center: (–2, –4)
Practice Test - Chapter 7
Write an equation for an ellipse with each set of
characteristics.
1. vertices (7, –4), (–3, –4); foci (6, –4), (–2, –4)
SOLUTION: The distance between the vertices is 2a.
2a = |7 − (–3)|
2
a = 5; a = 25
The distance between the foci is 2c.
2c = |6 – (–2)|
c =4
+
=1
3. MULTIPLE CHOICE What value must c be so
2
2
that the graph of 4x + cy + 2x – 2y – 18 = 0 is a
circle?
A –8
B –4
C 4
D8
SOLUTION: The graph of a second degree equation of the form
Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 is a circle if
B2 – 4AC < 0 and B = 0 and A = C. Since A = 4,
C must also equal 4.
The correct answer is C.
Write each pair of parametric equations in
rectangular form. Then graph the equation.
4. x = t – 5 and y = 3t – 4
= 2
center: (2, –4)
SOLUTION: +
=1
2. foci (–2, 1), (–2, –9); length of major axis is 12
SOLUTION: The distance between the foci is 2c.
2c = |1 − (–9)|
c =5
The length of the major axis is 2a.
2a = 12
2
a = 6; a = 36
Solve for t.
x =t−5
x +5 =t
Substitute for t.
y = 3t − 4
y = 3(x + 5) − 4
y = 3x + 15 − 4
y = 3x + 11
Make a table of values to graph y.
x
y
–6
–7
–4
–1
–2
5
0
11
Plot the (x, y) coordinates and connect the points to
form a smooth curve.
= –4
center: (–2, –4)
+
=1
3. MULTIPLE CHOICE What value must c be so
2
2
that Manual
the graph
of 4xby +Cognero
cy +
eSolutions
- Powered
circle?
A –8
2x – 2y – 18 = 0 is a
5. x = t2 – 1 and y = 2t + 1
SOLUTION: Solve for t.
Page 1
Practice Test - Chapter 7
5. x = t2 – 1 and y = 2t + 1
6. BRIDGES At 1.7 miles long, San Francisco’s
Golden Gate Bridge was the longest suspension
bridge in the world when it was constructed.
SOLUTION: Solve for t.
y = 2t + 1
y − 1 = 2t
equation for the design of the bridge. =t
Substitute for t.
2
x
=t −1
x
=
− 1
x +1 =
a. Suppose the design of the bridge can be
modeled by a parabola and the lowest point of the
cable is 15 feet above the road. Write an equation
for the design of the bridge.
b. Where is the focus located in relation to the
vertex?
SOLUTION: x +1
=
4(x + 1) = (y − 1)2
Make a table of values to graph y.
x
y
1
–1
0
–1, 3
3
–3, 5
8
–5, 7
Plot the (x, y) coordinates and connect the points to
form a smooth curve.
6. BRIDGES At 1.7 miles long, San Francisco’s
Golden Gate Bridge was the longest suspension
bridge in the world when it was constructed.
equation for the design of the bridge. a. Since the cable resembles a parabola that opens
up, the standard form of the equation that can be
2
used to model the cable is (x − h) = 4p (y − k). If
the parabola is centered on the y–axis, the vertex is
located at the point (0, 15). One point that lies on the
parabola is (2100, 500). Use the values of h, k, x and y to solve for p .
2
(x − h) = 4p (y − k)
2
(2100 − 0) = 4(p )(500 − 15)
4410000 = 1940p
2273.2 ≈ p
Use the values of h, k and p to write an equation for
the design of the bridge.
2
(x − h) = 4p (y − k)
2
(x − 0) = 4(2273.2)(y − 15)
2
x = 9092.8(y − 15)
b. The distance between the vertex and the focus is
p. Since p was found in part a. to be 2273.2 feet, the
focus is located 2273.2 feet above the vertex.
Write an equation for the hyperbola with the
given characteristics.
7. vertices (3, 0), (–3, 0); asymptotes
SOLUTION: a. Suppose the design of the bridge can be
modeled by a parabola and the lowest point of the
cable is 15 feet above the road. Write an equation
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Manual
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Cognero
for the
design
of thebybridge.
b. Where is the focus located in relation to the
vertex?
Because the y–coordinates of the vertices are the
same, the transverse axis is horizontal, and the
standard form of the equation is
= 1.
–
Page 2
2
= 9092.8(y − 15)
b. The distance between the vertex and the focus is
p. SinceTest
p was-found
in part 7a. to be 2273.2 feet, the
Practice
Chapter
focus is located 2273.2 feet above the vertex.
x
Write an equation for the hyperbola with the
given characteristics.
7. vertices (3, 0), (–3, 0); asymptotes
Using the values of h, k, a, and b, the equation for
the hyperbola is
–
= 1.
8. foci (8, 0), (8, 8); vertices (8, 2), (8, 6)
SOLUTION: Because the x–coordinates of the vertices are the
same, the transverse axis is vertical, and the
SOLUTION: Because the y–coordinates of the vertices are the
same, the transverse axis is horizontal, and the
standard form of the equation is
= 1.
–
standard form of the equation is
–
= 1.
The center is the midpoint of the segment between
the vertices, or (0, 0). So, h = 0 and k = 0. The
slopes of the asymptotes are ±
2
is equal to slope
. So, the positive
, where b = 2, b = 4, a = 3,
The center is the midpoint of the segment between
the foci, or (8, 4). So, h = 8 and k = 4. You can find
c by determining the distance from a focus to the
center. One focus is located at (8, 0) which is 4 units
from (8, 4). So, c = 4. You can find a by determining
the distance from a vertex to the center. One vertex
is located at (8, 2) which is 2 units from (8, 4). So, a
2
= 2 and a = 4.
Now you can use the values of c and a find b.
2
and a = 9.
Using the values of h, k, a, and b, the equation for
the hyperbola is
–
2
2
2
2
2
2
b =c –a
b =4 – 2
2
b = 16 – 4
= 1.
2
b = 12
b =
8. foci (8, 0), (8, 8); vertices (8, 2), (8, 6)
SOLUTION: Because the x–coordinates of the vertices are the
same, the transverse axis is vertical, and the
Using the values of h, k, a, and b, the equation for
standard form of the equation is
The center is the midpoint of the segment between
the foci, or (8, 4). So, h = 8 and k = 4. You can find
c by determining the distance from a focus to the
center. One focus is located at (8, 0) which is 4 units
from (8, 4). So, c = 4. You can find a by determining
the distance from a vertex to the center. One vertex
is located at (8, 2) which is 2 units from (8, 4). So, a
2
= 2 and a = 4.
Now you can use the values of c and a find b.
2
b =c –a
9. 7(x′ – 3) = (y′ )2, θ = 60º
SOLUTION: 2
7(x′ – 3) = (y′ ) , θ =
Use the rotation formulas for x′ and y′ to find the
equation of the rotated conic in the xy–plane.
x′ = x cos θ + y sin θ
2
x+
y
2
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2
= 1.
Write an equation for each conic in the xy–
plane for the given equation in x′y′ form and the
given value of θ.
x′ =
2
–
–
= 1.
the hyperbola is
2
b =4 – 2
2
b = 16 – 4
y′ = y cos θ − x sin θ
Page 3
=
Using the values of h, k, a, and b, the equation for
Practice
Test is- Chapter –7
the hyperbola
= 1.
Write an equation for each conic in the xy–
plane for the given equation in x′y′ form and the
given value of θ.
10. SOLUTION: 9. 7(x′ – 3) = (y′ )2, θ = 60º
Use the rotation formulas for x′ and y′ to find the
equation of the rotated conic in the xy–plane.
x′ = x cos θ + y sin θ
2
7(x′ – 3) = (y′ ) , θ =
Use the rotation formulas for x′ and y′ to find the
equation of the rotated conic in the xy–plane.
x′ = x cos θ + y sin θ
x+
x′ =
y
y′ = y cos θ − x sin θ
y
y′ =
y′ = y cos θ − x sin θ
y−
x+
y′ =
= 1, θ =
+
SOLUTION: x′ =
= 1, θ =
+
y−
x
Substitute these values into the original equation. x
Substitute these values into the original equation.
Graph the hyperbola given by each equation.
11. 10. The equation is in standard form, with h = 0 and k =
= 1, θ =
+
Use the rotation formulas for x′ and y′ to find the
equation of the rotated conic in the xy–plane.
x′ = x cos θ + y sin θ
x+
= 1
SOLUTION: SOLUTION: x′ =
–
= 1, θ =
+
y
2
2
4. Because a = 64 and b = 25, a = 8 and b = 5.
The values of a and b can be used to find c.
2
2
c2 = a + b
c2 = 64 + 25
c=
or about 9.43
Use h, k, a, b, and c to determine the characteristics
of the hyperbola.
y′ = y cos θ − x sin θ
orientation: In the standard form of the equation, the
y–term is being subtracted. Therefore, the
Page 4
orientation of the hyperbola is horizontal.
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y′ =
y−
x
center: (h, k) = (0, 4)
c=
c2 = 4 + 36
or about 9.43
Use h, k, a, b, and c to determine the characteristics
of the hyperbola.
Practice
Test - Chapter 7
orientation: In the standard form of the equation, the
y–term is being subtracted. Therefore, the
orientation of the hyperbola is horizontal.
center: (h, k) = (0, 4)
vertices: (h ± a, k) = (–8, 4) and (8, 4)
foci: (h ± c, k) = (–9.43, 4) and (9.43, 4)
asymptotes:
c=
or about 6.32
Use h, k, a, b, and c to determine the characteristics
of the hyperbola.
orientation: In the standard form of the equation, the
x–term is being subtracted. Therefore, the
orientation of the hyperbola is vertical.
center: (h, k) = (–6, –3)
vertices: (h, k ± a) = (–6, –5) and (–6, –1)
foci: (h, k ± c) = (–6, –9.32) and (–6, 3.32)
asymptotes:
Graph the center, vertices, foci, and asymptotes.
Then make a table of values to sketch the hyperbola.
x
y
–12
–1.59, 9.59
0.25, 7.75
–10
10
0.25, 7.75
12
–1.59, 9.59
12. –
= 1
Graph the center, vertices, foci, and asymptotes.
Then make a table of values to sketch the hyperbola.
x
y
–10
–5.40, –0.60
–8
–5.11, –0.89
–4
–5.11, –0.89
–2
–5.40, –0.60
13. MULTIPLE CHOICE Which ellipse has the
greatest eccentricity?
SOLUTION: The equation is in standard form, with h = –6 and k
2
2
= –3. Because a = 4 and b = 36, a = 2 and b = 6.
The values of a and b can be used to find c.
2
2
c2 = a + b
c2 = 4 + 36
c=
or about 6.32
Use h, k, a, b, and c to determine the characteristics
of the hyperbola.
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orientation: In the standard form of the equation, the
x–term is being subtracted. Therefore, the
orientation of the hyperbola is vertical.
SOLUTION: Page 5
The greater the eccentricity, the more stretched the
graph will appear. Graph D is more circular than the
Practice Test - Chapter 7
13. MULTIPLE CHOICE Which ellipse has the
greatest eccentricity?
stretched. Graph A has vertices at (–8, 0) and (8, 0).
Graph B has vertices at (–6, 0) and (6, 0). Graph C
has vertices at (0, –10) and (10, 0).
The correct answer is C.
Write an equation for and graph a parabola with
the given focus F and vertex V.
14. F(2, 8), V(2, 10)
SOLUTION: SOLUTION: The greater the eccentricity, the more stretched the
graph will appear. Graph D is more circular than the
other three graphs, so it cannot be the answer. The
other three graphs have similar co–vertices located
at either (–2, 0) and (2, 0) or (0, –2) and (0, 2) and
all have centers at the origin. Therefore, the graph
that has vertices farthest from the origin will have
the greatest eccentricity and will be the most
stretched. Graph A has vertices at (–8, 0) and (8, 0).
Graph B has vertices at (–6, 0) and (6, 0). Graph C
has vertices at (0, –10) and (10, 0).
The correct answer is C.
Because the focus and vertex share the same x–
coordinate, the graph is vertical. The focus is (h, k +
p), so the value of p is 8 − 10 or –2. Because p is
negative, the graph opens down.
Write the equation for the parabola in standard form
using the values of h, p , and k.
4p (y – k) = (x – h)2
4(–2)(y – 10) = (x – 2)2
–8(y − 10) = (x − 2)2
2
The standard form of the equation is (x − 2) = –8(y
− 10).
Graph the vertex and focus. Then make a table of
values to graph the parabola.
Write an equation for and graph a parabola with
the given focus F and vertex V.
14. F(2, 8), V(2, 10)
SOLUTION: Because the focus and vertex share the same x–
coordinate, the graph is vertical. The focus is (h, k +
p), so the value of p is 8 − 10 or –2. Because p is
negative, the graph opens down.
Write the equation for the parabola in standard form
using the values of h, p , and k.
4p (y – k) = (x – h)2
4(–2)(y – 10) = (x – 2)2
–8(y − 10) = (x − 2)2
2
The standard form of the equation is (x − 2) = –8(y
− 10).
Graph the vertex and focus. Then make a table of
values to graph the parabola.
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15. F(2, 5), V(–1, 5)
SOLUTION: 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 2 – (–1) or 3. Because p is
positive, the graph opens to the right.
Write the equation for the parabola in standard form
using the values of h, p , and k.
4p (x – h) = (y – k)2
4(3)[x – (–1)] = (y – 5)2
12(x + 1) = (y − 5)2
2
The standard form of the equation is (y − 5) = 12(x
+ 1).
Graph the vertex and focus. Then make a table Page
of 6
values to graph the parabola.
Practice Test - Chapter 7
Graph the ellipse given by each equation.
15. F(2, 5), V(–1, 5)
16. SOLUTION: 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 2 – (–1) or 3. Because p is
positive, the graph opens to the right.
Write the equation for the parabola in standard form
using the values of h, p , and k.
4p (x – h) = (y – k)2
4(3)[x – (–1)] = (y – 5)2
12(x + 1) = (y − 5)2
2
The standard form of the equation is (y − 5) = 12(x
+ 1).
Graph the vertex and focus. Then make a table of
values to graph the parabola.
+
= 1
SOLUTION: The ellipse is in standard form. The values of h and
k are 5 and –3, so the center is at (5, –3).
a=
b=
or 7
or 3
c=
≈ 6.32
orientation: horizontal
vertices: (–2, –3), (12, –3)
covertices: (5, –6), (5, 0)
17. (x + 3)2 +
= 1
SOLUTION: Graph the ellipse given by each equation.
16. +
= 1
SOLUTION: The ellipse is in standard form. The values of h and
k are 5 and –3, so the center is at (5, –3).
a=
b=
The ellipse is in standard form. The values of h and
k are –3 and –6, so the center is at (–3, –6).
a=
b =1
= 9
c=
≈ 8.94
orientation: vertical
vertices: (–3, –15), (–3, 3)
covertices: (–4, –6), (–2, –6)
or 7
or 3
c=
≈ 6.32
orientation: horizontal
vertices: (–2, –3), (12, –3)
covertices: (5, –6), (5, 0)
18. CAMPING In many U.S. parks, campers must
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secure food and provisions from bears and other
animals. One method is to secure food using a bear
bag, which is done by tossing a bag tied to a rope
over a tall tree branch and securing the rope to the
Page 7
tree. Suppose a tree branch is 30 feet above the
ground, a person 20 feet from the branch throws the
bag from 5 feet above the ground.
= t(40) sin 60 − 18. CAMPING In many U.S. parks, campers must
secure food
provisions from
Practice
Testand- Chapter
7 bears and other
animals. One method is to secure food using a bear
bag, which is done by tossing a bag tied to a rope
over a tall tree branch and securing the rope to the
tree. Suppose a tree branch is 30 feet above the
ground, a person 20 feet from the branch throws the
bag from 5 feet above the ground.
a. Will a bag thrown at a speed of 40 feet per
second at an angle of 60º go over the branch?
b. Will a bag thrown at a speed of 45 feet per
second at an angle of 75º go over the branch?
2
(32)t + 5
Graph the equation for the vertical position and the
line y = 30. The curve will intersect the line in two
places. The second intersection represents the bag
as it is moving down toward the branch. Use 5:
intersect function on the CALC menu to find the
second point of intersection with y = 30. The value is
about 1.89 seconds.
SOLUTION: a. To determine whether the bag will go over the
branch, you need the horizontal distance that the bag
has traveled when the height of the bag is 30 feet.
First, write a parametric equation for the vertical
position of the bag.
y
= tv0 sin θ − 2
gt + h 0
2
= t(40) sin 60 − (32)t + 5
Graph the equation for the vertical position and the
line y = 30. The curve does not intersect the line.
Therefore, the height of the bag never reaches 30
feet and the bag will not make it over the branch.
Determine the horizontal position of the bag at 1.89
seconds.
x = tv0 cos θ
= 1.89(45) cos 75
≈ 22.01
The bag will travel a distance of about 22.01 feet
before the height reaches 30 feet a second time.
Since the bag is being thrown from a distance of 20
feet from the tree, the bag will make it over the
branch.
Use a graphing calculator to graph the conic
given by each equation.
19. x2 – 6xy + y 2 – 4y – 8x = 0
SOLUTION: Graph the equation by solving for y.
2
2
x – 6xy + y – 4y – 8x
2
2
y + (–6x − 4)y + x − 8x
=0
=0
b. To determine whether the bag will go over the
branch, you need the horizontal distance that the bag
has traveled when the height of the bag is 30 feet.
First, write a parametric equation for the vertical
position of the bag.
y
= tv0 sin θ − 2
gt + h 0
= t(40) sin 60 − 2
(32)t + 5
Graph the equation for the vertical position and the
line y = 30. The curve will intersect the line in two
places. The second intersection represents the bag
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as it is moving down toward the branch. Use 5:
intersect function on the CALC menu to find the
second point of intersection with y = 30. The value is
[–10, 10] scl: 1 by [–10, 10] scl: 1
Page 8
2
2
20. x + 4y – 2xy + 3y – 6x + 5 = 0
The bag will travel a distance of about 22.01 feet
before the height reaches 30 feet a second time.
Since the bag is being thrown from a distance of 20
feet from
the tree,
the bag will
Practice
Test
- Chapter
7 make it over the
branch.
Use a graphing calculator to graph the conic
given by each equation.
19. x2 – 6xy + y 2 – 4y – 8x = 0
SOLUTION: Graph the equation by solving for y.
2
2
x – 6xy + y – 4y – 8x
2
2
y + (–6x − 4)y + x − 8x
=0
=0
[–10, 10] scl: 1 by [–10, 10] scl: 1
20. x2 + 4y 2 – 2xy + 3y – 6x + 5 = 0
SOLUTION: Graph the equation by solving for y.
2
2
x + 4y – 2xy + 3y – 6x + 5 = 0
2
2
4y + (–2x + 3)y + x − 6x + 5 = 0
[–10, 10] scl: 1 by [–10, 10] scl: 1
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