CHAPTER 3
3.1
Conics
The Cone and Conic Sections
Practice
Section 3.1 Page 135 Question 1
The generator of a cone is the line that is rotated about a point.
Section 3.1 Page 135 Question 3
The term “right circular cone” defines a cone whose axis of symmetry is at right angles (perpendicular) to
its base.
Applications and Problem Solving
Section 3.1 Page 135 Question 4
To create a circle, hold the axis of symmetry of the flashlight at a right angle to the table top.
To create an ellipse, hold the axis of symmetry of the flashlight at an angle between 0◦ and 45◦ to the table
top.
To create a parabola, hold the axis of symmetry of the flashlight at an angle of 45◦ to the table top.
To create a hyperbola, hold the axis if symmetry of the flashlight at an angle of between 45◦ and 90◦ to the
table top.
Section 3.1 Page 135 Question 5
The plane is parallel to the base when a circle is generated.
The plane is parallel to the generator when a parabola is generated.
The plane is parallel to the central axis when a hyperbola is generated.
The plane is not parallel to any part of the cone when an ellipse is generated.
Section 3.1 Page 135 Question 6
The parabola is symmetric about a line through its vertex, exactly halfway between corresponding points
on the two branches.
The ellipse and hyperbola are symmetric about a line through the vertices, and about a line perpendicular
to the first line, halfway between the vertices.
Section 3.1 Page 135 Question 7
The ellipse and circle are closed figures. The parabola and hyperbola are open figures.
Section 3.1 Page 135 Question 8
Answers may vary.
a) A circle can be thought of an ellipse in which the major and minor axes are the same length.
b) The hyperbola has two pieces whereas the parabola has one. The hyperbola has asymptotes whereas the
parabola does not.
Section 3.1
a) circle
b) parabola
c) ellipse
Page 135
Question 9
Section 3.1 Page 135 Question 10
Slices through the vertex yield degenerate conic sections. A degenerate ellipse or circle is a point. A
degenerate parabola is a line. A degenerate hyperbola is a pair of intersecting lines.
Section 3.1 Page 135 Question 11
a) If A, B, and C were all 0, the equation would not be a second-degree relation.
b) i) parabola ii) circle
iii) hyperbola
iv) ellipse
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
109
3.2
Look for a Pattern
Applications and Problem Solving
Section 3.2 Page 137 Question 1
Section 3.2 Page 137 Question 2
Use the form (a, b, c, d) to save space. For four The number 4 can be written in the following 8 ways:
steps, the possibilities are (1,1,1,1), (1,1,2), (1,2,1),
4=4
(2,1,1), and (2,2). For five steps the possibilities are
(1,1,1,1,1), (1,1,1,2), (1,1,2,1), (1,2,1,1), (2,1,1,1),
=3+1
and (1,2,2). For six steps the possibilities are
=1+3
(1,1,1,1,1,1), (1,1,1,1,2), (1,1,1,2,1), (1,1,2,1,1),
=2+2
(1,2,1,1,1), (2,1,1,1,1), (1,1,2,2), (1,2,1,2), (2,1,1,2),
=2+1+1
(2,1,2,1), (2,2,1,1), (1,2,2,1), and (2,2,2).
=1+2+1
Number of Steps Number of Possibilities
=1+1+2
1
1
2
2
=1+1+1+1
3
3
4
5
Since the number 2 can be written in 21 ways, the
5
8
number 3 can be written in 22 ways, and the number 4 can be written in 23 ways, the pattern suggests
6
13
that the number n can be written in 2n−1 ways. The
This is a Fibonacci sequence. The first 20 terms are, number 17 can be written in 216 or 65 536 ways.
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610,
987, 1597, 2584, 4181, 6765, 10 946. For 20 steps
there 10 946 possibilities.
Section 3.2
Page 137
Range
1−9
10 − 99
100 − 999
1000 − 9999
10 000 − 99 999
100 000 − 999 000
Question 3
Numbers
9
90
900
9000
90 000
899 001
Digits Per Number
1
2
3
4
5
6
Total
Total Digits
9×1=9
90 × 2 = 180
900 × 3 = 2700
9000 × 4 = 36 000
90 000 × 5 = 450 000
899 001 × 6 = 5 394 006
5 882 895
A total of 5 882 895 digits would be written.
Section 3.2 Page 137 Question 4
Consider the standard form of the sequence of powers of 17.
171 , 172 , 173 , 174 , 175 , 176 , 177 , . . . = 17, 289, 4913, 83 521, 1 419 857, 24 137 569, 410 338 673, . . .
Inspection of the units digits reveals the periodic sequence 7, 9, 3, 1, 7, 9, 3, . . .. Since 133 = 33 × 4 + 1, the
units digit of 17133 will be the same as 171 or 7.
Section 3.2 Page 137 Question 5
Let S be the sum of the series.
²
³ ²
³ ²
³
²
³
1
1 2
1 2 3
1 2 3 4
1
99
2
S= +
+
+
+ ... +
+
+ +
+ + +
+
+ ... +
2
3 3
4 4 4
5 5 5 5
100 100
100
1 3 6 10
4950
= + + +
+ ... +
2 3 4
5
100
99
1 2 3 4
= + + + + ... +
2 2 2 2
2
1
= (1 + 2 + 3 + 4 + . . . + 99)
2
1
= (4950)
2
= 2475
110
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
Section 3.2
Page 137
Question 6
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720
7! = 5040
1! + 2! = 1 + 2 = 3
1! + 2! + 3! = 3 + 3! = 9
1! + 2! + 3! + 4! = 9 + 4! = 33
1! + 2! + 3! + 4! + 5! = 33 + 5! = 152
All factorials greater than or equal to 5! end in a zero.
The sum of each factorial is added to a number whose
unit digit is 3. Thus, the sums from 1! + 2! + 3! + 4!
onward end in a 3.
Section 3.2 Page 137 Question 7
The first 19 sums consume the first 1 + 2 + 3 + . . . + 19 or 190 numbers.
S20 = 191 + 192 + 193 + . . . + 210
= (190 + 1) + (190 + 2) + (190 + 3) + . . . + (190 + 20)
= 20(190) + 1 + 2 + 3 + . . . + 20
= 3800 + 210
= 4010
3.3
The Circle
Practice
Section 3.3
Section 3.3
Page 141
Question 1
Section 3.1
(x − 0)2 + (y − 0)2 = 52
x2 + y 2 = 81
x2 + y 2 = 25
Page 141
Question 5
Section 3.3
Page 141
Question 9
Section 3.3
Question 13
(x − (−4))2 + (y − (−5))2 = 52
Question 17
x2 + y 2 = 121
2
2
(x − 0) + (y − 0) = 11
2
The centre is at (0, 0) and the radius is 11.
Question 11
(x − 2)2 + (y − 8)2 = 100
Section 3.3
Page 141
Question 15
√
(x − 5)2 + (y − (−4))2 = ( 6)2
(x + 4)2 + (y + 5)2 = 25
Page 141
Page 141
(x − 2)2 + (y − 8)2 = 102
(x + 2)2 + (y − 5)2 = 9
Section 3.3
Question 7
x2 + y 2 = 5
(x − (−2))2 + (y − 5)2 = 32
Page 141
Page 141
√
(x − 0)2 + (y − 0)2 = ( 5)2
x2 + y 2 = 2.25
Section 3.3
Question 3
(x − 0)2 + (y − 0)2 = 92
(x − 0)2 + (y − 0)2 = 1.52
Section 3.3
Page 141
(x − 5)2 + (y + 4)2 = 6
Section 3.3
Page 141
Question 19
3x2 + 3y 2 − 27 = 0
x2 + y 2 = 9
(x − 0)2 + (y − 0)2 = 32
The centre is at (0, 0) and the radius is 3.
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
111
Section 3.3
Page 141
Question 21
Section 3.3
4x2 + 4y 2 − 25 = 0
25
x2 + y 2 −
=0
4
25
x2 + y 2 =
4
² ³2
5
(x − 0)2 + (y − 0)2 =
2
Page 141
Question 23
(x + 3)2 + (y − 1)2 = 81
(x − (−3))2 + (y − 1)2 = 92
The centre is at (−3, 1) and the radius is 9.
5
.
2
Question 25
The centre is at (0, 0) and the radius is
Section 3.3
Page 141
(x − 6)2 + (y + 4)2 = 9.61
(x − 6)2 + (y − (−4))2 = 3.12
The centre is at (6, −4) and the radius is 3.1.
Section 3.3
Page 141
Question 27
(x − h)2 + (y − k)2 = r2
Substitute (h, k) = (8, 2) and (x, y) = (5, 0).
(5 − 8)2 + (0 − 2)2 = r2
9 + 4 = r2
r2 = 13
2
(x − 8) + (y − 2)2 = 13
Standard form:
x2 − 16x + 64 + y 2 − 4y + 4 − 13 = 0
x2 + y 2 − 16x − 4y + 55 = 0
General form:
Section 3.3
Page 141
Question 29
(x − h)2 + (y − k)2 = r2
Substitute (h, k) = (2, 3) and (x, y) = (7, 2).
(7 − 2)2 + (2 − 3)2 = r2
25 + 1 = r2
Standard form:
General form:
r2 = 26
(x − 2)2 + (y − 3)2 = 26
x2 − 4x + 4 + y 2 − 6y + 9 − 26 = 0
x2 + y 2 − 4x − 6y − 13 = 0
Section 3.3 Page 141 Question 31
A circle centered at (−6, −5) that is tangent to the y-axis passes through the point (0, −5).
(x − h)2 + (y − k)2 = r2
Substitute (h, k) = (−6, −5) and (x, y) = (0, −5).
(0 − (−6))2 + (−5 − (−5))2 = r2
36 + 0 = r2
r2 = 36
(x + 6)2 + (y + 5)2 = 36
Standard form:
x2 + 12x + 36 + y 2 + 10y + 25 − 36 = 0
General form:
112
x2 + y 2 − 12x + 10y + 25 = 0
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
Section 3.3 Page 141 Question 33
The centre of the circle is the midpoint of the endpoints of a diameter.
²
x1 + x2 y1 + y2
,
2
2
³
²
1 + (−3) 4 + (−6)
=
,
2
2
²
³
−2 −2
=
,
2 2
The radius of the circle is half the length of the
diameter.
1p
(x2 − x1 )2 + (y2 − y1 )2
2
1p
(−3 − 1)2 + (−6 − 4)2
=
2
1√
16 + 100
=
2
√
116
=
√2
2 29
=
√2
r = 29
³
r=
= (−1, −1)
The standard equation of the circle centered at (−1, −1) with a radius of
Section 3.3 Page 141 Question 35
The centre of the circle is the midpoint of the endpoints of a diameter.
√
29 is (x + 1)2 + (y + 1)2 = 29.
The radius of the circle is half the length of the
diameter.
1p
(x2 − x1 )2 + (y2 − y1 )2
2
1p
(−7 − 3)2 + (−2 − 10)2
=
2
1√
=
100 + 144
= (−2, 4)
2
√
244
=
√2
2 61
=
√2
r = 61
√
The standard equation of the circle centered at (−2, 4) with a radius of 61 is (x + 2)2 + (y − 4)2 = 61.
²
x1 + x2 y1 + y2
,
2
2
Section 3.3
³
²
3 + (−7) 10 + (−2)
,
2
2
²
³
−4 8
=
,
2 2
=
Page 141
³
Question 37
x2 − 7x + y 2 + 7y = 17.75
x2 − 7x + (−3.5)2 + y 2 + 7y + (3.5)2 = 17.75 + (−3.5)2 + (3.5)2
(x − 3.5)2 + (y + 3.5)2 = 42.25
(x − 3.5)2 + (y + 3.5)2 = (6.5)2
The centre is at (3.5, −3.5) and the radius is 6.5.
Section 3.3
Page 141
r=
Section 3.3
Page 141
Question 39
x2 + 8 + y 2 − 8y = 0
x2 + y 2 − 8y = −8
x2 + y 2 − 8y + 42 = −8 + 42
x2 + (y − 4)2 = 8
√
(x + 0)2 + (y − 4)2 = (2 2)2
√
The centre is at (0, 4) and the radius is 2 2.
Question 41
5x2 + 5y 2 = 100
x2 + y 2 = 20
y 2 = 20 − x2
p
y = ± 20 − x2
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
113
Section 3.3
Page 141
Question 43
x2 + 6x + y 2 − 4y = 37
x2 + 6x + 9 + y 2 − 4y + 4 = 37 + 9 + 4
(x + 3)2 + (y − 2)2 = 50
(y − 2)2 = 50 − (x + 3)2
p
y − 2 = ± 50 − (x + 3)2
p
y = 2 ± 50 − (x + 3)2
Section 3.3
Page 141
Question 45
x2 + 8x + y 2 + 10y + 13 = 0
x2 + 8x + 16 + y 2 + 10y + 25 + 13 = 16 + 25
(x + 4)2 + (y + 5)2 = 28
(y + 5)2 = 28 − (x + 4)2
p
y + 5 = ± 28 − (x + 4)2
p
y = −5 ± 28 − (x + 4)2
Applications and Problem Solving
Section 3.3 Page 141 Question 46
c) Answers will vary. The programs find all points that are the length of the radius away from the centre.
Section 3.3
Page 141
Question 47
Coin
Radius (cm)
Centre
Equation
Toonie
2.800 ÷ 2 = 1.4
(0, 5)
x2 + (y − 5)2 = 1.96
(3.5, 3.5)
(x − 3.5)2 + (y − 3.5)2 = 1.755 625
Loonie 2.650 ÷ 2 = 1.325
6
4
2.713 ÷ 2 = 1.3565 (−3.5, 3.5)
25/
c
2.388 ÷ 2 = 1.194
(1.5, −4.5)
(x − 1.5)2 + (y + 4.5)2 = 1.425 636
10/
c
1.803 ÷ 2 = 0.9015
(3.5, −3)
(x − 3.5)2 + (y + 3)2 = 0.812 702 25
5/
c
2.120 ÷ 2 = 1.06
(−1.5, −4.5)
(x + 1.5)2 + (y + 4.5)2 = 1.1236
1/
c
1.905 ÷ 2 = 0.9525
(−3.5, −3)
(x + 3.5)2 + (y + 3)2 = 0.907 256 25
Section 3.3
Page 141
2
(x + 3.5)2 + (y − 3.5)2 = 1.840 092 25
50/
c
–4
–2
–2
2
4
–4
Question 48
The centre of both circles is at (0, 5). The radius of
the inner ring is 1.7 ÷ 2 or 0.85 cm.
6
5.5
x2 + (y − 5)2 = 0.852
x2 + (y − 5)2 = 0.7225
5
4.5
4
–1 –0.5
0
0.5
1
Section 3.3 Page 141 Question 49
The presence of a Bxy term indicates that the conic has been rotated around its centre. The circle doesn’t
have this term because it possesses rotational symmetry. A = C = 1 because the circle is symmetric about
its centre (no stretching in any one direction).
114
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
Section 3.3 Page 142 Question 50
Define the image points under the translation as (x, y) → (x + 4, y − 1) = (u, v).
Expressing (x, y) in terms of (u, v) leads to the following equations.
x=u−4
y =v+1
y
(1)
(2)
Substituting (1) and (2) into the given equation leads
to the equation of the image circle.
x2 + y 2 = 49
(u − 4)2 + (v + 1)2 = 49
-1
x
4
(3)
Once the translation is complete, (3) can be expressed
in terms of x and y as (x − 4)2 + (y + 1)2 = 49.
For the general case, define the image points under the translation as (x, y) → (x + h, y + k) = (u, v), leading
to x = u − h and y = v − k. Substitution of these equations into the given equation leads to the general
equation of the image circle.
x2 + y 2 = r 2
(u − h)2 + (v − k)2 = r2
(4)
Once the translation is complete, (4) can be expressed in terms of x and y as (x − h) + (y − k) = r2 .
2
2
Section 3.3 Page 142 Question 51
Section 3.3 Page 142 Question 52
2
2
The equation of the image is (x − 4) + (y − 2) = 9. The equation of the image is (x + 5)2 + (y + 7)2 = 9.
Section 3.3 Page 142 Question 53
Section 3.3 Page 142 Question 54
2
2
The equation of the image is (x + 3) + (y − 8) = 9. The equation of the image is (x − 1)2 + (y + 6)2 = 9.
Section 3.3 Page 142 Question 55
The translation can be defined as (x, y) → (x − 3, y + 4) = (u, v). Given the original point, (x, y) = (m, n),
the coordinates of the image can be expressed as (u, v) = (m − 3, n + 4). The original coordinates of the
image point, (u, v) = (a, b), can be determined.
(u, v) = (a, b)
(x − 3, y + 4) = (a, b)
x−3=a
x=a+3
(1)
y+4=b
y =b−4
(2)
The original coordinates (a, b) were (a + 3, b − 4).
Section 3.3 Page 142 Question 56
The centre of the red circle is at (0, 0). The diameter, d, of the red circle can be expressed as the difference
of the distance between the centres of two large circles in non-adjacent quadrants and two radii. Consider
the centres, A(4, 4) and C(−4, −4). Let |AC| represent the distance between A and C and r represent the
radius of the larger circles.
d = |AC| − 2r
p
= (4 − (−4))2 + (4 − (−4))2 − 2(4)
√
= 64 + 64 − 8
√
=8 2−8
√
√
The radius of the red circle is (8 2 − 8) ÷ 2 or 4 2 − 4. The equation of the red circle is
√
x2 + y 2 = (4 2 − 4)2
√
x2 + y 2 = 16( 2 − 1)2
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
115
Section 3.3 Page 142 Question 57
Since the centre of the circle, C, lies on the line y = −2x, the coordinates can be expressed as (x, −2x). The
centre of the circle is equidistant from all points on the circle. Define the points A(1, 0) and B(−1, −2).
|CA| = |CB|
p
(x − 1)2 + (−2x − 0)2 = (x − (−1))2 + (−2x − (−2))2
p
(x − 1)2 + (−2x)2 = (x + 1)2 + (−2x + 2)2
x2 − 2x + 1 + 4x2 = x2 + 2x + 1 + 4x2 − 8x + 4
4x = 4
x=1
Substitution of x = 1 into the equation¬ y = −2x¬ yields y = −2. Since the coordinates of points C and A
have the same x-value, the radius, r, is ¬yA − yC ¬ or 2. The equation of the circle is (x − 1)2 + (y + 2)2 = 4.
Section 3.3
Page 142
Question 58
A plot of the two paths on the same set of axes reveals
two intersection points.
8
6
4
2
–6 –4 –2
–2
2 4 6 8 10
–4
Section 3.3
Page 142
Question 59
²
³
²
³
3
1
and B 2,
. Since
Assume (h, k) is the centre of the translated circle. Let the points be A 2,
2
2
¬
¬
¬y − y ¬ = 1, A and B are the endpoints of a diameter of the circle. Since A and B have the same
A
B
x-coordinates the y-value of the centre of this circle, k, is the y-value of the midpoint of A and B.
k=
3
2
+
2
1
2
2
2
=1
=
Thus, the centre of the translated circle is at (2, 1). Thus, the circle centered at the origin was translated
right 2 units and up 1 unit.
Section 3.3
Page 142
Question 60
² ³2
1
The circles with their centres on the x-axis can be expressed as (x ± 2 ± 1) + y =
.
2
² ³2
1
2
2
.
The circles with their centres on the y-axis can be expressed as x + (y ± 2 ± 1) =
2
2
These eight circles can be expressed as (x + a)2 + y 2 =
116
2
1
1
and x2 + (y + a)2 = where a = ±3 or ± 1.
4
4
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
Section 3.3
Page 142
Question 61
Since the centre of each circle is on the identity line, y = x, the coordinates are of the form (h, k) = (x, x).
The value for x can be expressed in terms of the radius, r, through the use of the Pythagorean Theorem.
x2 + x2 = r2
2x2 = r2
r2
x =
2
r
x= √ ; x>0
2
º
»
1 √ √ √
Since r ∈ {1, 2, 4, 8}, x ∈ √ , 2, 2 2, 4 2 . The
2
equations of the circles are
14
2
À
√ !2 À
√ !2
2
2
+ y−
x−
2
2
√
√
(x − 2)2 + (y − 2)2
√
√
(x − 2 2)2 + (y − 2 2)2
√
√
(x − 4 2)2 + (y − 4 2)2
Section 3.3
Page 142
12
10
8
6
4
=1
2
=4
–2
–2
= 16
2
4
6 x 8 10 12 14
= 64
Question 62
9
The radius of the largest circle will have a radius equal to half the side length of the square or . Since the
2
81
centre is at (−1, 3), the equation of the circle is (x + 1)2 + (y − 3)2 =
.
4
Section 3.3
Page 142
Question 63
The coordinates of equally-spaced points on a circle can most easily be determined through trigonometry.
Using the transformation (x, y) = (r cos θ, r sin θ), where r is the radius of the circle and θ is the angle of
360◦
rotation, the positions of the 8 stones can be determined. The angle of rotation to the first stone, A, is
12
or 30◦ .
À√
!
y
3 1
◦
◦
A(cos 30 , sin 30 ) = A
,
1
2 2
C
B
À √ !
1
3
B(cos 60◦ , sin 60◦ ) = B
,
2 2
À
√ !
A
D
0.5
1
3
◦
◦
C(cos 120 , sin 120 ) = C − ,
2 2
À √
!
3 1
◦
◦
D(cos 150 , sin 150 ) = D −
,
2 2
O
-0.5
0.5
À √
!
-1
1 x
3
1
E(cos 210◦ , sin 210◦ ) = E −
,−
2
2
À
√ !
H
1
3
-0.5
F(cos 240◦ , sin 240◦ ) = F − ,
E
2 2
À
√ !
1
3
◦
◦
G(cos 300 , sin 300 ) = G
,−
-1
2
2
G
F
À√
!
3 1
H(cos 330◦ , sin 330◦ ) = H
,−
2
2
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
117
3.4
The Ellipse
Practice
Section 3.4 Page 150 Question 1
a) The centre is located at (4, 2).
b) The length of the major axis is 7 − 1 or 6 units.
The length of the minor axis is 3 − 1 or 2 units.
c) Determine the c-value.
c2 = a2 − b2
p
c = ± 32 − 12
√
=± 8
√
= ±2 2
√
The foci are located at (4 ± 2 2, 2).
Section 3.4 Page 150 Question 5
The equation can be rewritten as
Section 3.4 Page 150 Question 3
a) The centre is located at (2, −4).
b) The length of the major axis is −2 − (−6) or 4
7 1
units. The length of the minor axis is − or
2 2
3 units.
c) Determine the c-value.
c2 = a2 − b2
s² ³
² ³2
2
4
3
c=±
−
2
2
r
16 9
c=±
−
4
4
r
7
c=±
4
√
7
c=±
2
À
√ !
7
The foci are located at 2, −4 ±
.
2
Section 3.4
Page 150
Question 7
The equation can be rewritten as
(y − 0)2
(x − 0)2
+
2
8
62
(y − 8)2
(x − 3)2
+
2
3
102
a) The centre is located at (0, 0).
b) The length of the major axis is 2(8) or 16 units.
The length of the minor axis is 2(6) or 12 units.
c) Determine the c-value.
a) The centre is located at (3, 8).
b) The length of the major axis is 2(10) or 20 units.
The length of the minor axis is 2(3) or 6 units.
c) Determine the c-value.
c2 = a2 − b2
p
c = ± 82 − 62
√
= ± 28
√
= ±2 7
c2 = a2 − b2
√
c = ± 100 − 9
√
= ± 91
The foci are located at (3, 8 ±
√
91).
√
The foci are located at (±2 7, 0).
Section 3.4
Page 150
Question 9
9x2 + 16y 2 = 144
9x2
16y 2
144
+
=
144
144
144
x2
y2
+
=1
16
9
(y − 0)2
(x − 0)2
+
=1
42
32
118
a) The centre is located at (0, 0).
b) The length of the major axis is 2(4) or 8 units.
The length of the minor axis is 2(3) or 6 units.
c) Determine the c-value.
c2 = a2 − b2
√
c = ± 16 − 9
√
=± 7
√
The foci are located at (± 7, 0).
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
Section 3.4
Page 150
Question 11
Section 3.4
Page 150
Question 13
Section 3.4
Page 150
Question 15
Section 3.4
Page 150
Question 17
Section 3.4
Page 150
Question 19
4x2 + (y + 1)2 = 9
(y + 1)2 = 9 − 4x2
p
y + 1 = ± 9 − 4x2
p
y = −1 ± 9 − 4x2
Section 3.4 Page 150 Question 21
The centre is at (3, −2), so h = 3 and k = −2. Plot the other points and sketch the ellipse.
1
0
–1
–2
–3
–4
–5
The major axis is parallel to the x-axis and 10 − (−4) or 14 units in length. So, a = 7. The minor axis is
parallel to the y-axis and 1 − (−5) or 6 units in length. So, b = 3.
(x − 3)2
(y − (−2))2
+
=1
72
32
(y + 2)2
(x − 3)2
+
=1
49
9
9(x − 3)2 + 49(y + 2)2 = 9 × 49
9(x2 − 6x + 9) + 49(y 2 + 4y + 4) = 441
9x2 − 54x + 81 + 49y 2 + 196y + 196 − 441 = 0
9x2 + 49y 2 − 54x + 196y − 164 = 0
MATHPOWERTM12, Western Edition, Solutions
(Standard Form)
(General Form)
Chapter 3
119
Section 3.4 Page 150 Question 23
Since the sum of the focal radii is 10, 2a = 10 or a = 5. The centre of the ellipse, (h, k), is the midpoint of
the line segment joining the foci.
²
³
0+0 0+8
,
(h, k) =
2
2
(x − 0)2
(y − 4)2
= (0, 4)
+
=1
32
52
The value for c is 8 − 4 or 4. Determine
x2
(y − 4)2
+
= 1 (Standard Form)
the b-value.
9
25
25x2 + 9(y − 4)2 = 9 × 25
b2 = a2 − c2
p
25x2 + 9(y 2 − 8y + 16) = 225
b = 52 − 42
2
25x + 9y 2 − 72y + 144 − 225 = 0
√
= 25 − 16
25x2 + 9y 2 − 72y − 81 = 0 (General Form)
√
= 9
=3
Section 3.4
Page 150
Section 3.4
Question 25
9x2 − 9x + 25y 2 − 50y = 197.75
3(x2 + 2x + 1) + (y 2 − 8y + 16) = 11 + 3 + 16
3(x + 1)2 + (y − 4)2 = 30
(x + 1)2
(y − 4)2
+
=1
10
30
(y − 4)2
(x + 1)2
√
+ √
=1
(1)
( 10)2
( 30)2
√
30, and b =
Question 27
9x2 + 25y 2 − 9x − 50y − 197.75 = 0
3x2 + y 2 + 6x − 8y − 11 = 0
3x2 + 6x + y 2 − 8y = 11
From (1), h = −1, k = 4, a =
Page 150
√
10.
9(x2 − x + 0.25) + 25(y 2 − 2y + 1) = 197.75 + 2.25 + 25
9(x − 0.5)2 + 25(y − 1)2 = 225
(x − 0.5)2
(y − 1)2
+
=1
25
9
(y − 1)2
(x − 0.5)2
+
=1
2
5
32
From (1), h = 0.5, k = 1, a = 5, and b = 3.
c2 = a2 − b2
√
c = ± 25 − 9
= ±4
c2 = a2 − b2
√
c = ± 30 − 10
√
= ±2 5
a) The centre is at (−1, 4).
√
b) The length of the major axis
√ is 2 30. The
length of the minor axis is√2 10.
c) The foci are at (−1, 4 ± 2 5).
a) The centre is at (0.5, 1).
b) The length of the major axis is 2(5) or 10. The
length of the minor axis is 2(3) or 6.
c) The foci are at (0.5 ± 4, 1) or (−3.5, 1) and
(4.5, 1).
Applications and Problem Solving
Section 3.4 Page 150
√
a) y = ± 16 − x2
120
Question 28
(1)
√
√
b) y = ± 16 − 2x2 and y = ± 16 − 3x2
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
c) y = ±
r
2x2
16 −
and y = ±
3
p
e) y = ±2
√
√
16 − x2
16 − x2
d) y = ±
and y = ±
2
3
r
3x2
16 −
4
p
16 − x2 and y = ±3
16 − x2
f ) If A > 1, the graph of Ax2 + y 2 = 16 is the
same height as, but narrower than, the graph
of x2 + y 2 = 16. If 0 < A < 1, the graph of
Ax2 + y 2 = 16 is the same height as, but wider
than, the graph of x2 + y 2 = 16. If C > 1, the
graph of x2 + Cy 2 = 16 is the same width as,
but shorter than, the graph of x2 + y 2 = 16.
If 0 < C < 1, the graph of x2 + Cy 2 = 16 is
the same width as, but taller than, the graph
of x2 + y 2 = 16.
g) y = ±4
h) x = ±4
i) In g), the graph gets wider and wider as A approaches 0, and eventually breaks into two horizontal lines.
Similarly in h), the graph gets taller and taller as C approaches 0, and eventually breaks into two vertical
lines.
Section 3.4
Page 151
Question 29
a) Let P(x, y) be a point on the ellipse F1 (c, 0) and F2 (−c, 0).
PF1 + PF2 = 2a
p
p
(x − c)2 + (y − 0)2 + (x − (−c))2 + (y − 0)2 = 2a
p
p
(x − c)2 + y 2 + (x + c)2 + y 2 = 2a
b) From a),
p
(x − c)2 + y 2 = 2a −
Square both sides.
p
(x + c)2 + y 2
(x − c)2 + y 2 = 4a2 − 4a
p
(x + c)2 + y 2 + (x + c)2 + y 2
p
x2 − 2cx + c2 = 4a2 − 4a (x + c)2 + y 2 + x2 + 2cx + c2
p
4a (x + c)2 + y 2 = 4a2 + 4cx
p
a (x + c)2 + y 2 = a2 + cx
a2 ((x + c)2 + y 2 ) = a4 + 2a2 cx + c2 x2
2 2
2
a x + 2a cx + a2 c2 + a2 y 2 = a4 + 2a2 cx + c2 x2
x2 (a2 − c2 ) + a2 y 2 = a2 (a2 − c2 )
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
121
x2 (a2 − c2 )
a2 y 2
a2 (a2 − c2 )
+ 2 2
= 2 2
2
2
2
2
a (a − c ) a (a − c )
a (a − c2 )
x2
y2
+ 2
=1
2
a
a − c2
c) Since a2 = b2 + c2 , b2 = a2 − c2 . From b),
x2
y2
+ 2
=1
2
a
a − c2
x2
y2
+
=1
a2
b2
Section 3.4
Page 151
Question 30
(x − h)2
(y − k)2
+
=1
2
a
b2
Multiply both sides by a2 b2 .
b2 (x − h)2 + a2 (y − k)2 = a2 b2
b2 x2 − 2b2 hx + b2 h2 + a2 y 2 − 2a2 ky + a2 k 2 = a2 b2
2 2
2 2
b x + a y − 2b2 hx − 2a2 ky + (b2 h2 + a2 k 2 − a2 b2 ) = 0
A = b2 and C = a2 , so A and C both have the same sign (positive). If a 6= b, then A 6= C.
Section 3.4
Page 151
Question 31
Answers may vary. The graphs of 3x2 + y 2 = 12 and
4x2 + y 2 = 12 appear below.
Section 3.4
a)
Page 151
Question 32
b) Circle: x-intercepts: ±1; y-intercepts: ±1
Ellipse: x-intercepts: ±4; y-intercepts: ±1
c) The given ellipse is a horizontal expansion by a
factor of 4 of the circle.
d) (x, y) → (4x, y).
e)
Circle: x-intercepts: ±1; y-intercepts: ±1
Ellipse: x-intercepts: ±1; y-intercepts: ±3
The given ellipse is a vertical expansion by a factor of
3 of the circle. (x, y) → (x, 3y).
122
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
Circle: x-intercepts: ±1; y-intercepts: ±1
f)
Ellipse: x-intercepts: ±3; y-intercepts: ±2
The given ellipse is a horizontal expansion by a factor
of 3 and a vertical expansion of factor 2 of the circle.
(x, y) → (3x, 2y).
g) (x, y) → (ax, by)
Section 3.4 Page 151 Question 33
The unit circle has been expanded horizontally by
a factor of 2 and expanded vertically by a factor
of 5.
x2
y2
+
=1
4
25
Section 3.4 Page 151 Question 34
The unit circle has been expanded horizontally by
a factor of 2, translated to the right 5 units and
upward 7 units.
Section 3.4 Page 151 Question 35
The unit circle has been expanded vertically by a
factor of 2 and translated to the left 4 units.
Section 3.4 Page 151 Question 36
The unit circle has been expanded horizontally by
a factor of 2, expanded vertically by a factor of 3,
translated to the left 3 units and upward 6 units.
(x + 4)2 +
(x − 5)2
+ (y − 7)2 = 1
4
y2
=1
4
(y − 6)2
(x + 3)2
+
=1
4
9
Section 3.4 Page 151 Question 38
The unit circle has been expanded horizontally by
a factor of 4, expanded vertically by a factor of
3
, translated to the right 5 units and downward 6
2
units.
(x − 5)2
4(y + 6)2
+
=1
16
9
Section 3.4 Page 151 Question 37
The unit circle has been expanded horizontally by
5
a factor of , expanded vertically by a factor of 2,
2
translated to the right 4 units and upward 2 units.
4(x − 6)2
(y − 2)2
+
=1
25
4
Section 3.4
Page 151
Question 39
x2
y2
From the information given, a = 6 and b = 5. The equation of the ellipse is
+
= 1. The equation of
36 25
r
x2
the semi-ellipse is y = 5 1 − .
36
The minimum clearance over the road is the y-value of the points where x = ±4.
r
42
y(4) = 5 1 −
36
r
36 − 16
=5
36
r
5
=5
9
.
= 3.727
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
123
The minimum clearance over the road is approximately 3.73 m. The height of the tallest person that can
walk down the middle of the sidewalk is the y-value of the points where x = ±5.
r
52
y(5) = 5 1 −
36
r
36 − 25
=5
36
r
11
=5
36
.
= 2.76
The height of the tallest person is approximately 2.76 m. No. That would be an extremely tall person.
Section 3.4 Page 152 Question 40
a) From the information given, a = 230 ÷ 2 or 115 m and b = 190 ÷ 2 or 95 m. The equation of the ellipse
x2
y2
+
= 1.
can be expressed as
1152
952
b) From the information given, a = 95 and b = 60. The equation of the semi-ellipse can be expressed as
x2
y2
+ 2 = 1; y ≥ 0.
2
95
60
c) The distance from the point on the stage to the roof is y + 30, where (40, y) is a point on the ellipse and
y > 0.
402
y2
+
=1
952
602
²
³
402
2
2
y = 60 1 − 2
95
s²
³
402
y = 60
1− 2
95
.
= 54.4
The height is approximately 54.4 + 30 or 84.4 m.
124
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
Section 3.4 Page 152 Question 41
Let b be the length of the semi-minor axis in Ellipse 1. Let c be the length of the semi-minor axis in Ellipse
3. Let d be the length of the semi-minor axis in Ellipse 4. The equation of Ellipse 1 is of the form
x2
y2
+
=1
(1)
b2
25
The semi-minor axis of Ellipse 2 is the semi-major axis of Ellipse 3.
The equation of Ellipse 2 is of the form
x2
y2
+ 2 =1
(2)
2
b
3
À
!
√
3 3
Substitute 2,
into (2) and solve for b.
2
° √ ±2
3 3
2
2
2
+
=1
4
b2
32
27
4
+ 4 =1
b2
9
2
3
4
+
=
1
b2
4
1
4
=
–4
–2
2
4
2
b
4
2
b = 16
–2
The equation of Ellipse 3 is of the form
x2
y2
+
=1
2
32
–4
À √
! c
4 2
Substitute
, 1 into Ellipse 3.
3
° √ ±2
4 2
3
12
+
=1
(3)
c2
32
32
1
+ =1
2
9c
9
8
32
=
2
9c
9
c2 = 4
The equation of Ellipse 4 is of the form
x2
y2
+
=1
c2
d2
x2
y2
+ 2 =1
4
d
√
Substitute ( 2, 1) into Ellipse 4.
√
( 2)2
12
(4)
+ 2 =1
4
d
1
1
+ 2 =1
2 d
1
1
=
2
d
2
d2 = 2
The equations of the ellipses are
x2
y2
x2
y2
x2
y2
x2
y2
+
= 1,
+
= 1, +
= 1, and
+
= 1.
16 25
16
9
4
9
4
2
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
125
Section 3.4 Page 152 Question 42
The width required for the four letters and spacing is
4(0.4) + 3(0.2) or 2.2 m. The height of the letters is 1.0 m.
Centring the name at (0, 0) places the coordinates of the
top right corner of the letters at P(1.1, 0.5). The name
will fit if P is in the interior of the ellipse.
P(1.1,0.5)
1.12 + 1.96(0.5)2 = 1.21 + 0.49
= 1.7
< 1.96
The name will fit.
Section 3.4 Page 152
6 F1 PQ= 6 F2 PR
Question 43
R
P
Q
F2
F1
Section 3.4 Page 152 Question 44
Let (x, y) be one point on the rectangle, where x > 0 and y > 0. Thus, the other points are (−x, y), (−x, −y),
and (x, −y). The dimensions of the rectangle are x − (−x) or 2x and y − (−y) or 2y. Since (x, y) is on the
x2
y2
ellipse
+
=1
25
9
r
x2
y =3 1−
25
3p
=
25 − x2
5
The area of the rectangle is given by
A = (2x)(2y)
² p
³
6
2
= 2x
25 − x
5
Square each side.
144 2
x (25 − x2 )
25
144 4
=−
(x − 25x2 )
25 À
² ³2 ² ³2 !
144
25
25
4
2
=−
x − 25x +
−
25
2
2
²
³2
144
25
=−
+ 900
x2 −
25
2
A2 =
Thus, A2 has a maximum of 900 when x2 =
area of the rectangle is 30 m2 .
126
√
25
. Thus, A has a maximum of 900 or 30 m2 . The maximum
2
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
Section 3.4 Page 152 Question 45
Let ²
P be the point ³
of intersection of the inscribed square and the upper semi-ellipse. The coordinates of P
bp 2
2
are x,
a − x . Since the square is symmetric about the origin, P also lies on the identity line y = x.
a
This property can be used to solve for x in terms of a and b.
bp 2
a − x2
a
b2
x2 = 2 (a2 − x2 )
²
³ a
2
b
x2 1 + 2 = b2
a
a2 b2
x2 = 2
a + b2
S
x=
b
P
a
(1)
R
Q
Result (1) can be interpreted as the area of the inscribed square lying in the first quadrant. The area of
4a2 b2
.
square PQRS is then 2
a + b2
Section 3.4 Page 152 Number Power
7 − 6 + 5 = 6
9 − 1 − 2 = 6
8 ÷ 4 × 3 = 6
3.5
The Hyperbola
Practice
Section 3.5 Page 159 Question 1
a) The centre is at (0, 0).
b) The transverse axis is horizontal and 4 units in length. The conjugate axis is vertical and 2 units in
length.
c) The vertices are located at (±2, 0).
1
d) The slopes of the asymptotes are ± .
2
Section 3.5 Page 160 Question 3
a) The centre is at (−3, 3).
b) The transverse axis is vertical and 8 units in length. The conjugate axis is horizontal and 2 units in
length.
c) The vertices are located at (−3, 7) and (−3, −1).
d) The slopes of the asymptotes are ±4.
Section 3.5 Page 160 Question 5
a) The centre is at (12, 2).
b) The transverse axis is horizontal and 12 units in length. The conjugate axis is vertical and 8 units in
length.
c) The vertices are located at (6, 2) and (18, 2).
2
d) The slopes of the asymptotes are ± .
3
Section 3.5 Page 160 Question 7
(y − 0)2
(x − 1)2
−
= 1.
The equation of the hyperbola can be expressed as
2
12
82
a) The centre is at (1, 0).
b) The transverse axis is horizontal and 2 × 12 or 24 units in length. The conjugate axis is vertical and 2 × 8
or 16 units in length.
c) The vertices are located at (1 − 12, 0) or (−11, 0) and (1 + 12, 0) or (13, 0).
2
d) The slopes of the asymptotes are ± .
3
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
127
Section 3.5
Page 160
Question 9
The equation of the hyperbola can be expressed as
(y + 4)2
x2
− 2 = 1.
2
7
10
a) The centre is at (0, −4).
b) The transverse axis is vertical and 2 × 7 or 14 units in length. The conjugate axis is horizontal and 2 × 10
or 20 units in length.
c) The vertices are located at (0, −4 − 7) or (0, −11) and (0, −4 + 7) or (0, 3).
7
d) The slopes of the asymptotes are ± .
10
Section 3.5 Page 160 Question 11
y2
(x − 3)2
= 1.
The equation of the hyperbola can be expressed as 2 −
14
82
a) The centre is at (3, 0).
b) The transverse axis is vertical and 2 × 14 or 28 units in length. The conjugate axis is horizontal and 2 × 8
or 16 units in length.
c) The vertices are located at (3, 0 − 14) or (3, −14) and (3, 0 + 14) or (3, 14).
7
d) The slopes of the asymptotes are ± .
4
Section 3.5 Page 160 Question 13
(x + 2)2
(y + 6)2
= 1.
−
The equation of the hyperbola can be expressed as
2
3
162
a) The centre is at (−2, −6).
b) The transverse axis is vertical and 2 × 3 or 6 units in length. The conjugate axis is horizontal and 2 × 16
or 32 units in length.
c) The vertices are located at (−2, −6 − 3) or (−2, −9) and (−2, −6 + 3) or (−2, −3).
3
d) The slopes of the asymptotes are ± .
16
Section 3.5 Page 160 Question 15
The centre is at (2, 3). The transverse axis is horizontal and 4 units in length. The conjugate axis is horizontal
and 6 units in length. The standard equation of the hyperbola is
(x − 2)2
(y − 3)2
=1
−
2
2
32
(x − 2)2
(y − 3)2
−
=1
4
9
Multiply both sides by 36 to determine the general equation.
9(x − 2)2 − 4(y − 3)2 = 36
9(x2 − 4x + 4) − 4(y 2 − 6y + 9) − 36 = 0
9x2 − 36x + 36 − 4y 2 + 24y − 36 − 36 = 0
9x2 − 4y 2 − 36x + 24y − 36 = 0
128
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
Section 3.5 Page 160 Question 17
The centre is at (2, 2). The transverse axis is vertical and 4 units in length. The conjugate axis is
horizontal and 4 units in length.
(x − 2)2
(y − 2)2
=1
−
2
2
22
Section 3.5 Page 161 Question 19
The centre is at (h, k) = (4, 0). The transverse
axis is horizontal and 4 units in length, so a = 2.
b
3
Since the slope of one asymptote is
= , then
a
2
b = 3.
(x − 4)2
(y − 0)2
=1
−
2
2
32
Standard form:
(y − 2)2
(x − 2)2
−
=1
4
4
Multiply both sides by 4.
(y − 2)2 − (x − 2)2 = 4
y 2 − 4y + 4 − x2 + 4x − 4 − 4 = 0
General form:
Section 3.5
Standard form:
(x − 4)2
y2
−
=1
4
9
Multiply both sides by 36.
9(x − 4)2 − 4y 2 = 36
2
9x − 72x + 144 − 4y 2 − 36 = 0
x2 − y 2 − 4x + 4y + 4 = 0
General form:
9x2 − 4y 2 − 72x + 108 = 0
Page 161
Section 3.5 Page 161 Question 23
The centre is at (h, k) = (6, −8). The transverse
axis is vertical and a = −8 − (−20) or 12. Since
a
12
the slope of one asymptote is = 3, then b =
b
3
or 4.
Question 21
The centre is at (h, k) = (1, 2). The transverse
axis is vertical and 4 units in length, so a = 2.
Since one focus is at (1, 15), then c = 15 − 2 or 13.
Since a2 + b2 = c2 , then b2 = 132 − 22 or 165.
(x − 1)2
(y − 2)2
−
=1
2
2
165
Standard form:
(y − 2)2
(x − 1)2
−
=1
4
165
Multiply both sides by 660.
165(y − 2)2 − 4(x − 1)2 = 660
2
165y − 660y + 660 − 4x2 + 8x − 4 − 660 = 0
General form:
4x2 − 165y 2 − 8x + 660y + 4 = 0
Section 3.5
Page 161
(x − 6)2
(y − (−8))2
−
=1
2
12
42
Standard form:
(y + 8)2
(x − 6)2
−
=1
144
16
Multiply both sides by 144.
(y + 8)2 − 9(x − 6)2 = 144
2
2
y + 16y + 64 − 9x + 108x − 324 − 144 = 0
General form:
9x2 − y 2 − 108x − 16y + 404 = 0
Question 25
(y + 2)2
(x − 1)2
−
=1
25
4
Multiply both sides by 100.
4(x − 1)2 − 25(y + 2)2 = 100
4(x − 1)2 − 100
(y + 2)2 =
r 25
4(x − 1)2 − 100
y+2=±
25
r
4(x − 1)2 − 100
y = −2 ±
25
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
129
2
The slopes of the asymptotes are ± and both pass through the centre (1, −2). Using the point-slope
5
equation for a line, the equations of the asymptotes are
y − y1 = m(x − x1 )
2
y − (−2) = − (x − 1)
5
2
2
y+2=− x+
5
5
2
8
y =− x−
5
5
y − y1 = m(x − x1 )
2
y − (−2) = (x − 1)
5
2
2
y+2= x−
5
5
12
2
y = x−
5
5
Use these equations to graph the hyperbola and its asymptotes, using a graphing calculator.
Section 3.5
Page 161
Question 27
(y − 2)2
(x − 2)2
−
=1
9
4
Multiply both sides by 36.
4(y − 2)2 − 9(x − 2)2 = 36
36 + 9(x − 2)2
(y − 2)2 =
r 4
36 + 9(x − 2)2
y−2=±
4
r
36 + 9(x − 2)2
y =2±
4
3
The slopes of the asymptotes are ± and both pass through the centre (2, 2). Using the point-slope equation
2
for a line, the equations of the asymptotes are
y − y1 = m(x − x1 )
3
y − 2 = − (x − 2)
2
3
y−2=− x+3
2
3
y =− x+5
2
y − y1 = m(x − x1 )
3
y − 2 = (x − 2)
2
3
y−2= x−3
2
3
y = x−1
2
Use these equations to graph the hyperbola and its asymptotes, using a graphing calculator.
Section 3.5
Page 161
Question 29
(x − 1)2
− 4y 2 = 1
25
Multiply both sides by 25.
(x − 1)2 − 100y 2 = 25
(x − 1)2 − 25
y2 =
r 100
(x − 1)2 − 25
y=±
100
130
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
1
1
and both pass through the centre (1, 0). Using the point-slope
The slopes of the asymptotes are ± 2 or ±
5
10
equation for a line, the equations of the asymptotes are
y − y1 = m(x − x1 )
1
y − 0 = − (x − 1)
10
1
1
y =− x+
10
10
y − y1 = m(x − x1 )
1
y−0=
(x − 1)
10
1
1
y=
x−
10
10
Use these equations to graph the hyperbola and its asymptotes, using a graphing calculator.
Section 3.5
Page 161
Question 31
4y 2 − 25(x − 2)2 = 100
4y 2 = 100 + 25(x − 2)2
1 + 25(x − 2)2
y2 =
r 4
100 + 25(x − 2)2
y=±
4
5
The slopes of the asymptotes are ± and both pass through the centre (2, 0). Using the point-slope equation
2
for a line, the equations of the asymptotes are
y − y1 = m(x − x1 )
5
y − 0 = − (x − 2)
2
5
y =− x+5
2
Section 3.5
Page 161
y − y1 = m(x − x1 )
5
y − 0 = (x − 2)
2
5
y = x−5
2
Question 33
x2 − 4y 2 − 6x − 8y − 11 = 0
(x2 − 6x) − 4(y 2 + 2y) = 11
(x2 − 6x + 9) − 4(y 2 + 2y + 1) = 11 + 9 − 4
(x − 3)2 − 4(y + 1)2 = 16
(y + 1)2
(x − 3)2
−
=1
16
4
(x − 3)2
(y + 1)2
=1
−
2
4
22
a) The centre is at (3, −1).
b) Since a = 4, the vertices are at (3 − 4, −1) or (−1, −1) and (3 + 4, −1) or (7, −1).
1
c) The slopes of the asymptotes are
and both pass through the centre (3, −1). Using the point-slope
2
equation for a line, the equations of the asymptotes are
y − y1 = m(x − x1 )
1
y − (−1) = − (x − 3)
2
2y + 2 = −x + 3
x + 2y − 1 = 0
y − y1 = m(x − x1 )
1
y − (−1) = (x − 3)
2
2y + 2 = x − 3
x − 2y − 5 = 0
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
131
Section 3.5
Page 161
Question 35
x2 − 10x − 36y 2 + 216y = 335
(x2 − 10x) − 36(y 2 − 6y) = 335
(x2 − 10x + 25) − 36(y 2 − 6y + 9) = 335 + 25 − 324
(x − 5)2 − 36(y − 3)2 = 36
(x − 5)2
(y − 3)2
−
=1
36
1
(x − 5)2
(y − 3)2
=1
−
2
6
12
a) The centre is at (5, 3).
b) Since a = 6, the vertices are at (5 − 6, 3) or (−1, 3) and (5 + 6, 3) or (11, 3).
1
c) The slopes of the asymptotes are
and both pass through the centre (5, 3). Using the point-slope
6
equation for a line, the equations of the asymptotes are
y − y1 = m(x − x1 )
1
y − 3 = (x − 5)
6
6y − 18 = x − 5
x − 6y + 13 = 0
y − y1 = m(x − x1 )
1
y − 3 = − (x − 5)
6
6y − 18 = −x + 5
x + 6y − 23 = 0
Applications and Problem Solving
Section 3.5 Page 161 Question 37
Let the foci be defined as F1 (0, c) and F2 (0, −c), and the general point on the hyperbola as P(x, y).
a)
For y > 0, PF2 > PF1 .
p
(y −
(−c))2
For y < 0, PF1 > PF2 .
p
(y −
c)2
p
0)2
c)2
PF2 − PF1 = 2a
+ (x − 0)2 = 2a
+ (x −
− (y −
p
p
2
(y + c) + x2 − (y − c)2 + x2 = 2a
0)2
p
(−c))2
(1)
PF1 − PF2 = 2a
+ (x − 0)2 = 2a
+ (x −
− (y −
p
p
(y − c)2 + x2 − (y + c)2 + x2 = 2a
b) From (1),
p
(y + c)2 + x2 = 2a +
Square both sides.
p
(y − c)2 + x2
(y + c)2 + x2 = 4a2 + 4a
p
(y − c)2 + x2 + (y − c)2 + x2
p
Square both sides.
y 2 + 2cy + c2 = 4a2 + 4a (y − c)2 + x2 + y 2 − 2cy + c2
p
a2 − cy = −a (y − c)2 + x2
a4 − 2a2 cy + c2 y 2 = a2 ((y − c)2 + x2 )
a4 − 2a2 cy + c2 y 2 = a2 y 2 − 2a2 cy + a2 c2 + a2 x2
(c2 − a2 )y 2 − a2 x2 = a2 (c2 − a2 )
Divide both sides by a2 (c2 − a2 ).
y2
x2
=1
− 2
2
a
c − a2
132
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
(2)
c)
a2 + b2 = c2
b2 = c2 − a2
Substitute (3) into (2).
y2
x2
− 2 =1
2
a
b
Section 3.5
Page 161
(3)
Question 38
(y − k)2
(x − h)2
−
=1
a2
b2
Multiply both sides by a2 b2 .
b2 (x − h)2 − a2 (y − k)2 = a2 b2
b2 (x2 − 2xh + h2 ) − a2 (y 2 − 2ky + k 2 ) − a2 b2 = 0
b2 x2 − 2b2 hx + b2 h2 − a2 y 2 + 2a2 ky − a2 k 2 − a2 b2 = 0
b2 x2 − a2 y 2 − 2b2 hx + 2a2 ky + (b2 h2 − a2 k 2 − a2 b2 ) = 0
where A = b2 , C = −a2 , D = −2b2 h, E = 2a2 k, and F = b2 h2 − a2 k 2 − a2 b2 .
Section 3.5 Page 161 Question 39
a) The coefficients of the x2 - and y 2 -terms are equal but have opposite signs.
b)
i) x2 − y 2 = 5
ii) x2 − y 2 = −4
iii) y 2 − x2 = 8
iv) y 2 − x2 = −2
c) The asymptotes are perpendicular.
Section 3.5
a)
Page 162
Question 40
b) The y-intercepts are ±1. There are no x-intercepts.
c) The second hyperbola can be defined as a horizontal
expansion by a factor of 3 of the rectangular hyperbola.
d) (x, y) → (3x, y)
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
133
e)
The y-intercepts are ±4. There are no x-intercepts.
The second hyperbola can be defined as a vertical
expansion by a factor of 4 of the rectangular hyperbola.
(x, y) → (x, 4y)
f)
The y-intercepts are ±3. There are no x-intercepts.
The second hyperbola can be defined as a vertical
expansion by a factor of 3 and a horizontal expansion by a factor of 2 of the rectangular hyperbola.
(x, y) → (2x, 3y)
g) The y-intercepts are ±a. There are no x-intercepts.
The second hyperbola can be defined as a vertical expansion by a factor of a and a horizontal expansion by
a factor of b of the rectangular hyperbola.
(x, y) → (bx, ay)
y2
x2
h) The first form can be expressed as 2 − 2 = 1. The transformation (x, y) → (ax, by) will map the first
1
1
form to the second.
Section 3.5
a)
Page 162
i) xy = 8 ⇐⇒ y =
8
x
Question 41
ii) xy = −5 ⇐⇒ y = −
5
x
iii) xy = 10 ⇔ y =
10
x
b) The asymptotes lie on the coordinate axes. The equations are y = 0 and x = 0.
c) Since the asymptotes are perpendicular, these are examples of rectangular hyperbolas.
Section 3.5 Page 162 Question 42
Since the sum of the focal distances is 2a = 6, then a = 3. Since the foci are located at (2, 7)
² and (2, −3),
³
7 − (−3)
7 + (−3)
c=
or 5. The centre (h, k) is the midpoint of the segment joining the foci, (h, k) = 2,
2
2
or (h, k) = (2, 2).
b2 = c2 − a2
= 52 − 32
= 25 − 9
= 16
The equation describing the path of the pod is
134
(y − 2)2
(x − 2)2
= 1.
−
9
16
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
For questions 43 to 45, the expansion factors are determined by comparing the slopes of the asymptotes with
slopes of ±1.
Section 3.5 Page 162 Question 43
Since the transverse axis is vertical, use the unit rectangular hyperbola y 2 − x2 = 1. The distance from the
centre at (h, k) = (−4, 3) to the vertices is 4 units. To obtain the image, perform a vertical expansion by a
factor of 4, followed by a translation to the left 4 units and upward 3 units. The equation of the image is
(x − h)2
(y − k)2
−
=1
a2
b2
(y − 3)2
(x − (−4))2
−
=1
2
4
12
(y − 3)2
− (x + 4)2 = 1
16
Section 3.5 Page 162 Question 44
Since the transverse axis is vertical, use the unit rectangular y 2 − x2 = 1. The distance from the centre at
(h, k) = (4, 6) to the vertices is 1 unit. To obtain the image, perform a horizontal expansion by a factor of
2, followed by a translation to the right 4 units and upward 6 units. The equation of the image is
(x − h)2
(y − k)2
=1
−
a2
b2
(y − 6)2
(x − 4)2
−
=1
2
1
22
(x − 4)2
(y − 6)2 −
=1
4
Section 3.5 Page 162 Question 45
Since the transverse axis is horizontal, use the unit rectangular x2 − y 2 = 1. The distance from the centre
at (h, k) = (4, −6) to the vertices is 1 unit. To obtain the image, perform a translation to the right 4 units
and downward 6 units. The equation of the image is
(y − k)2
(x − h)2
−
=1
2
a
b2
(y − (−6))2
(x − 4)2
−
=1
12
12
2
2
(x − 4) − (y + 6) = 1
Section 3.5 Page 162 Question 46
The symmetry of the four points requires that the hyperbola be centered at the origin. Substitute the points
x2
y2
(x, y) = (±m, ±n) into 2 − 2 = 1 to reveal the requirements for a and b. A possible solution is
a
b
(±m)2
(±n)2
−
=1
a2
b2
2
2
m
n
− 2 =1
2
a
b
Multiply both sides by a2 b2 .
b2 m2 − a2 n2 = a2 b2
a2 (b2 + n2 ) = b2 m2
b2 m2
a2 = 2
b + n2
The points (x, y) = (±m, ±n) all lie on a hyperbola with equation
x2
b2 m 2
b2 +n2
−
y2
= 1.
b2
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
135
Section 3.5 Page 162 Question 47
Let d be the distance, s be the speed, and t be the time taken.
a)
d = st
500 = st
500
s=
t
b) Using t = 4 h,
500
4
= 125
s=
A speed of 125 km/h is required.
Section 3.5 Page 163 Question 48
Let P be the pressure in Pascals and V be the volume in litres.
1
a)
P ∝
V
1
P =k×
V
Substitute (P, V ) = (500, 34.2).
1
500 = k ×
34.2
k = 17 100
The function is P =
b)
17 100
.
V
17 100
V
17 100
800 =
V
17 100
V =
800
= 21.375
P =
The volume is 21.375 L.
Section 3.5
Page 163
Question 49
4x2 − 25y 2 − 100 = 0
25y 2 = 4x2 − 100
4x2 − 100
y2 =
√25
4x2 − 100
y=±
5
2p 2
x − 25
=±
5
136
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
Changing −25 to −16 yields y = ±
1p 2
x − 25.
2
Changing −25 to −9 yields y = ±
2p 2
x − 25.
3
Section 3.5
Page 163 Question 50
(y − 1)2
(x − 2)2
From the definition
= 1, values of a = 5 and b = 3 emerge. The length of the transverse
−
52
32
axis is 2(5) or 10. The length of the conjugate axis is 2(3) or 6. Since the two hyperbolas share the same
asymptotes, the lengths of the axes are reversed. The length of the transverse axis is 6 units and the length
of the conjugate axis is 10.
Section 3.5 Page 163
Answers may vary.
Question 51
Section 3.5
Page 163 Question 52
x2
A base model of y 2 − 2 = 1 can be used. The requirements of the arch would include the point P(2, −2.2).
b
Determine b2 .
a)
x2
b2
22
(−2.2)2 − 2
b
4
4.84 − 2
b
4
b2
y2 −
y
=1
x
=1
-1
Q(1.2, y )
=1
1.2
= 4.84 − 1
4
3.84
1
=
0.96
2
b2 =
One possible equation is y 2 −
x2
1
0.96
P(2,-2.2)
2
= 1 or y 2 − 0.96x2 = 1.
b) Define the point Q(1.2, y) where y < 0. Determine y.
y 2 − 0.96x2 = 1
Substitute x = 1.2 into (1).
(1)
y 2 − 0.96(1.2)2 = 1
y 2 − 1.3824 = 1
y 2 = 2.3824
.
y = ±1.54
Using y = −1.54, the height of the arch is −1 − (−1.54) or
approximately 0.54 m.
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
137
Section 3.5 Page 163 Question 53
a) Let (u, v) be the coordinates of the image of Hyperbola I under the mapping.
²
x
y
x
y
(x, y) → √ − √ , √ + √
2
2
2
2
x
y
u= √ −√
2
2
x
y
v=√ +√
2
2
Add (1) and (2) and solve for x.
2x
u+v = √
2
u+v
x= √
2
Subtract (1) from (2) and solve for y.
2y
v−u= √
2
v−u
y= √
2
³
= (u, v)
(1)
(2)
(3)
(4)
Substitute (3) and (4) into the equation of Hyperbola I.
³2 ²
³2
v−u
u+v
√
√
−
=1
2
2
(u + v)2
(v − u)2
−
=1
2
2
(u + v)2 − (v − u)2 = 2
(u + v + v − u)(u + v − v + u) = 2
(2v)(2u) = 2
²
2uv = 1
2uv − 1 = 0
Replace u and v with x and y.
2xy − 1 = 0
(5)
The general equation of Hyperbola II is 2xy − 1 = 0. The length ²of the transverse
³
²axis is 2 units.
³ The
1
1
1
1
length of the conjugate axis is 2 units. The vertices are located at √ , √
and − √ , − √ . The
2
2
2
2
value of B is 2.
b) Substitute (3) and (4) into (5) to determine the equation of Hyperbola III.
2
²
³²
³
u+v
v−u
√
√
−1=0
2
2
(u + v)(v − u) − 1 = 0
uv − u2 + v 2 − uv − 1 = 0
v 2 − u2 − 1 = 0
Replace u and v with x and y.
y 2 − x2 − 1 = 0
Express the equation of Hyperbola III in general form.
x2 − y 2 + 1 = 0
The length of the transverse axis is 2 units. The length of the conjugate axis is 2 units. The vertices are
located at (0, ±1). The value of B is 0.
138
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
c) Substitute (3) and (4) into x2 + y 2 = 1 to determine the equation of the image of the unit circle centred
at the origin.
³2 ²
³2
u+v
v−u
√
√
+
2
2
2
(u + v) + (v − u)2
2
u + 2uv + v 2 + v 2 − 2uv + u2
2u2 + 2v 2
²
2
=1
=2
=2
=2
2
u +v =1
Replace u and v with x and y.
x2 + y 2 = 1
The image under this transformation is the same as the original graph.
d) The transformation is rotating the points 45◦ counterclockwise around the origin.
e) Applying the transformation three times successively would put the vertices on the line y = −x. The
equation of the image would be 2xy + 1 = 0. Applying the transformation eight times successively would
leave the hyperbola unchanged. The equation of the image would remain x2 − y 2 = 1.
3.6
The Parabola
Practice
Section 3.6 Page 167 Question 1
The y-coordinate of the focus is 2 and the directrix
is y = −4.
2p = 2 − (−4)
=6
p=3
The vertex is at (h, k) = (0, 2 − 3) or (0, −1).
(x − h)2 = 4p(y − k)
(x − 0)2 = 4(3)(y − (−1))
x2 = 12(y + 1)
Section 3.6 Page 167 Question 5
The y-coordinate of the focus is 2 and the directrix
is y = −1.
2p = 2 − (−1)
=3
3
p=
2
²
³
²
³
3
1
The vertex is at (h, k) = 2, 2 −
or 2,
.
2
2
² ³²
³
3
1
(x − 2) = 4
y−
2
2
²
³
1
(x − 2)2 = 6 y −
2
Section 3.6 Page 167 Question 3
The x-coordinate of the focus is 4 and the directrix
is x = 1.
2p = 4 − 1
=3
3
p=
2
²
³
²
³
3
5
,0 .
The vertex is at (h, k) = 4 − , 0 or
2
2
² ³²
³
3
5
x−
2
2
²
³
5
y2 = 6 x −
2
(y − 0)2 = 4
Section 3.6 Page 167 Question 7
The x-coordinate of the focus is −5 and the directrix is x = 5.
2p = −5 − 5
= −10
p = −5
The vertex is at (h, k) = (−5 + 5, −1) or (0, −1).
(y − (−1))2 = 4(−5)(x − 0)
(y + 1)2 = −20x
2
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
139
Section 3.6 Page 168 Question 9
The vertex is at (h, k) = (4, 2) and p = −3.
(y − 2)2 = 4(−3)(x − 4)
Standard form:
(y − 2)2 = −12(x − 4)
y 2 − 4y + 4 = −12x + 48
General form:
Section 3.6 Page 168 Question 11
The vertex is at (h, k) = (0, 0) and p = −3.
(y − 0)2 = 4(−3)(x − 0)
Standard form:
y 2 = −12x
General form:
y 2 + 12x = 0
y 2 + 12x − 4y − 44 = 0
Section 3.6 Page 168 Question 13
The vertex is at (h, k) = (2, −6 + 2) or (2, −4) and
p = −2.
Section 3.6
Page 168
Question 15
y 2 − 12x = 0
(y − 0)2 = 4(3)(x − 0)
2
(x − 2) = 4(−2)(y − (−4))
Standard form:
Since p = 3 and the vertex is at (h, k) = (0, 0), the
focus is at (3, 0). The equation of the directrix is
x = 0 − 3 or x = −3. The equation of the axis
of symmetry is y = 0. The parabola opens to the
right.
(x − 2)2 = −8(y + 4)
x − 4x + 4 = −8y − 32
2
General form:
x2 − 4x + 8y + 36 = 0
Section 3.6
Page 168
Question 17
x2 + 4x = −2y − 10
x + 4x + 4 = −2y − 10 + 4
(x + 2)2 = −2(y + 3)
²
³
1
(y + 3)
(x + 2)2 = 4 −
2
2
²
³
²
³
1
1
7
Since p = − and the vertex is at (h, k) = (−2, −3), the focus is at −2, −3 −
or −2, − . The
2
2
2
1
5
equation of the directrix is y = −3 + or y = − . The equation of the axis of symmetry is x = −2. The
2
2
parabola opens downward.
Section 3.6 Page 168 Question 19
Since A and C are different but have the same sign, the conic is an ellipse.
Section 3.6 Page 168 Question 21
Since A and C are different and have opposite signs, the conic is a hyperbola.
Section 3.6 Page 168 Question 23
Since A is 0, the conic is a parabola.
140
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
Section 3.6 Page 168 Question 25
The vertex is at (−1, −2) and p = 2 − (−1) or 3.
The axis of symmetry is horizontal, so the model
(y − k)2 = 4p(x − h) is used.
(y − (−2))2 = 4(3)(x − (−1))
Standard form:
(y + 2)2 = 12(x + 1)
y 2 + 4y + 4 = 12x + 12
General form:
y 2 − 12x + 4y − 8 = 0
Section 3.6 Page 168 Question 27
The centre is at (2, −4), a = −2 − (−4) or 2, and
b = 1. The transverse axis is vertical so the model
(x − h)2
(y − k)2
−
= 1 is used.
a2
b2
(y − (−4))2
(x − 2)2
=1
−
2
2
12
Standard form:
(y + 4)2
− (x − 2)2 = 1
4
Multiply both sides by 4.
(y + 4)2 − 4(x − 2)2 = 4
y 2 + 8y + 16 − 4x2 + 16x − 16 − 4 = 0
General form:
4x2 − y 2 − 16x − 8y + 4 = 0
Section 3.6
Page 168 Question 29
²
²
³
³
3
3
The vertex is at −3, −
and p = −2 − −
2
2
1
or − . The axis of symmetry is vertical, so the
2
model (x − h)2 = 4p(y − k) is used.
²
³²
²
³³
1
3
(x − (−3))2 = 4 −
y− −
2
2
Standard form:
²
³
3
2
(x + 3) = −2 y +
2
2
x + 6x + 9 = −2y − 3
General form:
x2 + 6x + 2y + 12 = 0
Section 3.6
Page 168 Question 31
²
³
3
The centre is at −4, − , a = −4 − (−7) or 3,
2
1
and b = . The transverse axis is horizontal, so
2
(y − k)2
(x − h)2
= 1 is used.
the model
−
2
a
b2
¡
(y − − 32 )2
(x − (−4))2
−
=1
1 ¡2
32
2
Standard form:
¡2
y + 23
(x + 4)2
−
=1
1
9
4
Multiply both sides by 9.
²
³2
3
=9
(x + 4)2 − 36 y +
2
²
³
9
x2 + 8x + 16 − 36 y 2 + 3y +
=9
4
x2 + 8x + 16 − 36y 2 − 108y − 81 − 9 = 0
General form:
x2 − 36y 2 + 8x − 108y − 74 = 0
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
141
Section 3.6
Page 168
Question 33
x2 + y 2 + 10y + 9 = 0
x2 + (y 2 + 10y) = −9
Section 3.6
Page 168
9x2 − 16y 2 = −1
Multiply both sides by −1.
16y 2 − 9x2 = 1
y2
x2
−
1
1 =1
x2 + (y 2 + 10y + 25) = −9 + 25
x2 + (y + 5)2 = 16
x2
(y + 5)2
+
=1
16
16
16
12
10
8
y6
4
2
–12
0
–2
–4
–6
–8
–10
–12
–8 –6 –4
Section 3.6
Page 168
Question 35
9
1.5
1
y
0.5
2 4 6 8 10 12
x
–2
0
–1
1
x
2
–0.5
–1
–1.5
Question 37
y 2 − 2x − 8y + 22 = 0
10
8
y 2 − 8y = 2x − 22
y 2 − 8y + 16 = 2x − 22 + 16
(y − 4)2 = 2(x − 3)
² ³
1
(x − 3)
(y − 4)2 = 4
2
6
y
4
2
0
2
4
6
x
8
10
12
–2
Section 3.6
Page 168
Question 39
8
2
3x + 24x + 2y + 54 = 0
3x2 + 24x = −2y − 54
3(x2 + 8x) = −2y − 54
3(x2 + 8x + 16) = −2y − 54 + 48
3(x + 4)2 = −2y − 6
3(x + 4)2 = −2(y + 3)
2
(x + 4)2 = − (y + 3)
3
²
³
1
2
(x + 4) = 4 −
(y + 3)
6
142
MATHPOWERTM12, Western Edition, Solutions
6
4y
2
–12 –10 –8
–6
x
–4
–2
0 2
–2
–4
–6
–8
Chapter 3
4
Section 3.6
2
Page 168
Question 41
6
2
4x + 25y − 24x + 200y + 336 = 0
y
4x2 − 24x + 25y 2 + 200y = −336
4
2
4(x2 − 6x) + 25(y 2 + 8y) = −336
4(x2 − 6x + 9) + 25(y 2 + 8y + 16) = −336 + 36 + 400
4(x − 3)2 + 25(y + 4)2 = 100
–2 0
2
4
x6
8
10
–2
(x − 3)2
(y + 4)2
+
=1
25
4
–4
–6
Section 3.6
Page 168
Question 43
6
2
2
64x + 9y − 384x − 36y + 468 = 0
64x2 − 384x + 9y 2 − 36y = −468
2
4
y
2
64(x − 6x) + 9(y − 4y) = −468
2
2
64(x − 6x + 9) + 9(y 2 − 4y + 4) = −468 + 576 + 36
64(x − 3)2 + 9(y − 2)2 = 144
(x − 3)2
9
4
+
(y − 2)2
=1
16
–2
2
4
x
6
8
–2
Applications and Problem Solving
Section 3.6 Page 169 Question 44
a) Let the focus be F(h, k + p). Let the general point on the parabola be P(x, y). Let D(x, k − p) be the
point on the directrix in vertical alignment with P.
Square both sides.
p
PD = PF
p
(x − x)2 + (y − (k − p))2 = (x − h)2 + (y − (k + p))2
(y − (k − p))2 = (x − h)2 + (y − (k + p))2
y − 2y(k − p) + (k − p)2 = (x − h)2 + y 2 − 2y(k + p) + (k + p)2
2
b)
y 2 − 2y(k − p) + (k − p)2 = (x − h)2 + y 2 − 2y(k + p) + (k + p)2
−2y(k − p) + (k − p)2 = (x − h)2 − 2y(k + p) + (k + p)2
−2ky + 2py + k 2 − 2pk + p2 = (x − h)2 − 2ky − 2py + k 2 + 2pk + p2
4py − 4pk = (x − h)2
4p(y − k) = (x − h)2
Section 3.6
Page 169
Question 45
4p(y − k) = (x − h)2
4py − 4pk = x2 − 2hx + h2
x2 − 2hx − 4py + (h2 + 4pk) = 0
where A = 1, D = −2h, E = −4p, and F = h2 + 4pk.
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
143
Section 3.6
a)
Page 169
Question 46
b) To score the field goal, y > 3 at x = 38.
40
10y = −382 + 39(38)
10y = 38(−38 + 39)
10y = 38
30
y = 3.8
y20
Lui will score the field goal.
c) Determine y for x = 41.15.
10
10y = −41.152 + 39(41.15)
10
20
x
30
10y = −88.4725
y = −8.847 25
40
Lui would not have scored the field goal.
Section 3.6
Page 169
Question 47
a) Express the conic in terms of y.
x2 − 115x + 95y = 0
95y = −x2 + 115x
23
1
y = − x2 + x
95
19
A TRACE reveals that y > 2 when x = 110,
so the ball clears the wall.
b) The height of the ball at the wall is approximately 5.8 m. A player could not catch it.
Section 3.6 Page 169 Question 48
a) Place the lowest²point of the
The coordinates of the point representing the top of the
³ cable
² at the origin.
³
472 111
111
right tower are
,
or 236,
. Determine p.
2
2
2
x2 = 4py
²
³
111
2362 = 4p
2
55 696 = 222p
.
p = 250.88
The equation of the parabola is x2 = 4(250.88)y or approximately x2 = 1004y.
b) Determine y for the ordered pair (30, y).
x2 = 1004y
302 = 1004y
900
y=
1004
.
= 0.896
The length of the cable is approximately 0.90 m.
c) The bridge deck is not likely attached to the cable, but rather suspended from it. The equation defining
the cable should be translated vertically by an amount equal to the length of suspension required at the
lowest point of the cable.
144
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
Section 3.6 Page 169 Question 49
a) Place the centre of the hyperbola at the origin and
the transverse axis along the y-axis. Using a = 1
x2
y2
and the model
− 2 = 1, substitute the point
1
b
P(30, −21) to determine b.
y
x
-1
(−21)2
302
− 2 =1
1
b
900
441 − 2 = 1
b
900
= 440
b2
900
b2 =
440
45
=
22
Q(25, y )
30
One possible hyperbolic equation for the roof arches is y 2 −
b) Determine the y-coordinate for the point Q(25, y).
y2 −
30
P(30,-21)
22x2
= 1.
45
22(25)2
=1
45
13 750
y2 = 1 +
45
.
y = ±17.51
The height of the hyperbolic roof arch at 25 m is approximately −17.51 − (−21) or 3.49 m.
c) Place the vertex of the parabola at the origin and d) Place the centre of the ellipse at the origin and the
the axis of symmetry on the y-axis. Using the
transverse axis on the x-axis. Using a = 30 and
x2
y2
model x2 = 4py, substitute the point P(30, −20)
b = 20 gives the equation,
+
= 1.
to determine p.
900 400
302 = 4p(−20)
900 = −80p
p = −11.25
The equation of the parabola is x2 = 4(−11.25)y
or x2 = 45y. Determine the y-coordinate of the
point Q(25, y).
252 = −45y
625 = −45y
625
y=−
45
.
= −13.9
252
y2
+
=1
900 400
25
y2
+
=1
36 400
²
³
25
y = 400 1 −
36
r
11
y = ±20
36
.
= ±11.1
2
The height of the elliptical roof arch at 25 m is
approximately 11.1 m.
The height of the parabolic roof arch at 25 m is
approximately −13.9 − (−20) or 6.1 m.
e) Answers may vary.
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
145
3.7
Work Backward
Practice
Section 3.7
Page 173
Question 1
Let p be the original retail price of the jacket. The 35% sale discount reduces the cost to (1 − 0.35)%p or
(0.65)p. The 10% employee discount further reduces the cost to (1 − 0.1)(0.65)p or (0.9)(0.65)p. After the
GST is applied the cost increases to (1 + 0.07)(0.9)(0.65)p or (1.07)(0.9)(0.65)p.
(1.07)(0.9)(0.65)p = 203.43
203.43
p=
(1.07)(0.9)(0.65)
.
= 324.99
The original retail price of the jacket was $324.99.
Section 3.7 Page 173 Question 2
Find the losers of the game in reverse order.
End of Game
3
2
1
Start
Player A
$24
24 + 12 + 12 = $48
$24
$12
Player B
$24
$12
12 + 24 + 6 = $42
$21
Player C
$24
$12
$6
6 + 12 + 21 = $39
The loser of the first game (Player C) started with $39. The loser of the second game (Player B) started
with $21. The loser of the third game (Player A) started with $12.
Section 3.7 Page 173 Question 3
After giving his cousin Stephan part of the collection Mario must have had 2(74 + 1) or 150 comic books.
After giving Ella her share, Mario must have had 2(150 + 1) or 302 comic books. Mario’s original collection
numbered 2(302 + 1) or 606 comic books.
Section 3.7 Page 173 Question 4
The sum of the original set of numbers was 100(38) or 3800. After discarding the numbers 45 and 55, the
3700
74
remaining 98 numbers total 3800 − (45 + 55) or 3700. The resulting arithmetic mean is
or 37 .
98
98
Section 3.7 Page 173 Question 5
Let x2 represent the population in 1996, where x is an integer. Thus, the population in 1998 is x2 + 100,
which is 1 more than a perfect square. So, if y is another positive integer,
x2 + 100 = y 2 + 1
y 2 − x2 = 99
(y − x)(y + x) = 99
(1)
One solution for (1) could be
y−x=1
y + x = 99
which gives y = 50 and x = 49. The population in 2000 is x2 + 200. Test x = 49.
492 + 200 = 2601
= 512
Thus, the population in 1996 was 492 or 2401.
146
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
Section 3.7 Page 173 Question 6
Let x and y be the two numbers.
x+y =4
(1)
xy = −60
(2)
Square both sides of (1).
(x + y)2 = 42
x2 + 2xy + y 2 = 16
(3)
Substitute (2) into (3).
x2 + 2(−60) + y 2 = 16
x2 + y 2 = 136
(4)
Determine the required result.
1
1
y 2 + x2
+
=
x2
y2
x2 y 2
2
x + y2
=
(xy)2
(5)
Substitute (2) and (4) into (5).
136
(−60)2
136
=
3600
17
=
450
=
Section 3.7 Page 173 Question 7
Lena will play a total of 12 + 3 + 9 or 24 games. She has already won 12 games, which is 50% of 24. She
doesn’t have to win any more games.
Section 3.7
Page 173
Question 8
The adjacent Venn diagram can developed by interpreting the given information in the order presented.
English
Biology
a) The total of the entries in the diagram is 45. Since
50 students were polled, 50 − 45 or 5 students are
not taking any of English, biology, and math.
b) There are 4 students (circled) that can be seen to
be taking biology and math but not English.
5
8
7
4
11
4
6
Mathematics
Section 3.7 Page 173 Question 9
Answers may vary. A possible 6-step solution is FLOUR, FLOOR, FLOOD, BLOOD, BROOD, BROAD,
and BREAD.
Section 3.7
Page 173
Question 10
1 000 000 000 = 109
= (2 × 5)9
= 29 × 59
= 512 × 1 953 125
The two whole numbers are 512 and 1 953 125.
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
147
Review
Section Review
Page 176
Question 1
Section Review
Page 176
Question 2
Page 176
Question 4
a) hyperbola
b) circle
c) ellipse
Section Review
Page 176
Question 3
Section Review
(x − h)2 + (y − k)2 = r2
√
(x − (−1))2 + (y − (−3))2 = ( 7)2
(x − h)2 + (y − k)2 = r2
(x − 2)2 + (y − (−6))2 = 42
(x − 2)2 + (y + 6)2 = 16
Section Review
Page 176
(x + 1)2 + (y + 3)2 = 7
Question 5
(x − h)2 + (y − k)2 = r2
°p
±2
(8 − 5)2 + (0 − (−2))2
(x − 5)2 + (y − (−2))2 =
(x − 5)2 + (y + 2)2 = 32 + 22
Standard form:
(x − 5)2 + (y + 2)2 = 13
x − 10x + 25 + y 2 + 4y + 4 = 13
2
General form:
x2 + y 2 − 10x + 4y + 16 = 0
Section Review
Page 176
Question 6
(x − h)2 + (y − k)2 = r2
°p
±2
(−2 − (−5))2 + (2 − 6)2
(x − (−5))2 + (y − 6)2 =
(x + 5)2 + (y − 6)2 = 32 + (−4)2
Standard form:
(x + 5)2 + (y − 6)2 = 25
x2 + 10x + 25 + y 2 − 12y + 36 = 25
General form:
x2 + y 2 + 10x − 12y + 36 = 0
Section Review
Page 176
Question 7
²
³
2 + 2 −1 + 5
The centre of the circle is at (h, k) =
,
or (2, 2). Since the endpoints have the same x2
2
coordinates, the length of the diameter is 5 − (−1) or 6. The radius is 3.
(x − h)2 + (y − k)2 = r2
(x − 2)2 + (y − 2)2 = 32
Standard form:
(x − 2)2 + (y − 2)2 = 9
x2 − 4x + 4 + y 2 − 4y + 4 − 9 = 0
General form:
x2 + y 2 − 4x − 4y − 1 = 0
148
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
Section Review
Page 176
Question 8
²
³
−4 + 2 12 + 0
The centre of the circle is at (h, k) =
,
or (−1, 6).
2
2
(x − h)2 + (y − k)2 = r2
² ³2
6
2
2
(x − (−1)) + (y − 6) =
2
(x + 1)2 + (y − 6)2 = 32 + (−6)2
Standard form:
(x + 1)2 + (y − 6)2 = 45
x2 + 2x + 1 + y 2 − 12y + 36 − 45 = 0
General form:
x2 + y 2 + 2x − 12y − 8 = 0
Section Review
Page 176
Question 9
x2 + y 2 + 9x − 8y + 4 = 0
(x2 + 9x) + (y 2 − 8y) = −4
²
³
81
81
x2 + 9x +
+ (y 2 − 8y + 16) = −4 +
+ 16
4
4
²
³2
9
−16 + 81 + 64
+ (y − 4)2 =
x+
2
4
²
³2
9
129
+ (y − 4)2 =
x+
2
4
À√
!2
²
³2
9
129
x+
+ (y − 4)2 =
2
2
√
²
³
9
129
The centre of the circle is at − , 4 and the radius is
.
2
2
Section Review
Page 176
Question 10
x2 + y 2 + 4x − 8 = 0
(x2 + 4x) + y 2 = 8
(x2 + 4x + 4) + (y − 0)2 = 8 + 4
°√ ±2
(x + 2)2 + (y − 0)2 =
12
The centre of the circle is at (−2, 0) and the radius is
√
√
12 or 2 3.
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
149
Section Review
Page 176
Question 11
(x + 3)2 + (y − 1)2 = 10
Section Review
Page 176
Question 12
x2 + y 2 − 10x + 4y + 17 = 0
2
2
(y − 1) = 10 − (x + 3)
p
y − 1 = ± 10 − (x + 3)2
p
y = 1 ± 10 − (x + 3)2
x2 − 10x + 25 + y 2 + 4y + 4 = 12
°√ ±2
(x − 5)2 + (y + 2)2 =
12
p
y = −2 ± 12 − (x − 5)2
Questions 13 and 14 use the generalization from Section 3.3, Question 50.
Section Review Page 176 Question 14
Question 13
2
2
The equation of the image is (x + 1) + (y − 3) = 16. The equation of the image is (x − 4)2 + (y + 6)2 = 16.
Section Review
Page 176
Section Review
Page 176
Question 15
(x − h)2 + (y − k)2 = r2
(x − (−2))2 + (y − 5)2 = 1202
(x + 2)2 + (y − 5)2 = 14 400
The earthquake epicentre could be located anywhere on the circle defined by (x + 2)2 + (y − 5)2 = 14 400.
Section Review
Page 176
Question 16
(x − h)2
(y − k)2
The standard form for a circle and ellipse is
+
= 1. The equation will define a circle if
2
a
b2
a = b.
Section Review Page 176 Question 17
The centre of the ellipse is at (6, −2). The length
of the major axis is 11 − 1 or 10. The length of
the minor axis is 1 − (−5) or 6.
Section Review
Page 176 Question 18
The centre of the ellipse is at (−2, −4). The length
of the major axis is 2 − (−10) or 12. The length
of the minor axis is 3 − (−7) or 10.
c2 = a2 − b2
² ³2 ² ³2
10
6
2
c =
−
2
2
c2 = a2 − b2
² ³2 ² ³2
12
10
c2 =
−
2
2
c2 = 25 − 9
= 16
c2 = 36 − 25
= 11
√
c = ± 11
c = ±4
The foci are at (6 ± 4, −2) or (10, −2) and (2, −2).
150
The foci are at √
(−2, −4 ±
and (−2, −4 − 11).
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
√
11) or (−2, −4 +
√
11)
Section Review Page 176 Question 19
The centre of the ellipse is at (2, −7). The length
of the major axis is 2(4) or 8. The length of the
minor axis is 2(1) or 2.
c2 = a2 − b2
² ³2 ² ³2
8
2
2
c =
−
2
2
Section Review
Page 176
Question 20
4x2 + 9y 2 = 36
x2
y2
+
=1
9
4
(y + 0)2
(x + 0)2
+
=1
32
22
The centre of the ellipse is at (0, 0). The length
of the major axis is 2(3) or 6. The length of the
minor axis is 2(2) or 4.
c2 = 16 − 1
= 15
√
c = ± 15
c2 = a2 − b2
√
√
The foci √
are at (2, −7 ± 15) or (2, −7 + 15) and
(2, −7 − 15).
c2 = 9 − 4
=5
√
c=± 5
√
The foci are at (0, ± 5).
Section Review Page 176 Question 21
Given c = 12 and b = 5 a value for a can be
determined.
a2 = b2 + c2
a2 = 52 + 122
= 169
a = ±13
The centre of the ellipse is at (h, k) = (0, 0).
(x − h)2
(y − k)2
+
=1
a2
b2
(y − 0)2
(x − 0)2
+
=1
2
13
52
Standard form:
x2
y2
+
=1
169 25
Multiply both sides by 4225.
25x2 + 169y 2 = 4225
General form:
25x2 + 169y 2 − 4225 = 0
Section Review Page 176 Question 22
The length of the semi-major axis is −1 − (−3) or
2. The length of the semi-minor axis is 2 − 1 or 1.
(y − k)2
(x − h)2
+
=1
2
a
b2
(x − 1)2
(y − (−3))2
+
=1
12
22
Standard form:
(y + 3)2
(x − 1)2 +
=1
4
Multiply both sides by 4.
4(x − 1)2 + (y + 3)2 = 4
4(x2 − 2x + 1) + y 2 + 6y + 9 = 4
4x2 − 8x + 4 + y 2 + 6y + 9 = 4
General form:
4x2 + y 2 − 8x + 6y + 9 = 0
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
151
Section Review
Page 176
Question 23
Section Review
4x2 + y 2 + 16x = 0
4(x2 + 4x) + y 2 = 0
3(x2 + 2x) + (y 2 + 4y) = 9
3(x2 + 2x + 1) + (y 2 + 4y + 4) = 9 + 3 + 4
3(x + 1)2 + (y + 2)2 = 16
Divide both sides by 16.
(x + 1)2
16
3
The centre of the ellipse is at (−2, 0). The length
of the major axis is 2(4) or 8. The length of the
minor axis is 2(2) or 4.
c2 = 16 − 4
= 12
√
c = ±2 3
√
The foci are at (−2, ±2 3).
Question 24
3x2 + y 2 + 6x + 4y − 9 = 0
4(x2 + 4x + 4) + y 2 = 0 + 16
4(x + 2)2 + y 2 = 16
Divide both sides by 16.
(x + 2)2
y2
+
=1
4
16
(y − 0)2
(x + 2)2
+
=1
22
42
c2 = a2 − b2
Page 176
+
y2
=1
16
(y + 2)2
(x + 1)2
=1
° √ ±2 +
42
4 3
3
The centre of the ellipse is at (−1, −2). The length
of the major axis
is 2(4)
8. The length of the
À √
! or √
8 3
4 3
or
.
minor axis is 2
3
3
c2 = a2 − b2
16
c2 = 16 −
3
32
=
3 √
√
4 2
3
c=± √ ×√
3
3
√
4 6
=
3
À
√ !
4 6
.
The foci are at −1, −2 ±
3
Section Review Page 176 Question 25
The new centre is at (−3, 0). The equation of the image is (x + 3)2 + 4y 2 = 9.
Section Review Page 176 Question 26
The new centre is at (6, 0). The equation of the image is (x − 6)2 + 4y 2 = 9.
Section Review Page 176 Question 27
Let (u, v) be the coordinates of the image of the transformation as defined by (x, y) → (5x, y) = (u, v).
u = 5x
(1)
v=y
(2)
From (1):
u
x=
(3)
5
From (2):
y=v
(4)
Substitute (3) and (4) into the equation of the ellipse.
° u ±2
+ 4v 2 = 9
5
u2
+ 4v 2 = 9
25
Multiply (5) by 25.
u2 + 100v 2 = 225
Replace (u, v) with (x, y).
(5)
x2 + 100y 2 = 225
152
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
Section Review Page 176 Question 28
The length of the major axis is 950 + 2(6337) + 230 or 13 854 km. The length of the semi-major axis is
a = 13 854 ÷ 2 or 6927 km. The position of the focus is c = 6927 − 6337 − 230 or 360 km.
y
b
b2 = a2 − c2
= 69272 − 3602
= 47 983 329 − 129 600
= 47 853 729
The equation defining the path of the satellite is
950 km
230 km
a
x
c
x2
y2
+
=1
a2
b2
2
2
x
y
+
=1
47 983 329 47 853 729
6337km
Section Review Page 177 Question 29
The standard form of an ellipse is a sum of squares yielding 1. The standard form of a hyperbola is a
difference of squares yielding 1.
Section Review
Page 177
Question 30
Section Review
Page 177
Question 31
16(x + 5)2 − 25(y + 3)3 = 400
y2
x2
−
=1
169 121
(x − 0)2
(y − 0)2
=1
−
132
112
(x + 5)2
(y + 3)2
−
=1
52
42
a) The centre is at (−5, −3).
a) The centre is at (0, 0).
b) The transverse axis is horizontal and of length 2(5)
b) The transverse axis is vertical and of length 2(13)
or 10. The conjugate axis is vertical and of length
or 26. The conjugate axis is horizontal and of
2(4) or 8.
length 2(11) or 22.
c) The vertices are located at (−5 ± 5, −3) or (0, −3)
and (−10, −3).
c) The vertices are located at (0, ±13).
13
4
d) The slopes of the asymptotes are ± .
d) The slopes of the asymptotes are ± .
11
5
Section Review Page 177 Question 32
From the centre (h, k) = (3, 1) and the vertex (1, 1), a value of a = 3 − 1 or 2 is determined. From the slope
b
4
2
= , a value of b = is determined. The appropriate model for a horizontal transverse axis is used.
2
3
3
(x − h)2
(y − k)2
=1
−
a2
b2
(x − 3)2
(y − 1)2
− ¡2 = 1
2
4
2
3
Standard form:
(x − 3)2
(y − 1)2
=1
−
16
4
9
Multiply both sides by 16.
4(x − 3)2 − 9(y − 1)2 = 16
4(x2 − 6x + 9) − 9(y 2 − 2y + 1) = 16
4x2 − 24x + 36 − 9y 2 + 18y − 9 − 16 = 0
General form:
4x2 − 9y 2 − 24x + 18y + 11 = 0
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
153
Section Review Page 177 Question 33
²
³
2 + (−4)
The centre is at
, 5 or (−1, 5). A transverse axis of 4 yields 2a = 4 or a = 2. From the coordinates
2
of the foci, the value for c is determined as 2c = 2 − (−4) or c = 3. Solve for b2 .
b2 = c2 − a2
= 32 − 22
=9−4
=5
Use the model for a horizontal transverse axis.
(x − (−1))2
(y − 5)2
−
=1
22
5
Standard form:
(x + 1)2
(y − 5)2
=1
−
4
5
Multiply both sides by 20.
5(x + 1)2 − 4(y − 5)2 = 20
5(x2 + 2x + 1) − 4(y 2 − 10y + 25) = 20
5x2 + 10x + 5 − 4y 2 + 40y − 100 = 20
General form:
5x2 − 4y 2 + 10x + 40y − 115 = 0
Section Review
Page 177
Question 34
xy = −25
25
y=−
x
Section Review
Page 177
Question 35
(x − 4)2 − 16(y + 5)2 = 4
(x − 4)2 − 4
16
p
(x − 4)2 − 4
y = −5 ±
4
1
The slopes of the asymptotes are ± and both pass
4
through the centre (4, −5). Using the point-slope
equation for a line, the equations of the asymptotes
can be determined.
(y + 5)2 =
y − y1 = m(x − x1 )
1
y − (−5) = − (x − 4)
4
4y + 20 = −x + 5
x + 4y + 15 = 0
154
y − y1 = m(x − x1 )
1
y − (−5) = (x − 4)
4
4y + 20 = x − 5
x − 4y − 25 = 0
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
Section Review
Page 177
Question 36
x2 − y 2 − 6 = 0
x2 − y 2 = 6
x2
y2
−
=1
6
6
(x − 0)2
(y − 0)2
√ ¡2 − √ ¡2 = 1
6
6
√
6
The centre is at (0, 0). The vertices are at (± 6, 0). The slopes of the asymptotes are ± √ or ±1. Their
6
equations are y = ±x.
√
Section Review
Page 177
Question 37
25y 2 − 9x2 − 100y − 72x − 269 = 0
25(y 2 − 4y) − 9(x2 + 8x) = 269
25(y 2 − 4y + 4) − 9(x2 + 8x + 16) = 269 + 100 − 144
25(y − 2)2 − 9(x + 4)2 = 225
(y − 2)2
(x + 4)2
−
=1
9
25
(x + 4)2
(y − 2)2
−
=1
2
3
52
The centre is at (−4, 2). The vertices are at (−4, 2±3) or (−4, 5) and (−4, −1). The slopes of the asymptotes
3
3
are ± . Determine the equation of the asymptote through (−4, 2) with a slope of .
5
5
3
(x − (−4))
5
5y − 10 = 3x + 12
y−2=
3x − 5y + 22 = 0
3
Determine the equation of the asymptote through (−4, 2) with a slope of − .
5
3
y − 2 = − (x − (−4))
5
5y − 10 = −3x − 12
3x + 5y + 2 = 0
Section Review Page 177
The value for p is 2.
Question 38
Section Review Page 177
The value for p is 2.
(y − k)2 = 4p(x − h)
(y − 3)2 = 4(2)(x − (−4))
Standard form:
Question 39
(x − h)2 = 4p(y − k)
(x − 5)2 = 4(2)(y − (−3))
Standard form:
2
(y − 3) = 8(x + 4)
y − 6y + 9 = 8x + 32
General form:
(x − 5)2 = 8(y + 3)
x − 10x + 25 = 8y + 24
General form:
y 2 − 8x − 6y − 23 = 0
x2 − 10x − 8y + 1 = 0
2
2
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
155
Section Review Page 177 Question 40
The model for a parabola with a vertical axis of symmetry is used.
(x − h)2 = 4p(y − k)
(x − 4)2 = 4p(y − 3)
(1)
Substitute the point (5, 2) in (1) to determine p.
(5 − 4)2 = 4p(2 − 3)
1 = −4p
1
p=−
4
Substitute (2) into (1).
²
³
1
(x − 4)2 = 4 −
(y − 3)
4
Standard form:
(x − 4)2 = −(y − 3)
(2)
x2 − 8x + 16 = −y + 3
General form:
x2 − 8x + y + 13 = 0
Section Review
Page 177
Question 41
Section Review
x2 − 12y = −12
x2 = 12y − 12
Page 177
Question 42
y 2 − 8x = 2y − 41
y 2 − 2y = 8x − 41
(x − 0)2 = 4(3)(y − 1)
y 2 − 2y + 1 = 8x − 41 + 1
(y − 1)2 = 8x − 40
(y − 1)2 = 4(2)(x − 5)
The parabola opens upward. The vertex is at
(0, 1). The focus is at (0, 1 + 3) or (0, 4). The
equation of the directrix is y = 1 − 3 or y = −2.
The equation of the axis of symmetry is x = 0.
The parabola opens to the right. The vertex is
at (5, 1). The focus is at (5 + 2, 1) or (7, 1). The
equation of the directrix is x = 5 − 2 or x = 3.
The equation of the axis of symmetry is y = 1.
Section Review Page 177 Question 43
a) The graph is an ellipse with a vertical major axis.
b) The graph is an ellipse with a horizontal major axis.
c) The graph is a pair of horizontal lines, y = ±12.
d) The graph is a hyperbola with vertices at (0, ±12).
Section Review Page 177 Question 44
From the dimensions supplied in centimetres, the parabola passes through the point (90, 24).
x2 = 4py
Substitute the point (90, 24) in (1) to determine p.
902 = 4p(24)
(1)
8100 = 96p
8100
p=
96
675
=
8
(2)
Substitute (2) into (1).
²
675
8
675
x2 =
y
2
x2 = 4
The focus is located at
156
²
³
y
³
675
0,
.
8
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
Exploring Math
Section Exploring Math Page 177
a) Nine possible equations with a = 1 are
Question 1
b) Use n = 2 for m = {2, 4, 6}.
y 8 = x2
y 8 = x4
y 8 = x6
y 10 = x2
y 10 = x4
y 10 = x6
y 12 = x2
c) Use m = 2 for n = {8, 10, 12}.
y 12 = x4
y 12 = x6
d) Values of a > 1 will stretch the graphs vertically.
e) y n = axm defines parabolas for (m, n) = (1, 2)
or (m, n) = (2, 1).
Section Exploring Math
Page 177
Question 2
a) Nine possible equations with b = 1 are
b) Use n = 4 for m = {1, 2, 3}.
x1 y 4 = 1
x2 y 4 = 1
x3 y 4 = 1
x1 y 5 = 1
x2 y 5 = 1
x3 y 5 = 1
x1 y 6 = 1
c) Use m = 1 for n = {4, 5, 6}.
2 6
x y =1
y3 y6 = 1
d) Values of b > 1 will stretch the graphs vertically.
e) xm y n = b defines a rectangular hyperbola for
m = n = 1.
Section Exploring Math Page 177 Question 3
See Scientific American, November 1997, pp. 68 − 73.
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
157
Chapter Check
Section Chapter Check Page 178 Question 1
Point the flashlight perpendicular to the plane surface to get a circle. Tilt the flashlight a little to get an
ellipse. Tilt it more to get a parabola. Tilt it even more to get one branch of a hyperbola.
Section Chapter Check
The graph is a circle.
Page 178
Question 2
(x − h)2 + (y − k)2 = r2
(x − 3)2 + (y − (−2))2 = 42
(x − 3)2 + (y + 2)2 = 16
Section Chapter Check Page 178 Question 3
The graph is a hyperbola. From the coordinates of the foci, it can be determined that c = 10. The difference
between the focal radii yields 2a = 12 or a = 6.
b2 = c2 − a2
= 102 − 62
= 100 − 36
= 64
The standard equation of the hyperbola can be determined.
(x − h)2
(y − k)2
=1
−
a2
b2
(x − 0)2
(y − 0)2
−
=1
2
6
64
x2
y2
−
=1
36 64
Section Chapter Check Page 178 Question 4
Comparison of the degrees of x and y suggests the graph is a parabola.
2x2 − 20x − y + 47 = 0
2(x2 − 10x) = y − 47
2(x2 − 10x + 25) = y − 47 + 50
2(x − 5)2 = y + 3
1
(x − 5)2 = (y + 3)
2
Section Chapter Check Page 178 Question 5
A stretched circle is an ellipse. Stretching the unit circle horizontally by a factor of 5 yields a = 5 and
preserves b = 1. The centre is at (h, k) = (−3, −8).
(x − h)2
(y − k)2
+
=1
2
a
b2
(y − (−8))2
(x − (−3))2
+
=1
52
12
(x + 3)2
+ (y + 8)2 = 1
25
158
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
Section Chapter Check Page 178 Question 6
The graph is an ellipse. From the coordinates of the foci, it can be determined that c = 4. The sum of the
focal radii yields 2a = 10 or a = 5.
b 2 = a2 − c 2
= 52 − 42
= 25 − 16
=9
The standard equation of the ellipse can be determined.
(x − h)2
(y − k)2
+
=1
a2
b2
(x − 0)2
(y − 0)2
+
=1
2
5
9
x2
y2
+
=1
25
9
Section Chapter Check Page 178 Question 7
The coefficients of the second-degree terms have the same sign but are not equal. The graph is an ellipse.
x2 + 225y 2 − 4x − 5 = 0
x2 − 4x + 225y 2 = 5
x2 − 4x + 4 + 225y 2 = 5 + 4
(x − 2)2 + 225y 2 = 9
(x − 2)2
y2
+ 1 =1
9
25
Section Chapter Check Page 178 Question 8
The coefficients of the second-degree terms have the same sign and are equal. The graph is a circle.
x2 + y 2 + 6x − 8y − 11 = 0
x2 + 6x + y 2 − 8y = 11
x2 + 6x + 9 + y 2 − 8y + 16 = 11 + 9 + 16
(x + 3)2 + (y − 4)2 = 36
Section Chapter Check Page 178 Question 9
The coefficients of the second-degree terms have different signs and are not equal. The graph is a hyperbola.
9x2 − 4y 2 − 180x − 56y + 740 = 0
9x2 − 180x − 4y 2 − 56y = −740
9(x2 − 20x) − 4(y 2 + 14y) = −740
9(x2 − 20x + 100) − 4(y 2 + 14y + 49) = −740 + 900 − 196
9(x − 10)2 − 4(y + 7)2 = −36
(y + 7)2
(x − 10)2
−
=1
9
4
Section Chapter Check Page 178 Question 10
Hyperbolas are the only conic associated with asymptotes. The centre is at (h, k) = (−2, −8). The position
of one vertex places the transverse axis vertically and yields a = −8 − (−20) or a = 12. From the slope of
one asymptote,
a
=6
b
12
=6
b
b=2
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
159
The equation of the hyperbola can be determined.
(y − k)2
(x − h)2
−
=1
2
a
b2
(y − (−8))2
(x − (−2))2
=1
−
122
22
2
2
(y + 8)
(x + 2)
−
=1
144
4
Section Chapter Check
Page 178
Question 11
1
.
4
From the diagram, the centre is at (h, k) = (−7, 0) and p is
(y − k)2 = 4p(x − h)
² ³
1
2
(y − 0) = 4
(x − (−7))
4
Standard form:
y2 = x + 7
General form:
y2 − x − 7 = 0
Section Chapter Check Page 178 Question 12
This rectangular hyperbola is of the form xy = k. Since the graph contains the point (1, 5), substitution will
reveal k.
xy = k
(1)(5) = k
k=5
Substitute (2) into (1) to obtain the standard form.
(1)
(2)
xy = 5
General form:
xy − 5 = 0
Section Chapter Check
Page 178
Question 13
²
³
1 1
. The semi-major axis is vertical and 5 units in length.
From the diagram, the centre is at (h, k) = − ,
2 2
The semi-minor axis is 3 units in length.
(x − h)2
(y − k)2
+
=1
2
b
a2
¡¡2
¡¡2
x − − 12
y − 12
+
=1
32
52
Standard form:
x + 12
9
¡2
y − 12
+
25
¡2
=1
Multiply both sides by 225.
²
³2
²
³2
1
1
+9 y−
= 225
25 x +
2
2
²
³
²
³
1
1
25 x2 + x +
+ 9 y2 − y +
= 225
4
4
25
9
25x2 + 25x +
+ 9y 2 − 9y + = 225
4
4
17
2
2
25x + 25x + 9y − 9y +
= 225
2
General form:
50x2 + 18y 2 + 50x − 18y − 433 = 0
160
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
Section Chapter Check Page 178 Question 14
The centre of the circle is located at (h, k) = (−2 − 2). The radius is r = 2.
(x − h)2 + (y − k)2 = r2
(x − (−2))2 + (y − (−2))2 = 22
Standard form:
(x + 2)2 + (y + 2)2 = 4
x2 + 4x + 4 + y 2 + 4y + 4 = 4
General form:
x2 + y 2 + 4x + 4y + 4 = 0
Section Chapter Check
Page 178
Question 15
²
³
9
, −1 . The length of the horizontal semi-transverse
The centre of the hyperbola is located at (h, k) =
2
9
7
axis is a = 8 − or a = . The length of the semi-conjugate axis is b = 1 − (−1) or 2 units.
2
2
(x − h)2
(y − k)2
−
=1
2
a
b2
¡2
x − 92
(y − (−1))2
=1
7 ¡2 −
22
2
Standard form:
x−
49
4
¡
9 2
2
−
(y + 1)2
=1
4
Multiply both sides by 196.
²
³2
9
− 49(y + 1)2 = 196
16 x −
2
²
³
81
16 x2 − 9x +
− 49(y 2 + 2y + 1) = 196
4
16x2 − 144x + 324 − 49y 2 − 98y − 49 = 196
General form:
16x2 − 49y 2 − 144x − 98y + 79 = 0
Section Chapter Check Page 178 Question 16
The centre of the parabola is located at (h, k) = (−6, −2) and p = −4.
(x − h)2 = 4p(y − k)
(x − (−6))2 = 4(−4)(y − (−2))
Standard form:
(x + 6)2 = −16(y + 2)
x2 + 12x + 36 = −16y − 32
General form:
x2 + 12x + 16y + 68 = 0
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
161
Section Chapter Check
Page 178
Question 17
9x2 + y 2 + 36x = 9
9x2 + 36x + y 2 = 9
9(x2 + 4x + 4) + y 2 = 9 + 36
9(x + 2)2 + y 2 = 45
(y + 0)2
(x + 2)2
+
=1
5
45
The coordinates of the centre are (−2, 0).
Section Chapter Check
Page 178
Question 18
x2 + y 2 − 2x + 4y = 5
x2 − 2x + y 2 + 4y = 5
2
(x − 2x + 1) + (y 2 + 4y + 4) = 5 + 1 + 4
(x − 1)2 + (y + 2)2 = 10
The coordinates of the centre are (1, −2).
Section Chapter Check
Page 178
Question 19
9x2 − 4y 2 − 8y + 32 = 0
9x2 − 4(y 2 + 2y) = −32
9x2 − 4(y 2 + 2y + 1) = −32 − 4
9x2 − 4(y + 1)2 = −36
x2
(y + 1)2
−
=1
9
4
The coordinates of the centre are (0, −1).
Section Chapter Check
Page 178
Question 20
3x2 + 12x − 4y − 12 = 0
3(x2 + 4x) = 4y + 12
3(x2 + 4x + 4) = 4y + 12 + 12
3(x2 + 4x + 4) = 4y + 24
4
(x + 2)2 = (y + 6)
3
The coordinates of the vertex are (−2, −6).
Section Chapter Check
Page 178
Question 21
(x + 2)2
(y − 1)2
+
=1
9
36
4(x + 2)2 + (y − 1)2 = 36
4(x2 + 4x + 4) + y 2 − 2y + 1 − 36 = 0
4x2 + 16x + 16 + y 2 − 2y − 35 = 0
4x2 + y 2 + 16x − 2y − 19 = 0
A = 4, C = 1
162
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
Section Chapter Check
Page 178
Question 22
9
y − 6 = − (x + 2)2 + 6
4
9
y − 12 = − (x + 2)2
4
4y − 48 = −9(x2 + 4x + 4)
4y − 48 = −9x2 − 36x − 36
9x2 + 36x + 4y − 12 = 0
A = 9, C = 0
Section Chapter Check
Page 178
Question 23
(y + 4)2
(x − 2)2
=1
−
16
9
9(y + 4)2 − 16(x − 2)2 = 144
9(y 2 + 8y + 16) − 16(x2 − 4x + 4) − 144 = 0
9y 2 + 72y + 144 − 16x2 + 64x − 64 − 144 = 0
16x2 − 9y 2 − 64x − 72y + 64 = 0
A = 16, C = −9
Section Chapter Check
Page 178
Question 24
x2
(y + 5)2
+
=1
16
16
x2 + (y + 5)2 = 16
x2 + y 2 + 10y + 25 − 16 = 0
x2 + y 2 + 10y + 9 = 0
A = 1, C = 1
Section Chapter Check Page 178 Question 25
The graph for the downward-opening parabola must contain the point (30, −40). Determine p.
(x − h)2 = 4p(y − k)
(30 − 0)2 = 4p(−40 − 0)
900 = −160p
900
p=−
160
45
=−
8
²
³
45
45
y or x2 = − y.
The equation of the path is x2 = 4 −
8
2
Section Chapter Check Page 178 Question 26
Position the centre of the ellipse at the origin. From the given information, a = 200 ÷ 2 or 100 and b = 50.
x2
y2
+
a2
b2
2
x
y2
+ 2
2
100
50
Substitute x = 100 − 30 or 70 into (1) and solve for y.
4900
y2
+
10 000 2500
4900 + 4y 2
4y 2
=1
=1
(1)
=1
= 10 000
= 5100
y 2 = 1275
√
y = ±5 51
The height of the domed stadium 30 m from one of the ends is approximately 35.7 m.
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
163
Using the Strategies
Section Problem Solving Page 179 Question 1
Define the points on the parabola, x = y 2 : P(y 2 , y)
and Q(y 2 , −y), where y > 0.
y
P( y 2, y )
PO = PQ
p
(y 2 − 0)2 + (y − 0)2 = y − (−y)
p
y 4 + y 2 = 2y
O
y 4 + y 2 = 4y 2
y 4 − 3y 2 = 0
y 2 (y 2 − 3) = 0
y = 0 or ±
x
Q(y 2,- y)
√
3
The two other coordinates of the equilateral triangle are P(3,
√
√
3) and Q(3, − 3).
Section Problem Solving Page 179 Question 2
Let n be the number of perfect squares.
√
√
n = 4 000 000 − 1 000 000 − 1
= 2000 − 1000 − 1
= 999
There are 999 perfect squares between 1 000 000 and 4 000 000.
Section Problem Solving Page 179 Question 3
Since 937 people purchased at least one item and 426 purchased at least 2 items, then 937 − 426 or 511
people purchased exactly 1 item.
Since 426 people purchased at least two items and 314 purchased at least 3 items, then 426 − 314 or 112
people purchased exactly two items.
Since 314 people purchased at least three items and exactly 282 people people purchased all four, then
314 − 282 or 32 people purchased exactly 3 items.
Section Problem Solving
Page 179
Question 4
The formula for the volume, V , of a right circular cone with base radius, r, and height, h, is V =
Determine an expression for r2 in the original pile.
1 2
πr h.
3
V = 85
1 2
πr (6) = 85
3
2πr2 = 85
85
r2 =
2π
Add 15 m3 to the pile. Let Vnew represent the new volume and ∆h represent the increase in height.
Vnew = 85 + 15
1 2
πr (6 + ∆h) = 100
3
Substitute (1) into (2).
² ³
1
85
π
(6 + ∆h) = 100
3
2π
600
∆h =
−6
85
.
= 1.059
The increase in height is approximately 1.06 m.
164
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
(1)
(2)
Section Problem Solving Page 179 Question 5
Make a tetrahedron using one triangle as the base and the other toothpicks as edges of the 3-dimensional
object.
Section Problem Solving Page 179 Question 6
Since the centre, C, of the inscribed circle lies on the line y = x, define the coordinates, C(h, h). Let P be
the point of tangency between the circles. Since P is also on the line y = x, define its coordinates as P(x, x).
OP = r
p
y
2
x + x2 = r
r
x2 + x2 = r 2
2x2 = r2
r
x= √
2
√
r 2
=
2
À √
√ !
r 2 r 2
The coordinates of P are
,
.
2
2
P
h
h
r-h
O
C
h
Q
r
R
x
Since4OCQ ∼ 4OPR, a value for h can be determined.
OC
OP
=
OQ
OR
r−h
r
= r
√
h
2
r−h √
= 2
h
√
r − h = 2h
√
h( 2 + 1) = r
√
r
2−1
h= √
×√
2+1
2−1
√
= ( 2 − 1)r
Determine the equation of the inscribed circle.
(x − h)2 + (y − h)2 = h2
√
√
√
(x − ( 2 − 1)r)2 + (y − ( 2 − 1)r)2 = (( 2 − 1)r)2
√
√
√
(x − ( 2 − 1)r)2 + (y − ( 2 − 1)r)2 = (2 − 2 2 + 1)r2
√
√
√
(x − ( 2 − 1)r)2 + (y − ( 2 − 1)r)2 = (3 − 2 2)r2
√
√
√
The equation of the inscribed circle is (x − ( 2 − 1)r)2 + (y − ( 2 − 1)r)2 = (3 − 2 2)r2 .
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
165
Section Problem Solving Page 179 Question 7
Consider the 2 by 2 square centred around the origin. Let P(x, y) be the general point on the locus and A,
B, C, and D be the vertices of the square.
y
P( , )
xy
A
D
1
1
B
x
C
PA2 + PB2 + PC2 + PD2 = 16
(x − (−1))2 + (y − 1)2 + (x − (−1))2 + (y − (−1))2 + (x − 1)2 + (y − (−1))2 + (x − 1)2 + (y − 1)2 = 16
(x + 1)2 + (y − 1)2 + (x + 1)2 + (y + 1)2 + (x − 1)2 + (y + 1)2 + (x − 1)2 + (y − 1)2 = 16
2((x + 1)2 + (y − 1)2 + (y + 1)2 + (x − 1)2 ) = 16
(x + 1)2 + (y − 1)2 + (y + 1)2 + (x − 1)2 = 8
x2 + 2x + 1 + y 2 − 2y + 1 + y 2 + 2y + 1 + x2 − 2x + 1 = 8
2x2 + 2y 2 + 4 = 8
2x2 + 2y 2 = 4
x2 + y 2 = 2
The locus of points is a circle circumscribed on the square.
Section Problem Solving
Page 179
Power
91
92
93
94
9n
9
99
Question 8
Standard Form
9
81
729
6561
variable
too large
Digits
1
2
3
4
n
99
9
There are 99 or 387 420 489 digits in the standard form of 99 .
Section Problem Solving Page 179 Question 9
Careful reading reveals that 15% of the students are moving from each of the three hospitals at the end
of each month, but 15%, or the same number, are arriving at each hospital. The number working at each
hospital remians the same, 400 each month.
166
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
Section Problem Solving Page 179 Question 10
√
The hypotenuse of the right triangle is 62 + 82 or 10 cm.
A
B
C
8
h1
8
8
8
r
6
6
Figure A is a right circular cone of
base radius 8 cm and height 6 cm.
Figure B is a right circular cone of
base radius 6 cm and height 8 cm.
1
π(8)2 (6)
3
= 128π
1
π(6)2 (8)
3
= 96π
VA =
VB =
h2
6
6
Figure C consists of two right circular cones of common base radius
and heights that total 10 cm. Let
r, h1 , and h2 stand for the lengths
between the marked points. The
value for r can be determined from
similar triangles.
r
8
=
6
10
r = 4.8
1 2
1
πr h1 + πr2 h2
3
3
1
= πr2 (h1 + h2 )
3
1
= π4.82 (10)
3
= 76.8π
VC =
The maximum possible volume is 128 cm3 .
Section Problem Solving
Page 179
Question 11
n = 5x + 5x + 5x + 5x + 5x
= 5(5x )
= 5x+1
¡5
n5 = 5x+1
= 55x+5
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
167
Section Problem Solving Page 179 Question 12
Let r be the length of the diagonal of the inner square.
The radius of the circle is then r. Determine the side
length of the square.
X
x2 + x2 = (2r)2
2x2 = 4r2
x2 = 2r2
P
Construct OP such that OP ⊥ XY. In 4XOP,
OP
OX
OP
OX =
sin 6 OXP
r
=
sin 30◦
= 2r
OP
tan 6 OXP =
PX
OP
PX =
tan 6 OXP
r
=
◦
tan
√ 30
= 3r
r
r
r
O
x
sin 6 OXP =
Y
Q
Since XQ = XO + OQ, the height of 4XYZ is 2r + r or 3r. Since QY = PX =
can be determined.
√
Atriangle
( 3r)(3r)
=
Asquare
2r2
√
3 3
=
2
√
The ratio of the area of the triangle to the area of the square is 3 3 : 2.
√
Z
3r, the ratio of the areas
Section Problem Solving Page 179 Question 13
Since 3 and 5 are factors of 15, the concern is the remainders obtained by dividing 7 by 3 and 5 respectively.
When 7 is divided by the 3, the remainder is 1. When 7 is divided by 5, the remainder is 2. The sum of the
remainders is 1 + 2 or 3.
Section Problem Solving
Page 179
Question 14
H
H
T
T
H
T
H
H
H
T
H
H
T
T
H
H
A minimum of four moves is required. One possible solution is outlined above.
Section Problem Solving Page 179 Question 15
Since the product
integers is 95 040, it is reasonable to assume that the middle factor is
√ of five consecutive √
.
approximately 5 95 040. The result, 5 95 040 = 9.8, suggests the middle factor could be 10. Since n is the
largest of the five integers, we test n = 12.
L.S. = 12(12 − 1)(12 − 2)(12 − 3)(12 − 4)
= 12(11)(10)(9)(8)
= 95 040
= R.S.
The value of n is 12.
168
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
Section Problem Solving
Page 179
Question 16
148
148
+ 148
444
A = 1, B = 4, C = 8.
MATHPOWERTM12, Western Edition, Solutions
Chapter 3
169