Chapter 7
A1
Glencoe Precalculus
DATE
PERIOD
Before you begin Chapter 7
Conic Sections and Parametric Equations
Anticipation Guide
A
D
A
7. The transverse axis of a hyperbola is longer than the
conjugate axis.
8. A circle is a special type of ellipse.
After you complete Chapter 7
D
6. The graph of a degenerate conic is a line, two intersecting
lines, or a point.
A
4. There are formulas to find the coordinates of a point on the
graph of a rotated conic section.
5. A conic section could be a triangle or a square.
D
D
3. The general form of the equation of a conic section is
y = mx + b.
A
2. The eccentricity of an ellipse is greater than 1.
3
Chapter Resources
10/24/09 3:59:34 PM
Glencoe Precalculus
• For those statements that you mark with a D, use a piece of paper to write an
example of why you disagree.
• Did any of your opinions about the statements change from the first column?
Chapter 7
STEP 2
A or D
1. The graph of y 2 = 4x is a parabola.
Statement
• Reread each statement and complete the last column by entering an A or a D.
Step 2
STEP 1
A, D, or NS
• Write A or D in the first column OR if you are not sure whether you agree or
disagree, write NS (Not Sure).
• Decide whether you Agree (A) or Disagree (D) with the statement.
• Read each statement.
Step 1
7
NAME
0ii_004_PCCRMC07_893808.indd Sec1:3
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Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
DATE
PERIOD
(3, -4)
(3, -1)
(h, k)
(h, k + p)
directrix:
axis of symmetry:
y = -7
x=3
y=k-p
x=h
1
-3 −
4
11
-3 −
12
11
-3 −
12
1
-3 −
4
y
0
y
x
2
)
Answers
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
005_020_PCCRMC07_893808.indd 5
Chapter 7
2
1
y = 3; x = 5 −
(6, 3); 6 −1 , 3 ;
(
3. (y − 3)2 = 2(x - 6)
y = -1; x = 1
(3, -1); (5, -1);
0
4
8
12
y
0
4
8
12
x
x
5
x = 3; y = -5
(3, -2); (3, 1);
1 (x − 3)2 = y + 2
4. −
)
(
12
x = -2; y = 0
(-2, 1); (-2, 2);
12
−2
2
0 4
y
4
8
x
12
8x
Lesson 7-1
11/9/09 10:07:10 PM
Glencoe Precalculus
−4
4
8
−4
−8 −4 0
4
8
For each equation, identify the vertex, focus, axis of symmetry, and directrix.
Then graph the parabola.
y
y
1. (y + 1)2 = 8(x - 3)
2. (x + 2)2 = 4(y - 1)
Exercises
6
4
2
0
x
Graph the vertex, focus, axis, and directrix of the parabola. Then make a table of values to
graph the general shape of the curve.
vertex:
focus:
The equation is in the form (x - h)2 = 4p(y - k), so h = 3 and k = -4. Use the values
of h, k, and p to determine the characteristics of the parabola.
The equation is in standard form and the squared term is x, which means that the parabola
opens vertically. Because 4p = 12, p = 3 and the graph opens upward.
Example
For (x - 3)2 = 12(y + 4), identify the vertex, focus, axis of symmetry,
and directrix. Then graph the parabola.
A parabola is the locus of all points in a plane
equidistant from a point called the focus and a line called the directrix. The standard
form of the equation of a parabola that opens vertically is (x - h)2 = 4p(y - k). When p is
negative, the parabola opens downward. When p is positive, it opens upward. The standard
form of the equation of a parabola that opens horizontally is (y - k)2 = 4p(x - h). When
p is negative, the parabola opens to the left. When p is positive, it opens to the right.
Parabolas
Study Guide and Intervention
Analyze and Graph Parabolas
7-1
NAME
Answers (Anticipation Guide and Lesson 7-1)
Parabolas
Study Guide and Intervention
DATE
−4
A2
−4
−4
y
4
8x
x
−8
Glencoe Precalculus
005_020_PCCRMC07_893808.indd 6
Chapter 7
y
4
8x
6
−8 −4 0
4
8
y
4
8x
2(x - 1) = (y - 1)2
−8
8
0
−4
−8
4
2
4x
Glencoe Precalculus
4. focus (1.5, 1); opens right; directrix x = 0.5
−8
8
(x - 1)2 = -4(y - 5)
−4
y
−12
−4
y
2. focus (1, 4); opens down; contains (-3, 1)
−4
−4 0
4
8
(x - 5)2 = -12(y - 3)
3. directrix y = 6; opens down; vertex (5, 3)
−8
4
8
12
(y - 5)2 = -12(x - 2)
1. focus (-1, 5) and vertex (2, 5)
Write an equation for and graph a parabola with the given
characteristics.
Exercises
Graph the vertex, focus, and parabola.
y - (-3)2 = 4(-5)(x - 1)
p = -5, h = 1, and k = -3
Simplify.
(y + 3)2 = -20(x - 1)
The standard form of the equation is (y + 3)2 = -20(x - 1).
4
−2 0
Write an equation for and graph a parabola with
focus (-4, -3) and vertex (1, -3).
Because the focus and vertex share the same y-coordinate, the graph is
horizontal. The focus is (h + p, k), so the value of p is -4 - 1 or -5. Because
p is negative, the graph opens to the left.
Write the equation for the parabola in standard form using the
values of h, p, and k.
Standard form
(y - k)2 = 4p(x - h)
Example
Specific characteristics can be used to
determine the equation of a parabola.
(continued)
PERIOD
10/24/09 4:05:41 PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Chapter 7
Equations of Parabolas
7-1
NAME
Parabolas
Practice
DATE
PERIOD
x
0
y
x
0
y
(y - 1)2 = 8x
x
4. focus (2, 1); opens right; contains (8, -7)
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
005_020_PCCRMC07_893808.indd 7
Chapter 7
x 2 = 8y
7
Lesson 7-1
3/13/09 4:23:17 PM
Glencoe Precalculus
6. SATELLITE DISH Suppose the receiver in a parabolic dish antenna is
2 feet from the vertex and is located at the focus. Assume that the vertex
is at the origin and that the dish is pointed upward. Find an equation that
models a cross section of the dish.
(x + 4)2 = -4(y - 2); (-4, 2); (-4, 1); x = -4; y = 3
5. Write x2 + 8x = -4y - 8 in standard form. Identify the vertex, focus, axis
of symmetry, and directrix.
0
y
(x + 2)2 = -4(y - 4)
3. vertex (-2, 4), focus (-2, 3)
x
vertex: (1, -3); focus: (-2, -3);
axis of symmetry: y = -3;
directrix: x = 4
2. y2 + 6y + 9 = 12 - 12x
Write an equation for and graph a parabola with the given
characteristics.
0
y
vertex: (1, 2); focus: (1, 4);
axis of symmetry: x = 1;
directrix: y = 0
1. (x - 1)2 = 8(y - 2)
Identify the vertex, focus, axis of symmetry, and directrix for each
equation. Then graph the parabola.
7-1
NAME
Answers (Lesson 7-1)
Chapter 7
Parabolas
A3
8
40 ft
10 ft
T
−30
−20
−10
0
y
20x
60 ft
22.5 ft
Glencoe Precalculus
b. Find the length of a supporting wire
that is 100 feet from the center.
x2 = 800 y
a. Write an equation for the
parabolic cable.
400 ft
10 ft
5. BRIDGES The cable for a suspension
bridge is in the shape of a parabola.
The vertical supports are shown in
the figure.
−20
5
b. Graph the equation.
x2 = -40 y
3/13/09 4:23:22 PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Glencoe Precalculus
005_020_PCCRMC07_893808.indd 8
Chapter 7
1 2
y=−
x
2
3. FLASHLIGHT A flashlight contains a
parabolic mirror with a bulb in the center
as a light source and focus. If the width
of the mirror is 4 inches at the top and
the height to the focus is 0.5 inch, find an
equation of the parabolic cross section.
21 ft
b. What is the maximum height that a
T-shirt reaches?
1
(x - 1)2 = - −
(y - 21)
16
a. Write the equation in standard form.
2. T-SHIRTS The cheerleaders at the high
school basketball game launch T-shirts
into the stands after a victory. The
launching device propels the shirts into
the air at an initial velocity of 32 feet per
second. A shirt’s distance y in feet above
the ground after x seconds can be
modeled by y = -16x2 + 32x + 5.
I
a. Write an equation that models
the parabola.
N
VERSI
Y
4 in.
x = 16y, where the values of x and y
are measured in inches. Find the
distance from the vertex to the focus
of this mirror.
2
PERIOD
4. ARCHWAYS The entrance to a college
campus has a parabolic arch above two
columns as shown in the figure.
Word Problem Practice
1. REFLECTOR The figure shows a
parabolic reflecting mirror. A cross
section of the mirror can be modeled by
7-1
DATE
U
NAME
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Enrichment
DATE
d
y-x-2
2
)
D(a, b)
P(x, y)
Chapter 7
9
y = -x - 2
F(1, 1)
m
x
Lesson 7-1
3/13/09 4:23:28 PM
Glencoe Precalculus
(0, 0), because (0, 0) is midway between point F and line d.
b. Use your answer from part a to find the coordinates of the vertex of
the parabola. Justify your answer.
Answers
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
005_020_PCCRMC07_893808.indd 9
0
y = x, because y = x contains F(1, 1) and is perpendicular to d.
4. a. Every parabola has an axis of symmetry. Find an equation for
the axis of symmetry of the parabola described above. Justify
your answer.
x2 - 2xy + y2 - 8x - 8y = 0
y
PERIOD
3. Use the coordinates of F, P, and D, along with (PD)2 = (PF )2 to find an
equation of the parabola with focus F and directrix d.
y-x-2
Add to get b = − .
2
a + b = -2. Subtract to get a = − .
x-y-2
2
From the equation for line m,
-a + b = -x + y. From the equation for d,
of point D are D(a, b) = − , − .
( x - y2 - 2
2. Use the equations for lines m and d to show that the coordinates
x - y + (b - a) = 0
1. Find an equation for line m.
Refer to the discussion above.
The diagram at the right shows a fixed point F (1, 1)
and a line d that has an equation y = -x - 2. If P (x, y)
satisfies the condition that PD = PF, then P is on a
parabola. Our objective is to find an equation for the
tilted parabola, which is the locus of all points that are
the same distance from (1, 1) and the line y = -x - 2.
First find an equation for the line m through P(x, y)
and perpendicular to line d at D(a, b). Using this
equation and the equation for line d, find the coordinates
(a, b) of point D in terms of x and y. Then use
(PD) 2 = (PF ) 2 to find an equation for the parabola.
Tilted Parabolas
7-1
NAME
Answers (Lesson 7-1)
DATE
Ellipses and Circles
Study Guide and Intervention
y
PERIOD
A01_A14_PCCRMC07_893808.indd 4
b
b
a
a
A4
vertical
(-2, 1)
(-2, 6), (-2, -4)
(-5, 1), (1, 1)
x = -2
y=1
y=k
x=h
25
−8
−8
−4
−4 0
4
8
y
25
x
4
8
Pdf Pass
y
005_020_PCCRMC07_893808.indd 10
−8
−4
−4 0
Chapter 7
−8
16
4
9
8x
(y - 1)2
(x + 3)2
3. − + − = 1
64
(y + 2)2
(x + 5)2
1. − + − = 1
25
64
−8
−4 0
y
9
4
25
8x
10
−8
−4
−4 0
y
4
8x
(y + 3)2
(x - 2)2
4. − + − = 1
−8
4
8
(y + 1)2
(x + 2)2
2. − + − = 1
(h ± b, k)
(h, k ± a)
(h, k)
Graph the ellipse given by each equation.
Exercises
orientation:
center:
vertices:
co-vertices:
major axis:
minor axis:
0
7
9
'
$
−8
y
4
8x
' 7
x
Glencoe Precalculus
−8
0
4
8
(y - 1)2
(x + 2)2
Graph the ellipse given by the equation − + − = 1.
The equation is in standard form. Use the values of h, k, a, and b to
determine the vertices and axes of the ellipse. Since a2 > b2, a2 = 25
and b2 = 9, or a = 5 and b = 3. Since a2 is the denominator of the
y-term, the major axis is parallel to the y-axis.
Example
a2 is in the denominator of the y-term. In both cases, c2 = a2 - b2.
case, a2 is in the denominator of the x-term. The standard form is
(y - k)2 (x - h)2
−
+−
= 1 when the major axis is vertical. In this case,
2
2
11/15/09 4:12:41 PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Chapter 7
The standard form of the equation of an ellipse is
(y - k)2
(x - h)2
+ −
= 1 when the major axis is horizontal. In this
−
2
2
An ellipse is the locus of
points in a plane such that the sum of the distances from two fixed points,
called foci, is constant.
Analyze and Graph Ellipses and Circles
7-2
NAME
DATE
Ellipses and Circles
Study Guide and Intervention
(continued)
PERIOD
4
2
2
Divide each side by 36.
( 2)
4
= 9, - −
=4
Factor.
(−62 )
2
= 64
Standard form
Factor.
(−162 )
Complete the square.
Original equation
5
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
005_020_PCCRMC07_893808.indd 11
Chapter 7
15
12
(y - 3)
(x + 1)2
− + − = 1; ellipse
2
5. 4x2 + 8x + 5y2 - 30y - 11 = 0
(y + 2)2 = -4(x - 3); parabola
3. 4x - 8 + y2 + 4y = 0
30
(y + 1)2
(x - 2)2
− + − = 1; ellipse
1. y2 + 2y + 6x2 - 24x = 5
11
3
Lesson 7-2
10/24/09 4:28:19 PM
Glencoe Precalculus
1
(x + 2)2 = - −
(y - 17); parabola
6. 6x2 + 24x + 2y - 10 = 0
(x + 2)2 + (y - 1)2 = 54; circle
4. x2 + 4x + y2 - 2y - 49 = 0
(x - 12)2 + (y + 1)2 = 169; circle
2. y2 + 2y + x2 - 24x = 14
Write each equation in standard form. Identify the related conic.
Exercises
Because only one term is squared, the graph is a parabola with vertex (8, 2).
(x − 8)2 = 8(y − 2)
(x − 8)2 − 8(y − 2) = 0
(x2 − 16x + 64) − 8y + 80 - 64 = 0
(x2 − 16x + ?) − 8y + 80 = 0
b. x2 - 16x - 8y + 80 = 0
x2 - 16x - 8y + 80 = 0
an ellipse with center (−3, 2).
(y – k)2
(x - h)2
Because the equation is of the form −
+−
= 1, the graph is
b2
a2
9
2
4(x + 3) + 9(y - 2) = 36
(y - 2)2
(x + 3)2
−+− =1
2
4(x2 + 6x + 9) + 9(y2 - 4y + 4) = -36 + 36 + 36
Example
Write each equation in standard form. Identify the
related conic.
a. 4x2 + 9y2 + 24x - 36y + 36 = 0
4x2 + 9y2 + 24x - 36y + 36 = 0
Original equation
4(x2 + 6x + ?) + 9(y2 - 4y + ?) = -36 + ? + ?
Complete the square.
If you are given the equation
for a conic section, you can determine what type of conic is represented using
the characteristics of the equation. The standard form of an equation for a
circle with center (h, k) and radius r is (x - h)2 + (y - k)2 = r2.
Determine Types of Conic Sections
7-2
NAME
Answers (Lesson 7-2)
Glencoe Precalculus
3/15/10 10:03:09 PM
Chapter 7
Ellipses and Circles
Practice
x
PERIOD
−4 0
4
y
4
x
2. 25x2 + 9y2 - 50x - 90y + 25 = 0
DATE
28
2
A5
16
12
2
25
16
64
√
55
−
8
9
(y + 2)2
(x + 1)2
10. − + − = 1
12
10/24/09 4:28:30 PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Glencoe Precalculus
005_020_PCCRMC07_893808.indd 12
Chapter 7
1.5 ft to the sides of the center
11. CONSTRUCTION A semi-elliptical arch is used to design a headboard
for a bed frame. The headboard will have a height of 2 feet at the center
and a width of 5 feet at the base. Where should the craftsman place the
foci in order to sketch the arch?
3
−
5
(y + 1)2
(x + 1)2
9. − + − = 1
Glencoe Precalculus
(x + 10)2 + (y - 5)2 = 121; circle
8. x2 + y2 + 20x - 10y + 4 = 0
9
2
(y - 3)
(x - 4)
− + − = 1; ellipse
6. 4x2 - 32x + 3y2 - 18y = -55
Determine the eccentricity of the ellipse given by each equation.
(x - 4)2 + (y - 12)2 = 169; circle
7. x2 + y2 - 8x - 24y = 9
(y - 2)2 + 4(x + 5); parabola
5. y2 - 4y = 4x + 16
Write each equation in standard form. Identify the related conic.
25
(y - 4)2
(x + 2)2
− + − =1
4. foci (-2, 1), (-2, 7); length of major axis 10 units
64
2
(y - 6)
(x + 4)
− + − =1
3. vertices (-12, 6), (4, 6); foci (-10, 6), (2, 6)
Write an equation for the ellipse with each set of characteristics.
0
y
1. 4x2 + 9y2 - 8x - 36y + 4 = 0
Graph the ellipse given by each equation.
7-2
NAME
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Ellipses and Circles
y2
368.64
y2
225
Answers
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
005_020_PCCRMC07_893808.indd 13
Chapter 7
≈13 ft
b. Find the height of the tunnel 10 feet
from the center.
400
x
−
+−=1
2
a. Write an equation that models
the ellipse.
40 ft
15 ft
3. TUNNEL The entrance to a tunnel is in
the shape of half an ellipse as shown in
the figure.
38.4 in.
the sign?
b. What is the maximum height of
576
x
−
+− =1
2
a. Write an equation for the ellipse if the
center of the sign is at the origin and
the major axis is horizontal.
2. SIGNS A sign is in the shape of an
ellipse. The eccentricity is 0.60 and the
length is 48 inches.
≈35 ft on either side of the
center on the major axis
b. Find the location of the foci.
y2
x2
−
+−=1
1764
529
a. Write an equation modeling the shape
of the room. Assume that it is centered
at the origin and that the major axis is
horizontal.
13
DATE
PERIOD
0y
ds
70
Ho
us
e
100
100
x
x2 + y2 = 40,000;
x2 + y2 = 810,000
Lesson 7-2
10/24/09 4:06:33 PM
Glencoe Precalculus
c. Write an equation for the circle of
water at the current time and an
equation for the circle when the water
reaches the house.
7 days
b. If the pond continues to overflow at
the same rate, how many days will it
take for the water to reach the house?
; 900 yd
−100
−100 0
y
a. Graph the circle that represents the
water, and find the distance from the
center of the pond to the house.
ds
0y
20
increases the radius 100 yards per
day, as shown below.
4. RETENTION POND A circular retention
pond is getting larger by overflowing and
flooding the nearby land at a rate that
Word Problem Practice
1. WHISPERING GALLERY A whispering
gallery at a museum is in the shape of an
ellipse. The room is 84 feet long and
46 feet wide.
7-2
NAME
Answers (Lesson 7-2)
⎪ by ⎥
ellipse
y 2
x 2
b. −
+ − =1
2
circle
A6
8
y
⎪−2x ⎥ + ⎪−3 ⎥
6
⎪−2x ⎥ ⎪ ⎥
8
=1
Glencoe Precalculus
005_020_PCCRMC07_893808.indd 14
Chapter 7
14
a rectangle that is 6 units long and
4 units wide, centered at the origin
approximate for greater and greater even,
whole-number values of n?
⎪⎥ ⎪⎥
y n
x n
+ − =1
3. What shape will the graph of −
2
3
c. a = 2, b = 3, n = 8
=1
b. a = 2, b = 3, n = 6
6
y
+ −
3
y
=1
4
⎪−2x ⎥ + ⎪−3 ⎥
a. a = 2, b = 3, n = 4
4
2. In each of the following cases, you are given
values of a, b, and n to use in the general equation.
Write the resulting equation. Then graph.
Sketch each graph on the grid at the right.
⎪3⎥ ⎪2⎥
2
⎪2⎥ ⎪2⎥
x
a. −
+ − =1
y
1. Consider two curves that are not superellipses.
Graph each equation on the grid at the right.
State the type of curve produced each time.
For even values of n greater than 2, the curves are called
superellipses.
x
for the family is ⎪−
a ⎥ + − = 1, with a ≠ 0, b ≠ 0, and n > 0.
n
The circle and the ellipse are members of an interesting family
of curves that were first studied by the French physicist and
mathematician Gabriel Lamé (1795-1870). The general equation
n
Enrichment
DATE
1
0
1
0
1
1
x
x
Glencoe Precalculus
y
y
PERIOD
3/13/09 4:23:56 PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Chapter 7
Superellipses
7-2
NAME
DATE
(y + 2)2
32
2
2
18
Take the square root of each side.
Divide each side by 18.
Subtract 32(x - 3)2 from each side.
Multiply each side by 576.
Original equation
PERIOD
3
)
2nd
)
1
)
(–)
÷ 18
[{]1 ,
—
2nd
2
ENTER
[}]
2nd
ZOOM
6.
[ √ ]
(
576
—
32
(
32
1–4. See students’ work.
18
32
(y + 6)2
(x - 3)2
4. − + − = 1
(y + 2)2
(x + 1)2
2. − + − = 1
18
32
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
005_020_PCCRMC07_893808.indd 15
Chapter 7
15
Sample answer: the graph will move up or down.
Lesson 7-2
3/13/09 4:24:01 PM
Glencoe Precalculus
6. Describe the effects of replacing y + 2 in Exercises 1 and 2 with (y ± c) for c > 0.
Sample answer: the graph will move right or left.
5. Describe the effects of replacing x - 3 in Exercises 3 and 4 with (x ± c) for c > 0.
18
32
(y - 2)2
(x - 3)2
3. − + − = 1
18
(x - 7)2
(y + 2)2
1. − + − = 1
Graph each equation on a graphing calculator.
Exercises
Like other graphs, there are families of ellipses. Changing certain values
in the equation of an ellipse creates a new member of that family.
—
Y=
The result is two equations. To graph both equations in Y1, replace ±
with {1, -1}. Be careful to include the proper parentheses or you will get
an error message.
√
576 - 32(x - 3)2
y=± − -2
576 - 32(x - 3)2
(y + 2)2 = −
18
18(y + 2) = 576 - 32(x - 3)
2
32(x - 3) + 18(y + 2) = 576
2
− + − =1
(x - 3)2
18
graphing calculator, you must first solve for y.
(x - 3)2
18
To graph an ellipse, such as − + − = 1, on a
(y + 2)2
32
Graphing Calculator Activity
Translations of Ellipses
7-2
NAME
Answers (Lesson 7-2)
Chapter 7
Hyperbolas
Study Guide and Intervention
DATE
PERIOD
b
b
a
y2
16
2
4
Simplify.
a = 4 and b = 2
A7
(0, 0)
vertices: (0, 4), (0, -4)
-5.65, 5.65
2
−8
4
4
16
4
8x
−8
4
−4 0
8x
3/18/09 11:56:28 PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Glencoe Precalculus
005_020_PCCRMC07_893808.indd 16
4
Glencoe Precalculus
−8
−4 0
−8
−8
−8
8x
8
−4
4
36
y
16
y
(x - 1)2
−4
y
8x
(y + 2)2
3. − - − = 1
4
b
a
y - k = ±−
(x - h)
(y - 3)2
(x + 2)2
2. − - − = 1
25
9
−8
−4 0
−4
−4 0
Chapter 7
−8
8
y2
x2
1. −
-−=1
25
36
Graph the hyperbola given by each equation.
Exercises
-4.5, 4.5
1
-4.5, 4.5
-4, 4
-1
0
y
-5.65, 5.65
x
-2
y
asymptotes: y = 2x, y = -2x
(h, k ± a)
8
foci:
(h, k)
Make a table of values to sketch the hyperbola.
center:
(0, √
20 ), (0, - √
20 )
Equation relating a, b, and c
Determine the characteristics of the hyperbola.
c2 = a2 + b2
c2 = 42 + 22
or about 4.47
c = √20
Because a2 = 16 and b2 = 4, a = 4 and b = 2. Use the values of a and b to
find the value of c.
(h, k ± c)
x
Graph the hyperbola given by the equation − - −
= 1.
The equation is in standard form. Both h and k are 0, so the center is at the
origin. Because the x-term is subtracted, the transverse axis is vertical. Use
the values of a, b, and c to determine the vertices and foci of the hyperbola.
Example
cases, a2 + b2 = c2.
(y - k)2 (x - h2)
−
-−
= 1 when the transverse axis is vertical. In both
2
2
a
Analyze and Graph Hyperbolas A hyperbola is the locus of all
points in a plane such that the difference of their distances from two foci is
constant. The standard form of the equation of a hyperbola is
(y - k)2
(x - h)2
−
- −
= 1 when the transverse axis is horizontal, and
2
2
7-3
NAME
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
DATE
Hyperbolas
Study Guide and Intervention
(continued)
PERIOD
hyperbola
greater than 0
Use the discriminant to identify each conic section.
parabola
equal to 0
2
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Answers
005_020_PCCRMC07_893808.indd 17
Chapter 7
parabola
7. 25x2 + 100x − 54y = −200
ellipse
5. 5x2 + 2xy + 4y2 + x + 2y + 17 = 0
hyperbola
3. -2x + 6xy + y - 4x - 5y + 2 = 0
2
circle
1. 4x2 + 4y2 - 2x - 9y + 1 = 0
17
hyperbola
Lesson 7-3
10/24/09 4:28:42 PM
Glencoe Precalculus
8. 16x2 + 100x − 54y2 = −100
parabola
6. x2 + 2xy + y2 + x + 10 = 0
hyperbola
4. x2 + 6xy + y2 - 2x + 1 = 0
ellipse
2. 10x2 + 6y2 - x + 8y + 1 = 0
Use the discriminant to identify each conic section.
Exercises
A is 12, B is 12, and C is 3. Find the discriminant.
B2 - 4AC = 122 - 4(12)(3) or 0
The discriminant is 0, so the conic is a parabola.
c. 12x2 + 12xy + 3y2 - 7x + 2y - 6 = 0
A is 5, B is 8, and C is -2. Find the discriminant.
B2 - 4AC = 82 - 4(5)(-2) or 104.
The discriminant is greater than 0, so the conic is a hyperbola.
b. 5x2 + 8xy - 2y2 + 4x - 3y + 10 = 0
A is 2, B is 0, and C is 6. Find the discriminant.
B2 - 4AC = 02 - 4(2)(6) or -48
The discriminant is less than 0, so the conic must be either a
circle or an ellipse. Because A ≠ C, the conic is an ellipse.
a. 2x2 + 6y2 - 8x + 12y - 2 = 0
Example
circle
ellipse
less than 0; B ≠ 0 or A ≠ C
Conic Section
less than 0; B = 0 and A = C
Discriminant
You can determine the type of conic when the
equation for the conic is in general form, Ax2 + Bxy + Cy2 + Dx + Ey + F = 0.
2
The discriminant, or B - 4AC, can be used to identify a conic when the
equation is in general form.
Identify Conic Sections
7-3
NAME
Answers (Lesson 7-3)
Hyperbolas
Practice
x
y2
16
(x – 1)2
0
4
y
2. − – − = 1
x
PERIOD
A8
16
9
(y - 1)2
x2
− -−
=1
4. foci (0, 6), (0, –4); length of transverse
axis 8 units
121
y2
5,759,600
Glencoe Precalculus
005_020_PCCRMC07_893808.indd 18
Chapter 7
parabola
9. 4x2 + 8xy + 4y2 + x + 11y + 10 = 0
ellipse
7. 5x2 + xy + 2y2 - 5x + 8y + 9 = 0
18
ellipse
Glencoe Precalculus
10. 2x2 + 4y2 - 3x - 6y + 2 = 0
hyperbola
8. 16x2 − 4y2 - 8x - 8y + 1 = 0
Use the discriminant to identify each conic section.
1,210,000
x
−
- − =1
2
6. ENVIRONMENTAL NOISE Two neighbors who live one mile apart hear
an explosion while they are talking on the telephone. One neighbor hears
the explosion two seconds before the other. If sound travels at 1100 feet
per second, determine the equation of the hyperbola on which the
explosion was located.
√
157
−
6
36
5. Determine the eccentricity of the hyperbola given by the equation
(y + 10)2
(x - 7)2
− - − = 1.
(y - 6)2
(x + 3)2
− - − =1
49
32
3. vertices (–10, 6), (4, 6);
foci (–12, 6), (6, 6)
Write an equation for the hyperbola with the given characteristics.
0
y
1. x2 - 4y2 - 4x + 24y - 36 = 0
DATE
3/13/09 4:24:16 PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Chapter 7
Graph the hyperbola given by each equation.
7-3
NAME
Hyperbolas
(x – 50)2
(y – 35)2
y
40
80
x
2
y
8
x
16
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
005_020_PCCRMC07_893808.indd 19
Chapter 7
−
√
317
14
c. Find the eccentricity.
11
14
y = ±−
x
b. Write the equations of the asymptotes.
−8
−16 −8 0
8
16
a. Graph the hyperbola.
121
x
by − - −
= 1.
y
196
2
2. SHADOWS A lamp projects light onto
a wall in the shape of a hyperbola.
The edge of the light can be modeled
(-14, 35), (114, 35)
b. Find the locations of the seismographs.
0
20
40
80
a. Graph the hyperbola.
− - − = 1, where the
1600
2500
seismographs are located at the foci.
DATE
PERIOD
(3, 4)
x
4
9
19
2
y2
52
y2
144
−8
−4
−16 −8 0
4
8
y
Sample answer:
8
b. Graph one hyperbola.
144
2
Lesson 7-3
10/24/09 4:06:46 PM
Glencoe Precalculus
16 x
52
x
x
−
- − = 1; − - −
=1
a. Write two equations of the hyperbola
in standard form if the center is at the
origin, given that the path contains
a transverse axis of 24 millimeters
with one focus 14 millimeters from
the center.
4. SHADOWS The path of the shadow
cast by the tip of a sundial is usually
a hyperbola.
12 ft
b. If each unit on the coordinate plane
represents 3 feet, what is the narrowest
vertical width of the play area?
(y - 4)2
(x - 3)2
− -− =1
a. Write an equation that models the
curved sides of the play area.
0
y
3. PARKS A grassy play area is in the shape
of a hyperbola, as shown.
Word Problem Practice
1. EARTHQUAKES The epicenter of an
earthquake lies on a branch of the
hyperbola represented by
7-3
NAME
Answers (Lesson 7-3)
Chapter 7
Enrichment
y2
b
DATE
-4
focus
A9
-2
-4
-2
0
-2
-4
y
20
4
focus
x
3/13/09 4:24:30 PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Glencoe Precalculus
005_020_PCCRMC07_893808.indd 20
Chapter 7
2
Glencoe Precalculus
PERIOD
The vertices recede to infinity and the branches become
flatter and farther from the center. As c approaches infinity,
the graphs tend to disappear.
4. Suppose b is held fixed and c approaches infinity. How does the
graph of the hyperbola change?
The vertices remain at (±a, 0), but the branches
become more vertical. The graphs approach
the vertical lines x = -a and x = a.
3. Suppose a is held fixed and c approaches infinity. How does the
graph of the hyperbola change?
The asymptotes remain the same, but the
branches become sharper near the vertices.
The graphs approach the lines y = x and y = -x.
2. Use a graphing calculator or computer to graph x2 - y2 = 1,
x 2 - y 2 = 0.1, and x 2 - y 2 = 0.01. (Such hyperbolas correspond
to smaller and smaller values of c.) Describe the changes in the
graphs. What shape do the graphs approach as c approaches 0?
Because 0 < a < c, as c approaches 0, a
approaches 0. So the x-coordinates of the foci
and vertices approach 0, which is the
x-coordinate of the center. Because the
y-coordinates are equal, the points become the same.
1. Write a convincing argument to show that as c approaches 0,
the foci, the vertices, and the center of the hyperbola become
the same point.
Refer to the hyperbola described above.
What happens to the shape of the graph as c grows
very large or very small?
b
y=±−
a x. Such a hyperbola is shown at the right.
(c, 0), where c 2 = a 2 + b 2, the vertices are at (-a, 0)
and (a, 0), and the asymptotes have equations
a
x
- −2 = 1. The foci are at (-c, 0) and
equation −
2
2
Recall that the equation of a hyperbola with center
at the origin and horizontal transverse axis has the
Moving Foci
7-3
NAME
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
DATE
PERIOD
2
√2
2
4
√2
2
2
√2
x - √2
y
2
√2
x + √2
y
2
− 2xy
√2
2
√2
x + √2
y
2
− 4y2
2
2
1
2
2
5
1
1
-−
(x)2 - 5 x y - −
(y)2 + −
=0
2
2
2
2
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Answers
021_032_PCCRMC07_893808.indd 21
21
Simplify.
binomials.
Expand the
Lesson 7-4
10/24/09 4:10:01 PM
Glencoe Precalculus
7(y)2 = 160; hyperbola
y - 2 = 0; circle
2 √2
Chapter 7
19(x)2 - 26 √
3 xy -
2. 8x2 - 5y2 = 40, θ = 30°
x +
(x)2 + (y)2 - 2 √2
4
π
1. x2 - 4x + y2 = 2, θ = −
Write each equation in the xy-plane for the given value of θ. Then
identify the conic.
Exercises
The equation of the hyperbola after the 45° rotation is 5(x)2 + 10xy+ (y)2 = 1.
2
1
1
1
(x)2 - xy + −
(y)2 - (x)2 + (y)2 - 2 (x)2 - 4xy - 2 (y)2 + −
=0
−
√2
x - √2
y
2
Replace x and y.
2
2
1
+−
=0
√
2
√
2
= − x + − y
y = x sin θ + y cos θ
( − ) − 2( − )( − ) − 4( − ) + − = 0
x2
4
π
π
sin −
= − and cos −
=−
Rotation equations for x and y
Substitute into the original equation.
√2
2
= − x − − y
x = x cos θ − y sin θ
The conic is a hyperbola because B2- 4AC > 0. Find the equations for x and y.
Example
Use θ = 45° to write the standard form of
1
x2 - 2xy - 4y2 + −
= 0 in the xy-plane. Then identify the conic.
An equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0
in the xy-plane can be rewritten as A(x)2 + C(y)2 + Dx + Ey + F = 0 in the xy-plane
by rotating the coordinate axes through an angle θ. The equation in the xy-plane can be
found using x = x cos θ - y sin θ and y = x sin θ + y cos θ.
Rotations of Conic Sections
Study Guide and Intervention
Rotation of Conic Sections
7-4
NAME
Answers (Lesson 7-3 and Lesson 7-4)
Rotations of Conic Sections
Study Guide and Intervention
DATE
2
(x)
(y)
Graph − + − = 1 if it has been rotated 30° from
36
64
2
2
√3
2
1
sin 30° = −
and cos 30° = −
Rotation equations for x and y
A10
Multiply.
x = 0 and y = 0
√3
2
=0
Glencoe Precalculus
−8
−4
y
021_032_PCCRMC07_893808.indd 22
Chapter 7
−8 −4
y'
4
−8
8x
22
y'
−8
−4 0
4
y
4
x'
8x
(y)2
(x)2
2. − - − = 1; 60°
4
9
x'
2
π
1. (x + 6)2 = 12(y + 2); −
Graph each equation at the indicated angle.
Exercises
The new center, vertices, and co-vertices can be used to sketch the
graph of the ellipse in the xy-plane.
4
Glencoe Precalculus
−8
−4
−8 −4 0
4
8
y
8x
√3
1
y=−
(0) + − (0)
2
2
2
1
=−
x + − y
y = x sin θ + y cos θ
, 3) (-6, 0) → (-3 √
(0, 8) → (−4, 4 √
3 ) (0, -8) → (4, -4 √
3 ) (6, 0) → (3 √3
3 , −3)
Likewise, convert the coordinates of the vertices and co-vertices
into xy-coordinates.
=0
√3
1
x = − (0) - −
(0)
2
2
Use the equations to convert the xy-coordinates of the center into
xy-coordinates.
1
= − x - −
y
√3
2
x = x cos θ - y sin θ
Find the equations for x and y for θ = 30°.
The equation represents an ellipse, and it is in standard form. Use the
center (0, 0); the vertices (0, 8), (0, -8); and the co-vertices (6, 0), (-6, 0)
to determine the center and vertices for the ellipse in the xy-plane.
its position in the xy-plane.
Example
2
When the equations of rotated conics are given
for the xy-plane, they can be graphed by finding points on the graph of the
conic and then converting these points to the xy-plane.
(continued)
PERIOD
10/24/09 4:29:05 PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Chapter 7
Graph Rotated Conics
7-4
NAME
Rotations of Conic Sections
Practice
DATE
PERIOD
(x)2
25
(y)2
4
-71x + 58 √
3 xy - 13y2 = 400
2
π
4. − - − = 1; θ = −
3(x)
4
(y)
4
2
(y)2
3
(x)2
9
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
021_032_PCCRMC07_893808.indd 23
Chapter 7
y'
y
x'
x
23
10. Graph (x + 1) 2 = -16(y + 3) if it has been rotated 45° from its position
in the xy-plane.
45°
(y)2
22
33.7°; − + − = 1
Lesson 7-4
10/24/09 4:10:10 PM
Glencoe Precalculus
8. 18x2 + 12xy + 13y2 - 198 = 0
(x)2
9
63°; − + − = 1
6. 13x2 − 8xy + 7y2 - 45 = 0
9. COMMUNICATIONS Suppose the orientation of a satellite
dish that monitors radio waves is modeled by the equation
4x 2 + 2xy + 4y2 + √
2 x - √
2 y = 0. What is the angle of
rotation of the satellite dish about the origin?
53°; (y)2 = 2x
7. 16x2 − 24xy + 9y2 - 30x - 40y = 0
45°; − + − = 1
2
5. x2 + xy + y2 = 2
Determine the angle θ through which the conic with each equation
should be rotated. Then write the equation in standard form.
y
x + 16 √
2 x + 2xy - 16 √2
+ y2 = 0
2
3. (x)2 = 16(y); θ = 45°
3
120 = 0; ellipse
hyperbola
Write an equation for each conic in the xy-plane for the given
equation in xy form and the given value of θ.
xy + 23(y)2 21(x)2 + 2 √3
2
2. 5x2 + 6y2 = 30; θ = 30°
1
1
−
(x)2 - −
(y)2 - 1 = 0;
2
4
π
1. xy = 1; θ = −
Write each equation in the xy-plane for the given value of θ. Then
identify the conic.
7-4
NAME
Answers (Lesson 7-4)
Chapter 7
A11
(x)2
16
y
(y)2
20
60˚
x'
x
x
24
)
45˚
x
x'
elliptical
Glencoe Precalculus
5. SHAPES The shape of a reflecting
mirror in a telescope can be modeled by
25x2 + 13xy + 2y2 = 100. Determine
whether the reflector is elliptical,
parabolic, or hyperbolic.
40° counterclockwise
4. LOGIC A hyperbola has been rotated
40° clockwise. Through what angle
must it be rotated to return it to its
original position?
y'
y
c. Draw the graph in the xy-plane.
-x
y = 3 √2
b. Find the equation for the axis of
symmetry in the xy-plane.
(
√
2 7 √2
- −, −
2
2
a. Find the vertex in the xy-plane.
3. ART Mimi is drawing a picture with
rotated parabolas. She wants to graph
(x - 3)2 = 12(y - 4) if it has been
rotated 45° from the xy-plane.
PERIOD
3/13/09 4:22:02 PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Glencoe Precalculus
021_032_PCCRMC07_893808.indd 24
0
y
b. Draw the graph.
17x2 + 2 √
3 xy + 19y2 - 320 = 0
a. Write an equation for the ellipse
formed by the gear in the xy-plane.
xy-plane is − + − = 1.
2. GEARS Suppose the equation of
an elliptical gear rotated 60° in the
y'
b. Draw the graph.
(x)2
√
3
3 2
− - − xy + −
y 32
16
32
√
3
1
− x - −
y = 0
2
2
a. Write an equation that models the
new orientation of the satellite dish.
when it is directly overhead. Later in the
day, the dish has rotated 60°.
8
1 2
x
modeled by the equation y = −
Chapter 7
DATE
Rotations of Conic Sections
Word Problem Practice
1. COMMUNICATIONS A satellite dish is
7-4
NAME
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Enrichment
DATE
1
0
y=x+6
y
1
y=
± √4x - x 2
y=6-x
y = 6 - x ± √4x - x 2
x
2. y = x ±
0
1
√
x
y
Answers
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
021_032_PCCRMC07_893808.indd 25
Chapter 7
See students’ graphs.
7 + 6x - x2 ellipse;
3. y = 2x ± √
1
#
#
y = ± √x
1
y = x ± √x
25
See students’ graphs.
x
Lesson 7-4
3/13/09 4:22:08 PM
Glencoe Precalculus
x
y=x
4. y = - 2x ± √
-2x parabola;
"
y = ± √4x - x 2
y = x + 6 ± √4x - x 2
"
PERIOD
parabola
Use a separate sheet of graph paper to graph these equations.
State the type of curve for each graph.
0
1
y
1. y = 6 - x ± √
4 - x2 ellipse
Graph each equation. State the type of curve for each graph.
Often, however, you can use the graphs of two
simpler equations to graph a more complicated
equation. For example, the graph of the ellipse in
the diagram at the right is obtained by adding the
y-coordinate of each point on the circle and the
y-coordinate of the corresponding point of the line.
Equations of parabolas, ellipses, and hyperbolas
that are “tipped” with respect to the x- and y-axes
are more difficult to graph than the equations you
have been studying.
Graphing with Addition of y-Coordinates
7-4
NAME
Answers (Lesson 7-4)
DATE
Parametric Equations
Study Guide and Intervention
PERIOD
A01_A14_PCCRMC07_893808.indd 12
-15
-11
-7
-3
-3
-2
-1
0
1
2
3
12
7
4
A12
x+2
4
13
9
5
1
19
12
7
4
3
y
−8
−16 −8 0
8
16
24
y
8
x+2
4
4
4
Pdf Pass
4
021_032_PCCRMC07_893808.indd 26
Chapter 7
−12
−8
−4
−4 0
y
8
12
16
6
20
x
1. x = t2 + 4 and y = −t - 3; -4 ≤ t ≤ 4
26
0
3
y
x
Glencoe Precalculus
2. x = −t and y = √t + 2; 0 ≤ t ≤ 8
Sketch the curve given by each pair of parametric equations over
the given interval.
Exercises
x2
x
5
The rectangular equation is y = −
+−
+−
.
16
4
4
4
Simplify.
2
x
x
5
=−
+−
+−
x+2
4
Substitute − for t in the equation for y.
Solve for t.
Square − .
16
16x
Parametric equation for x
= − +1
x2 + 4x + 4
16
⎛ x + 2 ⎞2
y=⎪−⎥ +1
⎝ 4 ⎠
−=t
x
-3
Plot the (x, y) coordinates for each t-value and
connect the points to form a smooth curve.
Write x = 4t - 2 and y = t2 + 1 in rectangular form.
4
0
3
t
y
19
x = 4t - 2
Example 2
x
-19
t
-4
Make a table of values for -4 ≤ t ≤ 4.
10/24/09 4:10:41 PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Chapter 7
Example 1
Sketch the curve given by the parametric equations
x = -3 + 4t and y = t2 + 3 over the interval -4 ≤ t ≤ 4.
Parametric equations are used to
describe the horizontal and vertical components of an equation. Parameters
are arbitrary values, usually time or angle measurement.
Graph Parametric Equations
7-5
NAME
DATE
(continued)
PERIOD
48˚
v0 = 35, θ = 48°, g = 32, and h0 = 2
Parametric equation for vertical position
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
021_032_PCCRMC07_893808.indd 27
Chapter 7
27
Lesson 7-5
11/15/09 4:17:20 PM
Glencoe Precalculus
2. Jerome hits a tennis ball at an initial velocity of 38 feet per second and an
angle of 42° with the horizontal from a height of 1.5 feet. How far away
will the ball land if it is not hit by his opponent? about 46.6 ft
1. Julie is throwing a ball at an initial velocity of 28 feet per second and an
angle of 56° with the horizontal from a height of 4 feet. How far away will
the ball land? about 25 ft
Exercises
Step 4 Determine the horizontal position of the ball at 1.7 seconds.
x = tv0 cos θ
Parametric equation for horizontal position
= 1.7(35) cos 48
v0 = 35, θ = 48°, and t = 1.7
≈ 39.8
Use a calculator.
The ball will travel about 39.8 feet before it hits the ground.
Step 3 Graph the equation for the vertical position and the line y = 0. Use
5: INTERSECT on the CALC menu of a calculator to find the point of
intersection of the curve with y = 0. The value is about 1.7 seconds.
You could also use 2: ZERO and not graph y = 0.
2
1
= t(35) sin (48) - −
(32)t2 + 2
2
1 2
gt + h0
y = tv0 sin θ - −
Step 2 Write a parametric equation for the vertical position of the ball.
2 ft
35 ft/sec
Step 1 Make a diagram of the situation.
Example
Luigi is kicking a soccer ball. He kicks the ball with
an initial velocity of 35 feet per second at an angle of 48° with the
horizontal. The ball is 2 feet above the ground when he kicks it. How
far will the ball travel horizontally before it hits the ground?
2
1 2
and the vertical position y by y = tv0 sin θ - −
gt + h0.
Parametric equations are often used to simulate
projectile motion. For an object launched at an angle θ with the horizontal
at an initial velocity v0, where g is the gravitational constant, t is time, and
h0 is the initial height, the horizontal distance x can be found by x = tv0 cos θ
Parametric Equations
Study Guide and Intervention
Projectile Motion
7-5
NAME
Answers (Lesson 7-5)
Glencoe Precalculus
3/16/10 9:37:55 PM
Chapter 7
Parametric Equations
Practice
DATE
PERIOD
y
6
10
14x
A13
2
25
x
−+−
=1
y2
16
6. y = 4 sin θ, x = 5 cos θ
y = -3(x - 5)
2
4. x = t + 5, y = -3t2
−12
−8
−4
y
t = −1
4
8x
t=0
2
t = −3
t = −5
0
t=3
t=1
t = −1
y
x
8. t = 4x - 1 for y = x2 + 2
28
Glencoe Precalculus
3/13/09 4:22:25 PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Glencoe Precalculus
021_032_PCCRMC07_893808.indd 28
Chapter 7
0.72 s; 48.43 ft
b. How long will it take the rocket to travel 10 feet horizontally from its
starting point? What will be its vertical distance at that point?
x = 80t cos 80°; y = 80t sin 80° - 16t2
a. Write parametric equations to represent the path of the rocket.
9. MODEL ROCKETRY Manuel launches a toy rocket from ground level
with an initial velocity of 80 feet per second at an angle of 80° with
the horizontal.
t=2
−8
t=1
−4 0
3
2-x
3-x
7. t = −
for y = −
2
Use each parameter to write the parametric equations that can
represent each equation. Then graph the equation, indicating the
speed and orientation.
y2
x2
−
+−=1
9
4
5. x = 3 sin θ, y = 2 cos θ
1
11
y=−
x-−
2
2
3. x = 2t + 3, y = t - 4
Write each pair of parametric equations in rectangular form.
−8
−4 0 2
−12
x
−24
24
−16
16
2
−4
8
2
2
t
2. x = 2t + 6 and y = - −
; -5 ≤ t ≤ 5
−8
−8 0
y
2
1. x = t2 + 1 and y = −t - 6; -5 ≤ t ≤ 5
Sketch the curve given by each pair of parametric equations over
the given interval.
7-5
NAME
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Parametric Equations
Answers
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
021_032_PCCRMC07_893808.indd 29
Chapter 7
vertical: y = t(28) sin (60)
- 16t2 + 5;
horizontal: x = t(28) cos (60)
4. BASKETBALL Mandy throws a
basketball with an initial velocity of
28 feet per second at an angle of 60° with
the horizontal. If Mandy releases the ball
from a height of 5 feet, write a pair of
equations to determine the vertical and
horizontal positions of the ball.
Yes, at 20 feet the ball is about
3.8 feet above the ground.
3. TENNIS Melinda hits a tennis ball with
an initial velocity of 42 feet per second at
an angle of 16° with the horizontal from
a height of 2 feet. She is 20 feet from the
net and the net is 3 feet high. Will the
ball go over the net?
8.97 ft
b. What is the maximum height
of the ball?
2.53 ft
a. How high above the ground will the
ball be when it gets to Sarah?
2. PLAYING CATCH Tom and Sarah are
playing catch. Tom tosses a ball to Sarah
at an initial velocity of 38 feet per second
at an angle of 28° from a height of 4 feet.
Sarah is 40 feet away from Tom.
1.98 m
29
DATE
PERIOD
34˚
Lesson 7-5
3/13/09 4:22:33 PM
Glencoe Precalculus
Sample answer: Yes; at 400 ft
the ball is 15.4 feet above the
ground so it will clear the
fence.
c. If the fence is 8 feet tall and 400 feet
from home plate, will the ball clear
the fence to be a home run? Explain.
73.36 ft
b. What is the maximum height the
ball will reach?
421.6 ft
a. How far will the ball travel
horizontally before it hits the
ground?
3 ft
120 ft/sec
6. BASEBALL Micah hit a baseball at an
initial velocity of 120 feet per second
from a height of 3 feet at an angle of 34°.
about 61.9 ft
b. Find the maximum height the ball
reaches.
x = 100t cos 39°,
y = 100t sin 39° - 16t2
a. Write parametric equations for the
flight of the ball.
5. GOLF Julio hit a golf ball with an
initial velocity of 100 feet per second at
an angle of 39° with the horizontal.
Word Problem Practice
1. PHYSICS A rock is thrown at an initial
velocity of 5 meters per second at an
angle of 8° with the ground. After
0.4 second, how far has the rock traveled
horizontally?
7-5
NAME
Answers (Lesson 7-5)
Enrichment
DATE
Time
x
A14
g
cos α
)
Glencoe Precalculus
021_032_PCCRMC07_893808.indd 30
Chapter 7
y = -0.013x 2 + 5.67x
1. a projectile fired at 80° to the
horizontal with an initial velocity
of 200 ft/s
30
Glencoe Precalculus
y = -0.00037x 2 + 0.84x
2. a projectile fired at 40° to the
horizontal with an initial velocity
of 150 m/s
Find the equation of the path of each projectile.
y = -0.00035x 2 + 0.18x
9.8
y=- −
x 2 + (tan 10°)x
2
2
( 2(120) cos 10° )
Example
Write the equation of a projectile fired at an angle of
10° to the horizontal with an initial velocity of 120 m/s.
where g is the acceleration due to gravity, 9.8 m/s 2 or 32 ft/s 2,
v 0 is the initial velocity, and α is the angle at which the
projectile is fired.
0
x 2+ (tan α)x,
y=- −
2
2
( 2v
The equation of the path of a projectile on the coordinate plane is
given by
The path assumes that gravity is the only force acting on the
projectile.
0
Height
y
The path of a projectile after it is launched is a parabola when graphed
on a coordinate plane.
PERIOD
3/13/09 4:22:37 PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Chapter 7
Coordinate Equations of Projectiles
7-5
NAME
Spreadsheet Activity
DATE
2
4
6
8
10
2
3
4
5
6
B
-4.75
-0.99302
2.935661
6.356741
8.676882
x
C
8.209921
9.428068
9.016016
7.045013
3.855863
y
y
15x
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
021_032_PCCRMC07_893808.indd 31
Chapter 7
= 6*SIN(A2*3) and
= -4*COS(A2)
31
3. Given the parametric equations x = 6 sin 3t and y = -4 cos t, write the
formulas to put into your spreadsheet if time is found in cell A2.
(0.97, -9.43)
Lesson 7-5
3/13/09 4:22:41 PM
Glencoe Precalculus
In the spreadsheet, the values in column A
represent t, and the values in columns B and C
represent the calculated x and y values of Saturn’s
position at time t.
2. Find the coordinates representing Saturn’s position after
23 years.
−15
0 5
5
15
PERIOD
Use the parametric equations for Saturn’s
position in its orbit in a spreadsheet. To
calculate the x position, enter the following
formula in B2: = 9.5*COS(PI()15*A2). To
calculate the y position, enter the
formula = 9.48*SIN(PI()15*A2) in C2. Use
the fill handle to paste these formulas into
the remaining cells in columns B and C.
1. Complete the spreadsheet to find the position of Saturn every
2 years from t = 2 to t = 32. Use the spreadsheet to make an
X-Y graph of Saturn’s position.
Exercises
t
1
A
You have learned that the motion of orbiting planets can be modeled
using parametric equations. A spreadsheet can be used to evaluate
parametric equations and to provide a graph of a planet’s orbit.
Parametric Equations
7-5
NAME
Answers (Lesson 7-5)
Chapter 7 Assessment Answer Key
(Lessons 7-1 and 7-2)
Page 33 2
(y + 1)
(x - 3) 2
− + − = 1;
16
4
1. ellipse
y
2.
(6, 9)
6
(14, -1)
0
−6
(-2, -1)
−12
(6, -1)
12
18 x
Quiz 3
(Lesson 7-4)
Page 34
ellipse;
3 xy +
31(x) 2- 10 √
2
21(y) - 144 = 0
(x + 2)
(y - 5) 2
−+− =1
11
D
5.
4.
Quiz 4 (Lesson 7-5)
Page 33
Page 34
y
8
1.
1.
−4
−8
2
(y + 1)
(x + 1)
− - − =1
3.
6 √2
−
11
4.
hyperbola
B
4.
H
0 x
x
0
7
3.
y
4
−12 −8 −4
9
F
D
Quiz 2 (Lesson 7-3)
2.
2.
11x 2- 10 √
3 xy +
2
y + 144 = 0
5.
2
B
14˚
3.
x = 2; y = 1
36
1.
2.
(6, -11)
2
4.
Page 35
hyperbola; x 2 2
1.4xy - y - 12 = 0
(2, -1); (2,-3);
3.
Mid-Chapter Test
2
2.
3.
(y - 1) 2 (x - 2) 2
− - − =1
5.
D
5.
Chapter 7
16
25
y2
49
x
−
+−=1
25
D
about 2.33 s
A15
x 2 = -32y
6.
y
x = 65t cos 35˚,
y = 65t sin 35˚- 16t 2;
4. yes
5.
Answers
Quiz 1
12
(-1, 6)
7.
−8
−4
8 (-1, 9)
4
0
(-1, 3)
8x
Glencoe Precalculus
Chapter 7 Assessment Answer Key
Vocabulary Test
Page 36
Form 1
Page 37
1.
Page 38
10.
H
11.
B
12.
G
13.
B
14.
J
15.
C
16.
F
17.
C
18.
F
B
2.
F
3.
C
true
false; greater than
4.
5.false; transverse axis
4.
G
6. false; parametric
5.
C
true
1.
2
2
2. false; a - b = c
2
3.
true
7.
false; perpendicular
8.
H
true
9.
10.
6.
a
false; ± −
7.
b
11. the set of all points
in a plane such that
the difference of
their distances from
two foci is constant
8.
12. a conic formed by
the intersection of a
double-napped right
cone and a plane
through the vertex of
the cone
9.
Chapter 7
A
F
D
B:
A16
y = 4x - 4
Glencoe Precalculus
Chapter 7 Assessment Answer Key
Page 40
10.
1.
4.
5.
6.
7.
13.
D
14.
J
15.
B
16.
F
G
G
Chapter 7
B
4.
H
5.
C
6.
G
8.
B:
J
11.
D
12.
F
13.
C
14.
H
15.
C
16.
J
17.
D
18.
J
G
A
B
G
B
10.
D
3.
7.
A
18.
9.
J
A
A
Page 42
D
2.
17.
8.
1.
H
12.
3.
J
B
11.
2.
Form 2B
Page 41
J
F
1
y = -−
x-2
2
9.
A17
Answers
Form 2A
Page 39
A
B:
y=x-3
Glencoe Precalculus
Chapter 7 Assessment Answer Key
Form 2C
Page 43
Page 44
(-4, 1)
1.
12.
(y + 3)2
(x + 2)2
3
9
−-− =1
2. (-4, 3), (-4, -1)
3.
(-4 ± 2 √
2 , 1),
13.
(-6, 1), y = 1
3)
(-4, 1 ± 2 √
27°
4.
2
7.
(x + 2)2 = -20(y - 1)
y2
4
x
−
+−=1
64
5.
6.
14.
t+3
x = −,
2
2
t + 6t + 13
y= −
4
2
(x')
− + (y')2 = 1
3
hyperbola
15.
y
16.
16
12
x = 100t cos 35°,
y = 100t sin 35°
+ 4 - 16t2
8.
8
4
−4
Sample answer:
x2 = -48y
9.
11.
8
12
17.
about 0.7 s
18.
about 32.1 ft
19.
hyperbola
20.
10.
0
x
(x - 2)2 +
(y - 7)2 = 225
145 )
(9, 6 ± √
8
y - 6 = ±−
(x - 9)
Chapter 7
9
B:
A18
(x - 2)2
(y + 5)2
9
4
−+−=1
Glencoe Precalculus
Chapter 7 Assessment Answer Key
Form 2D
Page 45
Page 46
(y + 2)2
1.
(3, -1)
2.
(3 ± √
5 , -1)
(6, -1), (0, -1),
(3, 1), (3, -3)
(2, 1), y = 1
13.
63°
4.
y2
x2
−+−=1
100
9
5.
(x - 2)2 = -16(y + 5)
14.
t-4
,
x=−
2
t - 8t + 24
y= −
4
2
6.
7.
(x')2
2
(y')2
26
hyperbola
15.
−+−=1
16
16.
Answers
3.
12.
(x + 1)2
-− =1
−
4
16
y
12
x = 110t cos 32°,
y = 110t sin 32°
+ 3 - 16t2
8.
8
4
−4
4
0
8
12
16x
−4
x2 = -60y
9.
10.
11.
17.
about 0.62 s
18.
about 34.6 ft
19.
ellipse
20.
(x + 6)2 +
(y + 9)2 = 121
(-3 ± √
5 , 2)
y - 2 = ± 2(x + 3)
Chapter 7
B:
A19
(x + 4)2
(y - 1)2
25
4
−+−=1
Glencoe Precalculus
Chapter 7 Assessment Answer Key
Form 3
Page 47
Page 48
(-3, 2)
1.
2.
12.
(x + 2)2
(y - 4)2
25
16
−-− =1
(-3, 2 ± √
5)
(-3, 5), (-3, -1),
3. (-1, 2), (-5, 2)
4.
hyperbola; 2°
5.
y2
x
−
+−=1
4
9
2
t+2
x = −,
3
6.
-t2 - 4t + 32
y= −
9
2
7(x' )
40
7.
13.
14.
15.
(2, -2), y = -2
(y + 5)2 = -40(x - 2)
hyperbola
y
16.
2
11(y' )
40
−+−=1
0 x
x = 38t cos 29°,
8.
y = 38t sin 29°
+ 4 - 16t2;
Sample answer: Yes,
because at 38 feet the
ball is only 4.1 ft above
the ground
17.
18.
about 0.68 ft
Sample answer:
Yes, when the ball
is 20 feet in front
of her it is at a
height of 0.94 feet.
√2
2
−
Sample answer:
125
y
x2 = - −
3
9.
10.
11.
19.
(x - 13)2 +
(y - 12)2 = 325
20.
(-1 ± 2 √
5 , -2)
y + 2 = ±2(x + 1)
Chapter 7
(x + 1)2
(y - 3)2
625
400
−+−=1
B:
A20
Glencoe Precalculus
Chapter 7 Assessment Answer Key
Page 49, Extended-Response Test
Sample Answers
1e. The graph is an ellipse since
1a. The equation is a circle if A = B.
Sample answer:
(-1) 2- 4(3)(2) < 0.
2
2
(x - 1) + (y - 2) = 4
3-2
= - 1, 2θ = 135˚,
cot 2θ = −
-1
θ ≈ 68˚
y
(1, 2)
x
0
2a. The graph of x 2 - 4y 2 = 0 is two lines
1
1
of slope −
and slope - −
that intersect
2
2
at the origin.
x 2 - 4y 2 = 0
x 2 = 4y 2
⎪x⎥ = 2⎪y⎥
1b. The equation is an ellipse if A ≠ B
and A and B have the same sign.
Sample answer:
x
-x
−
= y or −
=y
2
2
2b.
y
2
y
y = 2x
x
x
0
y = - 2x
2c. degenerate hyperbola
1c. The equation is a parabola when A
or B is zero, but not both.
Sample answer: y - 2 = 4(x + 1) 2
3. Sample answer: Most lamps with
circular shades shine a cone of light.
When this light cone strikes a nearby
wall, the resulting shape is a hyperbola.
The hyperbola is formed by the cone of
light intersecting the plane of the wall.
y
(reflected light)
(-1, 2)
0
x
1d. The equation is a hyperbola if A and
B have opposite signs. Sample answer:
(reflected light)
2
y
x2
−-−
=1
4
Chapter 7
1
A21
Glencoe Precalculus
Answers
(y - 1) 2
(x + 2)
− + − =1
4
1
⎪ x⎥
− = ⎪y⎥
2
Chapter 7 Assessment Answer Key
Standardized Test Practice
Page 50
1.
A
B
C
D
2.
F
G
H
J
3.
A
B
C
D
4.
F
G
H
A
B
C
D
6.
F
G
H
J
8.
A
F
B
G
Chapter 7
C
H
9.
A
B
C
D
10.
F
G
H
J
11.
A
B
C
D
12.
F
G
H
J
13.
A
B
C
D
14.
F
G
H
J
15.
A
B
C
D
16.
F
G
H
J
17.
A
B
C
D
18.
F
G
H
J
19.
A
B
C
D
J
5.
7.
Page 51
D
J
A22
Glencoe Precalculus
Chapter 7 Assessment Answer Key
Standardized Test Practice
(continued)
Page 52
2
y2
16
x
−
+− = 1
100
20.
y
21.
x
0
(x - 4)2 = 8(y - 9)
Answers
22.
26.57°,
5(x') - √
5 y' = 13
2
23.
24.
(1, -4)
25.
25.4 mi
log 6
4
−
log4 e
26.
27.
(-4, 1), (3, ∞)
x = 85t cos 34°,
y = 85t sin 34° +
3 - 16t2
28a.
about 38.3 ft
28b.
28c.
about 213.5 ft
Chapter 7
A23
Glencoe Precalculus