Analytic geometry. Textbook
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5.2. Second order curves on the plane |
141 |
k = 0 this angle is zero. If x1 = x2 the slope is undetermined, the angle of inclination is right.
a) obtuse; b) obtuse; c) acute; d) acute; e) right; f) obtuse.
33. d), e), g).
34. a) 54x + 53y 2 = 0; |
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x + 1312y 3 = 0; |
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c) 53x + |
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e)23 x + 12y 2 = 0; f) x cos 190 + y sin 190 4 = 0.
35.The equations have the form x cos + y sin p = 0, where
a) = 225 (5 =4), |
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b) = 135 (3 =4), |
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c) = 45 ( =4), |
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d) = 315 (7 =4), |
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p = 2 |
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p = 3 |
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e) = 240 (4 =3), |
p = 3; |
f) = 60 ( =3), |
p = 2; |
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g) = 30 ( =6), |
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h) = 150 (5 =6), |
p = 1. |
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36. |
The line passes through the point ( 3; 0), it has equation y = |
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2(x + 3), or 2x y + 6 = 0. |
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37. |
k = 1. |
38. |
k = 61. 39. k = 1. |
40. |
The straight lines Ax + By + 10 = 0 and 3x + y = 7 are parallel |
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if and only if A = B , B = A. The straight line intersects the x-axis at the |
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point at the point with y = 0, therefore with x = 7. This point is on the line Ax + By + 10 = 0, we get the relation 7A + 10 = 0, A = 107 , B = A3 = 1021.
5.2Second order curves on the plane
41. a) |
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C
x
Fig. 1. Center C(1; 3), radius 5
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Chapter 5. Answers and solutions |
b)y 
Cx
Fig. 2. Center C(3; 0), radius 3
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Fig. 3. Center C( 3; 2), radius 4 |
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d) |
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Fig. 4. Center C(1; 4), radius 5
5.2. Second order curves on the plane |
143 |
42. y
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Fig. 5. Intersection points (9=10; 3=10), (2; 3)
43. |
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Fig. 6. The straight line is the tangent, the tangency point ( 2; 1)
44.( 2; 0), (4; 0).
45.The equation can be reduced to the form
(x 3)2 + (y + 1)2 = 17;
p
it is the circle with the center (3; 1) and the radius 17 . Let (x; 3) be a point of the circle. Then (x 3)2 + 16 = 17, (x 3)2 = 1, x 3 = 1, x = 4 or x = 2. We get the points A(4; 3) and B(2; 3). The circle cuts the x-axis
144 Chapter 5. Answers and solutions
at such points (x; 0) that (x 3)2 + 1 = 17, (x 3)2 = 16, x 3 = 4, x = 7 or x = 1. We get the points C(7; 0) and D( 1; 0)
y 
BA
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Fig. 7. The circle and the obtained points
46.(x 2)2 + (y + 2)2 = 169, B, C, E are on the circle, A, D are inside the circle, F is outside the circle.
47.Let us denote A(5; 1), B(8; 3), C(2; 5). The points B and C are on a circle whose center is A if and only if jABj = jACj, in this case such value is the radius,
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5)2 + ( 5 |
( 1)2) =p |
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AC |
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3 + 4 |
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AB = (8 5)2 |
+ (3 ( 1)2) = 32 |
+ 42 |
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= 5; |
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the points lie on the circle mentioned above, the radius equals 5. |
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48. a) 3x + 4y 25 = 0; |
b) 2x 5y 49 = 0; |
c) 5x 3y + 34 = 0; |
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d) x y 3 2 = 0; e) 2x + y + 10 = 0; |
f) x + 6y + 37 = 0; |
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g)x 3y + 4 = 0; h) 10x 3y 24 = 0.
49.a) 3x 4y + 25 = 0, ( 3; 4), 3x 4y 25 = 0, (3; 4);
b)12x + 5y + 91 = 0, ( 84=13; 35=13), 12x + 5y 91 = 0, (84=13; 35=13);
c)4x 3y + 30 = 0, ( 24=5; 18=5);
d)2x + 3y + 13 = 0, ( 2; 3), 2x + 3y 13 = 0, (2; 3);
5.2. Second order curves on the plane |
145 |
e)x + 5y 52 = 0, (2; 10), 5x y + 52 = 0, ( 10; 2);
f)x 8 = 0, (8; 0), 3x + 4y 40 = 0, (24=5; 32=5).
50.5x + 3y = 34.
51.2x + y (x + 2) + 2(y + 1) 5 = 0, or x + 3y 5 = 0.
52.y
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Fig. 8. The circle (x 5)2 + (y 2)2 = 4
53.y
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The circle (x 4)2 + (y 3)2 = 16.
54. If a line is the tangent to a circle with the center at the origin then the the radius of this circle equals the distance of the origin from this line. So the distances of the given lines from the origin must be the same. They are
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j30j |
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q22 + (p |
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146 Chapter 5. Answers and solutions
therefore these lines are the tangents to the circle with the center at the origin and the the radius 5 units.
55. |
r = 4, (x 4)2 + (y 3)2 |
= 16. |
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56. |
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r = 2 10 , (x 3) |
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+ (y + 1) = 40. |
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57. |
r = 2, (x 2)2 + (y 4)2 |
= 4. |
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58. |
Rewriting the equation of the given straight line in the form kx |
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y + 10 = 0, we get that |
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p |
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= 6; |
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pk2 + 1 = |
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59. |
Representing the given equation of the circle in the form (x 4)2 + |
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(y + 6)2 = 64 we get that this circle has the center at the point C(4; 6) and the radius 8. We write equations of the lines parallel to the given line 5x 12y 14 = 0 in the form 5x 12y + C = 0. Such line is the tangent of the given circle if and only if its distance from the point C equals the radius of the circle. We obtain the equation
5 4 12 ( 6) + C = 8; |
jC + 92j = 8; |
C + 92 = 104; |
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C = 12 or |
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52 + ( 4)2 |
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C = |
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196. We get the equations 5x |
12y + 12 = 0, 5x |
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196 = 0. |
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60. a) |
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Fig. 9. y2 = 8x, focus F (2; 0), directrix d : x = 2
5.2. Second order curves on the plane |
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Fig. 10. y2 = 8x, focus F ( 2; 0), directrix d : x = 2
c) |
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Fig. 11. x2 = 24y, focus F (0; 6), directrix d : y = 6
d) |
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Fig. 12. x2 = 24y, focus F (0; 6), directrix d : y = 6
148 |
Chapter 5. Answers and solutions |
61.a) y2 = 18x; b) y2 = 16x; c) y2 = 20x.
62.a) x2 = 12y; b) y2 = 24x, x2 = 12y; c) x2 = 24y.
63.(9; 6).
64.a) 
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Fig. 13. Intersection points (4; 6), (25; 15) |
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Fig. 14. Tangency point (12; 6) |
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Fig. 15. Tangency point ( 3; 1) |
5.2. Second order curves on the plane |
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d) |
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Fig. 16. Intersection points (0; 0), (12; 12) |
e) |
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Fig. 17. Intersection points (4; 2) |
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Fig. 18. Intersection points (0; 0), (5; 5)
150 |
Chapter 5. Answers and solutions |
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Fig. 19. Intersection points (2; 3), ( 2; 3) |
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65. |
(3; 6). 66. (4; 2 5 ). |
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67. |
a) y2 = 12x; |
b) y2 = 10x 25; |
c) y2 = 16x; |
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d) x2 = 8y; e) x2 = 18y. |
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68. |
(18; 12). |
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69. |
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Fig. 20. Two intersection points (2; 6) and |
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