Analytic geometry. Textbook

second condition we get that 1444b2 27b2 = 1. Therefore b2 = 9, a2 = 4b2 = 36.

Answer: x362 y92 = 1.

109. The asymptotes of the hyperbola xa22 yb22 = 1 are de ned by the equations y = ab x. Therefore the angle between the asymptotes is 2

Answer: a) = 23 ( = 120 ). b) = 2 ( = 90 ).

110.r1 = 19, r2 = 9.

111.x 2y 12 = 0, x + 2y + 8 = 0.

e)x + y 1 = 0.

113.The standard equation of the hyperbola has the form x252 16y2 = 1, the straight lines parallel to the given line are de ned by the equations 2x 2y + C = 0. From the tangency condition we obtain that 22 25

( 2)2 16 = C2, C2 = 36, C = 6. We get equations of the tangents 2x 2y 6 = 0, or x y 3 = 0. To nd the tangency point for the tangent x y 3 = 0 we solve the system of equations

(

16x2 25y2 = 400; x y 3 = 0:

The solution is (25=3; 16=3). The second tangent is analysed similarly.

162 Chapter 5. Answers and solutions

Answer: x y 3 = 0, (25=3; 16=3); x y + 3 = 0, ( 25=3; 16=3).

114. The standard equation of the hyperbola has the form x242 32y2 = 1, the straight lines parallel to the given line are de ned by the equations 2x y +C = 0. From the tangency condition we obtain that 22 24 ( 1)2

32 = C2, C2 = 64, C = 8. We get equations of the tangents 2x y 8 = 0. To nd the tangency point for the tangent x y 8 = 0 we solve the system

of equations

(

4x2 3y2 = 96;

2x y 8 = 0:

The solution is (6; 4). The second tangent is analyzed similarly.

Answer: 2x y 8 = 0, (6; 4), 2x y + 8 = 0, ( 6; 4).

115. a) The point M(1; 4) is not the point of this hyperbola. The standard equation of the hyperbola has the form 18x=25 y92 = 1, the general form of straight lines passing through the point M(1; 4) is A(x 1) +

B(y + 4) = 0, or Ax + By + ( A + 4B) = 0. From the tangency condition

we get that

185 A2 9B2 = ( A + 4B)2;

or after simpli cation 13A2 + 40AB 125B2 = 0. If B = 0 we get from this equation that A = 0 which is impossible for equation of the line. Therefore

B 6= 0, we let B = 1 and obtain the equation 13A2 + 40A 125 = 0 with the solutions A = 5 and A = 25=13. In the case A = 5, B = 1 the equation has the form 5x + y + 9 = 0, or 5x y 9 = 0, the tangency point obtained from the system of equations

(

5x2 2y2 = 18;

5x y 9 = 0;

is the point (2; 1). In the case A = 25=13, B = 1 equation of the tangent after simpli cation takes the form 25x + 13y + 27 = 0 and the tangency point is ( 10=3; 13=3).

b) The point M(3; 9) is not the point of this hyperbola. The standard equation of the hyperbola has the form x92 y92 = 1, the general form of

straight lines passing through the point M(3; 9) is A(x 3) + B(y 9) = 0, or Ax + By (3A + 9B) = 0. From the tangency condition we get that

9A2 9B2 = (3A + 9B)2;

or after simpli cation 3AB + 5B2 = 0, B(3A + 5B) = 0. If B = 0 we let

A = 1 and get the equation x 3 = 0. To nd the corresponding tangency point we put this value in the equation of the hyperbola and get that y = 0, the tangency point is (3; 0). If 3A + 5B = 0 we let A = 5, B = 3 and get the equation 5x 3y + 12 = 0. To nd the tangency point we consider the

system of equations

(

x2 y2 = 9;

5x 3y + 12 = 0;

and obtain the desired point ( 15=4; 9=4).

Answer: a) 5x y 9 = 0, tangency point (2; 1), 25x + 13y + 27 = 0,

tangency point ( 10=3; 13=3);

b) x 3 = 0, tangency point (3; 0), 5x 3y + 12 = 0, tangency point

( 15=4; 9=4).

116. The standard equation of the hyperbola has the form x162 y42 = 1, equations of straight lines with equal intercepts have the following general

form x + y A = 0. Using the tangency condition we get that

In order to avoid calculations with radicals, we introduce new unknowns X p p

and Y by the formulas x = X 3 , y = Y 3 . We obtain the equations

(

3X2 12Y 2 = 16;

X + Y 2 = 0:

164 Chapter 5. Answers and solutions

From the second equation if follows that Y = 2 X, excluding Y using this relation from the rst equation after simpli cation we get the equation

117. a) 3x + 2y 6 = 0, 3x 2y 6 = 0; b) 3x + 2y + 6 = 0 (the point is on the hyperbola); c) the tangents do not exist (the point is inside the

hyperbola).

p

118. a) x + y 3 = 0; b) 2x + y 54 = 0.

119.x2 y2 = 1. 8 4

5.3Analytic geometry in the space

120.a) 9; b) 7; c) 6; d) 7; e) 9; f) 7.

121.x 3y + 4z + 5 = 0. 122. (2; 1; 1).

123. All planes which are parallel to the plane de ned by the equation 3x 6y 2z + 14 = 0 may be de ned by the equations of the form 3x

6y 2z + D = 0 with a suitable value D.

The distance between the given plane 3x 6y 2z + 14 = 0 and the desirable plane 3x 6y 2z + D = 0 is given by the formula

D 14 = 21 D = 35 or D = 7.

Answer: 3x 6y 2z + 35 = 0 and 3x 6y 2z 7 = 0.

124. In the table below the sign \|" denotes that the intercept is undetermined (the plane does not intersect the corresponding axis).

c)116 x 116 y + 117 z 3 = 0.

126.10. 127. a) x 4y + 5z + 15 = 0; b) 2x y z = 0. 128. d = 10.

129. The desired point has the coordinates (0; 0; z). Equating the distances from the given planes we get:

These equations have the following solutions z = 3, z = 52.

Answer: (0; 0; 3), 0; 0; 52 .

130. 4. 131. 3x 6y 2z + 35 = 0 and 3x 6y 2z 7 = 0.

132. Since the plane passes through the origin O its equation has the

form Ax + By + Cz = 0. We denote M1(3; 2; 1), M2(1; 4; 0). The normal

! !

vector of the plane is orthogonal to the vectors OM1 and OM2. Hence we can take their vector product as the normal vector of the plane,

The plane is de ned by the equation 4x + y + 14z = 0 or 4x y 14z = 0.

Answer: 4x y 14z = 0.

133. We write the general form of the planes parallel to the given plane:

p jD 4j = 3;

32 + 62 + 22

135.x + y + z 5 = 0.

136.a) x 1 = 0; b) 4y + z = 0; c) 9x z 54 = 0.

137.We perform the following transformation:

Therefore the x-intercept is 3, the y-intercept is 15, the z-intercept is 5.

138. Coordinates of the common point satisfy all these equation. We

We get that x = 2, y = 1, z = 1, the common point M(2; 1; 1).

139. We have three points M0(0; 0; 0), M1(3; 2; 1), M2(1; 4; 0). Equation of the plane has the form

x y z

3 2 1 = 0:

1 4 0

Expanding the determinant along the rst row we get:

3 2

+z = 4x y 14z:

with the normal vectors ~n1 = f2; 1; 1g, ~n2 = f1; 3; 1g. The desired straight line is parallel to each of these planes. Therefore the direction vector of this line is orthogonal to the normal vectors of these planes. We nd the vector product of the normal vectors:

~

~{ ~| k

~ ~n1 ~n2 = 2 1 1 = 4~{ ~| 7k:

1 3 2

The obtained vector can be taken as the direction vector of the desired line and we get its equation

144. The plane xOy is de ned by the equation z = 0. We add this equation to the equations of the line and solve the arising system of equa-

tions:

8

> 3x + y 3z = 10;

<

2x + 2y z = 4;

>

: z = 0;

3x + y = 10;

2x + 2y = 4;

x = 4, y = 2, z = 0.

The other cases are investigated similarly.

Answer: The points of intersection: the plane xOy | the point (4; 2; 0), the plane xOz | the point 23; 0; 83 , the plane yOz | the point 0; 25; 165 .

145. The distance of the point M1 from the straight line passing through the point M0 and having the direction vector ~e is found by the formula

!

d = jM0M1 ~e j: j~e j

In this case M0(2; 1; 0), M1(7; 9; 7), ~e = f4; 3; 2g,

4 3 2

p

Answer: 22 .

146. Parametric equations of the given line have the form x = t; y = 7 + 2t; z = 3 t:

To nd the value of the parameter t for the desired point we can use two approaches.

a) Let d be the distance of an arbitrary point of the given line from the point (3; 2; 6). Then

d2 = (t 3)2 + (2t 9)2 + (t + 3)2 =

6t2 36t + 99 = 6(t2 6t + 9) + 45 = 6(t 3)2 + 45:

This distance d and its square d2 take their minimal values at the same point, for d2 it is the point t = 3 since at this point the value 6(t 3)2 vanishes and at the other points this value is positive.

b) We consider the plane which passes through the point (3; 2; 6) and is perpendicular to the given straight line. We take the direction vector of the given straight line f1; 2; 1g as the normal vector of this plane. Equation of this plane takes the form

(x 3) + 2(y 2) (z + 6) = 0; or x + 2y z 1 = 0:

To nd the coordinates of the intersection point we replace here the values of the coordinates by the right sides of the parametric equations:

t + 2( 7 + 2t) (3 t) 1 = 0; 6t 18 = 0; t = 3:

Now we get the coordinates of the desired point (3; 1; 0).

Answer: (3; 1; 0).

147. Let (x; y; z) be the desired point. Equating the squares of the distances of this point from the given points we get:

(x 3)2 + (y 11)2 + (z 4)2 = (x + 5)2 + (y + 13)2 + (z + 2)2:

After simpli cation we get the equation

4x + 12y + 3z + 13 = 0:

Remark. It is possible to get the same equation using geometrical arguments. The locus of the points equidistant from two di erent points M1 and M2 is the plane passing through the midpoint of the segment

M1M2 and perpendicular to this segment. In the case under consideration

M1(3; 11; 4), M2( 5; 13; 2), the midpoint of the segment M1M2 is the

!

point ( 1; 1; 1). There takes place the equality M1M2 = f8; 24; 6g. We can take this vector or any nonzero proportional one as the normal vector of this plane. Taking the vector f4; 12; 3g we get the equation

4(x + 1) + 12(y + 1) + 3(z 1) = 0

170 Chapter 5. Answers and solutions

and after simpli cation the equation found above.

Adding the obtained equation to the equations of the given line we obtain

:

Its solution x = 2, y = 3, z = 5 gives the coordinates of the desired point

Answer: (2; 3; 5).

148. The distance between the parallel straight lines passing through the points M1 and M2 respectively and having the same direction vector ~e can be found from the equality (corollary of theorem 8, page 105)

!

d = jM1M2 ~e j:

!

In this case M1(2; 1; 0), M2(7; 1; 3), M1M2 = f5; 2; 3g, ~e = f3; 4; 2g,

3 4 2

Answer: d = 3.

149. The distance between the straight line `1 with the direction vector

~e1 passing through the point M1 and the straight line `2 with the direction vector ~e2 passing through the point M2 can be found by the formula (theorem 9, page 106)

!

d = jM1M2 ~e1~e2 j:

a) In this case ~e1 = f4; 3; 1g, M1(9; 2; 0), ~e2 = f2; 9; 2g,