Analytic geometry. Textbook

Fig. 23. The line is parallel to the symmetry axis, intersection at 12; 3

70. The table uses the following notations in the column \axis": +Ox means that the axis of the parabola is parallel to the axis Ox and has the same direction, Ox means that the axis of the parabola is parallel to the axis Ox and has the opposite direction. Similarly for the other axis.

72. c = 12. 73. x + y + 2 = 0, 2x + 5y + 25 = 0. 74. (9; 6).

75. 1) x + y + 3 = 0, tangency point of (3; 6), x y + 3 = 0, tangency point (3; 6);

2)3x y + 1 = 0, the tangency point 13; 2 ;

3)x 2y + 12 = 0, the tangency point (12; 12).

76. If a straight line and a parabola do not intersect then the shortest distance between these lines equals the distance between the given straight line and the tangent to the parabola that is parallel to the given line.

In the case under consideration the straight line and the parabola do not intersect. To prove it we consider the following system of equations

y2 = 64x; 4x + 3y + 46 = 0:

From the rst equation we get that x = y2 . Substituting this expression into

64

the second equation we get the quadratic equation y2 + 48y + 736 = 0 with negative discriminant. Such an equation has no real solutions. It means that two given lines do not intersect.

Fig. 24. Distance from a parabola to a straight line

The desired tangent to the parabola may be de ned by the equation 4x + 3y + C = 0. From the tangency condition pB2 = 2AC in the case

154 Chapter 5. Answers and solutions

p = 32, A = 4, B = 3 we get that C = 36. The distance between the straight lines 4x + 3y + 46 = 0 and 4x + 3y + 36 = 0 equals

p

2x 3y 10 = 0. Reducing the equation 2x 3y 10 = 0 to the intercept

point we solve the system of equations

(

7x2 + 3y2 = 28;

2x 3y 10 = 0:

We get from the second equation that x = 3y+10 and put this expression

2

into the rst equation:

7 3y + 10 2 + 3y2 = 28: 2

After simpli cation we obtain the following equation 25y2 +140y +196 = 0, therefore y = 14=5, x = 4=5.

For the tangent 2x 3y+10 = 0 we get the intercepts 5, 10=3, tangency point is ( 4=5; 14=5).

Answer: 2x 3y + 10 = 0, intercepts 5, 10=3, tangency point ( 4=5; 14=5); 2x 3y 10 = 0, intercepts 5, 10=3, tangency point (4=5; 14=5).

93. The desired tangent to the ellipse x2 + y2 = 1 has the equation

35 14

A(x 8) + B(y + 1) = 0, or Ax + By + ( 8A + B) = 0. The tangency

condition takes the form 35A2 + 14B2 = ( 8A + B)2, or after simpli cation 29A2 16AB 13B2 = 0. If B = 0 we get from this equation that A = 0 which is impossible for the equation of the line. Therefore B 6= 0, we let

B = 1 and obtain the equation 29A2 16A 13 = 0 with two solutions

A = 1 and A = 13=29. In the case A = 1, B = 1, the equation has the form x + y 7 = 0, the tangency point obtained from the system of equations

is the point (5; 2). In the case A = 13=29, B = 1 equation of the tangent after simpli cation takes the form 13x 29y 133 = 0 and the tangency point is (65=19; 58=19).

Answer: x + y 7 = 0, tangency point (5; 2); 13x 29y 133 = 0

is de ned by the equation x + y A = 0. From the tangency condition we get that 9 + 16 = A2, A = 5. We obtain the tangents x + y 5 = 0. Tond the tangency point for the tangent x + y 5 = 0 we solve the system of equations

(

16x2 + 9y2 = 144; x + y 5 = 0:

The solution is (9=5; 16=5). The second tangent is analyzed similarly

Answer: x + y 5 = 0, the tangency point (9=5; 16=5); x + y + 5 = 0, the tangency point ( 9=5; 16=5).

95.x 2y 8 = 0.

96.a) The straight lines passing through the point ( 6; 3) have the form

A(x + 6) + B(y 3) = 0 or Ax + By + (6A 3B) = 0. From the tangency condition we get that 15A2 + 9B2 = (6A 3B)2, or after simpli cation 7A2 12AB = 0, A(7A 12B) = 0. In the case A = 0 we obtain the equation y 3 = 0. We put this value into the equation of the ellipse and get that x = 0, the the tangency point (0; 3).

If 7A 12B = 0 we let A = 12, B = 7 and obtain the equation 12x + 7y + 51 = 0. In order to nd the tangency point we solve the system of equations

(

x2 + y2 = 1;

15 9

12x + 7y + 51 = 0:

From the second equation we nd that y = 12x7+51 and exclude the unknown y in the rst equation. After simpli cation we get the equation 289x2 + 2040x + 3600 = 0, or (17x + 60)2 = 0, therefore x = 60=17, y = 21=17.

b) The straight lines parallel to the line 2x y 17 = 0 have the general form 2x y + C = 0. From the tangency condition we get that 22 30+ ( 1)2 24 = C2, C2 = 144, C = 12. On order to nd the tangency point for the tangent 2x y 12 = 0 we we solve the system of equations

(

x2 + y2 = 1;

30 24

2x y 12 = 0:

From the second equation we nd that y = 2x 12 and exclude the unknown y in the rst equation. After simpli cation we get the equation x2 10x + 25 = 0, or (x 5)2 = 0, therefore x = 5, y = 2. In the case of the tangent 2x y + 12 = 0 we get that x = 5, y = 2.

c) The straight lines orthogonal to the line 13x + 12y 115 = 0 have the general form 12x 13y + C = 0. From the tangency condition we get that 122 169 + ( 13)2 25 = C2, C2 = 169(144 + 25) = 1692, C = 169. On order to nd the tangency point for the tangent 12x 13y 169 = 0 we we solve the system of equations

(

x2 + y2 = 1;

169 25

12x 13y 169 = 0:

From the second equation we nd that y = 12x 169 and exclude the unknown

13

y in the rst equation. After simpli cation we get the equation x2 24x +

144 = 0, or (x 12)2 = 0, therefore x = 12, y = 2513. In the case of the tangent 12x 13y + 169 = 0 we get that x = 12, y = 2513.

Answer. a) y = 3, tangency point (0; 3), 12x + 7y + 51 = 0, tangency point 6017; 2117 ;

b) 2x y 12 = 0, tangency point (5; 2), 2x y + 12 = 0, tangency point ( 5; 2);

c) 12x 13y 169 = 0, tangency point 12; 2513 , 12x 13y + 169 = 0, tangency point 12; 2513 .

97.(5; 4).

98.The focuses of the ellipse are the points F1( 4; 0) and F2(4; 0). Let

Ax + By + C = 0 be the equation of the desired tangent. The distances of this line from the focuses are

The focuses are from one side of any tangent of the ellipse. Therefore 4A+C and 4A + C have the same signs,

4A + C = 36A + 9C; 40A = 8C; C = 5A:

Substituting this value into the tangency condition 25A2 + 9B2 = C2 we get:

25A2 + 9B2 = 25A2; 9B2 = 0; B = 0:

Let A = 1, then C = 5 and we get the equation of the desired tangent

x + 5 = 0.

the tangency condition gives the relation 16A + 25B = 625. We obtain the following system of equations

The both solutions satisfy the conditions A > 0, B > 0, so we obtain two

To nd the tangency points we solve we solve the system consisting of equation of the ellipse together with equation of the tangent. For the rst ellipse this system has the form

101. a) Let Ax + By + C = 0 be equation of the desired tangent. From the tangency condition we obtain the following equations:

(

5A2 + 4B2 = C2; 4A2 + 5B2 = C2:

Taking the di erence of these equations we get that A2 = B2, therefore

C2 = 9B2. Let B2 = 1, then A2 = 1, C2 = 9. We obtain the solutions

A = 1, B = 1, C = 3 and eight equations x y 3 = 0 with all possible combinations of signs. Some pairs of these equations de ne the same lines, e.g. x+y +3 = 0 and x y 3 = 0. To exclude the redundant equations we may assume that A = 1. We get equations x y 3 = 0 with arbitrary combinations of signs, or x + y 3 = 0 and x y 3 = 0.

b) Here we get the system of equations: Taking the di erence of these equations we get that A2 = 4B2, therefore C2 = 25A2. Let B2 = 1, then

160 Chapter 5. Answers and solutions

A2 = 4, C2 = 25. As in the previous case we get four equations of di erent tangents 2x + y 5 = 0, 2x y 5 = 0.

Answer: a) x+y 3 = 0, x y 3 = 0; b) 2x+y 5 = 0, 2x y 5 = 0.

102. x 3y + 15 = 0.

107.a) ( 5; 0); b) e = 53; c) y = 43x, x = 95.

108.Let xa22 yb22 = 1 be equation of the desired hyperbola. Its asymptotes are de ned by the equations y = ab x, therefore in the case under