7.4 Solutions
1. plotting points with the orientation arrow and a graphing calculator
3. The arrows show the orientation, the direction of motion according to increasing values of t.
5. The parametric equations show the different vertical and horizontal motions over time.
7.
9.
11.
13.
15.
17.
19.
21.
23.
25.
27.
29.
31.
33.
35.
37.
39. There will be 100 back-and-forth motions.
41. Take the opposite of the x(t) equation.
43. The parabola opens up.
45. {x(t)=5costy(t)=5sint
47.
49.
51.
53. a=4,b=3,c=6,d=1
55. a=4,b=2,c=3,d=3
57.
59.
61. The y -intercept changes.
63. y(x)=−16(x15)2+20(x15)
65. {x(t)=64tcos(52∘)y(t)=−16t2+64tsin(52∘)
67. approximately 3.2 seconds
69. 1.6 seconds
71.
73.
Section 7.5 Solutions
1. plotting points with the orientation arrow and a graphing calculator
3. The arrows show the orientation, the direction of motion according to increasing values of t.
5. The parametric equations show the different vertical and horizontal motions over time.
7.
9.
11.
13.
15.
17.
19.
21.
23.
25.
27.
29.
31.
33.
35.
37.
39. There will be 100 back-and-forth motions.
41. Take the opposite of the x(t) equation.
43. The parabola opens up.
45. {x(t)=5costy(t)=5sint
47.
49.
51.
53. a=4,b=3,c=6,d=1
55. a=4,b=2,c=3,d=3
57.
59.
61. The y -intercept changes.
63. y(x)=−16(x15)2+20(x15)
65. {x(t)=64tcos(52∘)y(t)=−16t2+64tsin(52∘)
67. approximately 3.2 seconds
69. 1.6 seconds
71.
73.
Section 7.6 Solutions
1. The xy term causes a rotation of the graph to occur.
3. The conic section is a hyperbola.
5. It gives the angle of rotation of the axes in order to eliminate the xy term.
7. AB=0, parabola
9. AB=−4<0, hyperbola 11. AB=6>0, ellipse
13. B2−4AC=0, parabola
15. B2−4AC=0, parabola
17. B2−4AC=−96<0, ellipse
19. 7x′2+9y′2−4=0
21. 3x′2+2x′y′−5y′2+1=0
23. θ=60∘,11x′2−y′2+√3x′+y′−4=0
25. θ=150∘,21x′2+9y′2+4x′−4√3y′−6=0
27. θ≈36.9∘,125x′2+6x′−42y′+10=0
29. θ=45∘,3x′2−y′2−√2x′+√2y′+1=0
31. √22(x′+y′)=12(x′−y′)2
33. (x′−y′)28+(x′+y′)22=1
35. (x′+y′)22−(x′−y′)22=1
37. √32x′−12y′=(12x′+√32y′−1)2
39.
41.
43.
45.
47.
49.
51. θ=45∘
53. θ=60∘
55. θ≈36.9∘
57. [latex]-4\sqrt{6}
1. If eccentricity is less than 1, it is an ellipse. If eccentricity is equal to 1, it is a parabola. If eccentricity is greater than 1, it is a hyperbola.
3. The directrix will be parallel to the polar axis.
5. One of the foci will be located at the origin.
7. Parabola with e=1 and directrix 34 units below the pole.
9. Hyperbola with e=2 and directrix 52 units above the pole.
11. Parabola with e=1 and directrix 310 units to the right of the pole.
13. Ellipse with e=27 and directrix 2 units to the right of the pole.
15. Hyperbola with e=53 and directrix 115 units above the pole.
17. Hyperbola with e=87 and directrix 78 units to the right of the pole.
19. 25x2+16y2−12y−4=0
21. 21x2−4y2−30x+9=0
23. 64y2=48x+9
25. 96y2−25x2+110y+25=0
27. 3x2+4y2−2x−1=0
29. 5x2+9y2−24x−36=0
31.
33.
35.
37.
39.
41.
43. r=45+cosθ
45. r=41+2sinθ
47. r=11+cosθ
49. r=78−28cosθ
51. r=122+3sinθ
53. r=154−3cosθ
55. r=33−3cosθ
57. r=±2√1+sinθcosθ
59. r=±24cosθ+3sinθ