Mid-Chapter Quiz: Lessons 7-1 through 7-3 - Thurmond

Mid-Chapter Quiz: Lessons 7-1 through 7-3

Write an equation for and graph a parabola with the given focus F and vertex V 1. F(1, 5), V(1, 3)

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 5 - 3 or 2. Because p is positive, the graph opens up.

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 ? 3) = (x ? 1)2

8(y - 3) = (x - 1)2

The standard form of the equation is (x - 1)2 = 8(y ? 3). Graph the vertex and focus. Then make a table of values to graph the parabola.

2. F(5, ?7), V(1, ?7)

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 5 ? 1 or 4. 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(4)(x ? 1) = [y ? (?7)]2

16(x - 1) = (y + 7)2

The standard form of the equation is (y + 7)2 = 24(x - 1). Graph the vertex and focus. Then make a table of values to graph the parabola.

2. F(5, ?7), V(1, ?7)

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 5 ? 1 or 4. 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(4)(x ? 1) = [y ? (?7)]2

16(x - 1) = (y + 7)2

The standard form of the equation is (y + 7)2 = 24(x - 1). Graph the vertex and focus. Then make a table of values to graph the parabola.

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3. MULTIPLE CHOICE In each of the following, a parabola and its directrix are shown. In which parabola is the focus farthest from the vertex?

SOLUTION:

The distance between the focus and the vertex of a parabola is equal to the distance between the vertex and the directrix. The directrix appears to be 5 units from the vertex in the graph for choice D. This is farther than the 3 units for choice A, the 0.5 unit for choice B, and the 3 units for choice C.

The correct answer is D.

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4. DESIGN The cross?section of the mirror in the

Mid-Chapter Quiz: Lessons 7-1 through 7-3

3. MULTIPLE CHOICE In each of the following, a parabola and its directrix are shown. In which parabola is the focus farthest from the vertex?

from the vertex in the graph for choice D. This is farther than the 3 units for choice A, the 0.5 unit for choice B, and the 3 units for choice C.

The correct answer is D.

4. DESIGN The cross?section of the mirror in the flashlight design below is a parabola. a. Write an equation that models the parabola. b. Graph the equation.

SOLUTION: The distance between the focus and the vertex of a parabola is equal to the distance between the vertex and the directrix. The directrix appears to be 5 units from the vertex in the graph for choice D. This is farther than the 3 units for choice A, the 0.5 unit for choice B, and the 3 units for choice C.

The correct answer is D.

4. DESIGN The cross?section of the mirror in the flashlight design below is a parabola. a. Write an equation that models the parabola. b. Graph the equation.

SOLUTION: a. Sketch a coordinate plane on top of the mirror with the vertex located at the origin. Since the mirror has a width of 8 inches, 4 inches of the mirror will exist on each side of the x?axis. Using a depth of 3 inches and a width of 4 inches, the points (3, 4) and (3, ?4) must lie on the mirror. Since the parabola opens to the right, the standard form is 4p (x - h) = (y - k)2. Substitute the values of x, y, h, and k to solve for p .

4p (x - h) = (y - k)2 4p (3 - 0) = (4 - 0)2

12p = 16

p = or 1.33

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Substitute the values of h, k, and p to write an equation that models the parabola.

2

SOLUTION:

a. Sketch a coordinate plane on top of the mirror with the vertex located at the origin. Since the mirror has a width of 8 inches, 4 inches of the mirror will exist on each side of the x?axis. Using a depth of 3 inches and a width of 4 inches, the points (3, 4) and (3, ?4) must lie on the mirror. Since the parabola opens to the right, the standard form is 4p (x - h) = (y - k)2. Substitute the values of x, y, h, and k to solve for p .

4p (x - h) = (y - k)2 4p (3 - 0) = (4 - 0)2

12p = 16

p = or 1.33

Substitute the values of h, k, and p to write an equation that models the parabola.

4p (x - h) = (y - k)2

4 (x - 0) = (y - 0)2

x = y2

b. Use a table of values to graph the parabola.

x

y

0

0

2

?3.3

4

?4.6

6

?5.7

8

?6.5

Graph the ellipse given by each equation. Page 2

5.

+

= 1

Mid-Chapter Quiz: Lessons 7-1 through 7-3

Graph the ellipse given by each equation.

5.

+

= 1

SOLUTION:

The ellipse is in standard form. The values of h and k are ?4 and ?2, so the center is at (?4, ?2).

a = or 9

b = or 4

c =

8.1

orientation: horizontal

vertices: (?13, ?2), (5, ?2)

covertices: (?4, ?6), (?4, 2)

Write an equation for the ellipse with each set of characteristics. 7. vertices (9, ?3), (?3, ?3); foci (7, ?3), (?1, ?3)

SOLUTION:

The distance between the vertices is 2a. 2a = |9 - (?3)| a = 6; a2 = 36 The distance between the foci is 2c. 2c = |7 ? (?1)| c =4

b = b2 = 20

= 4.47

= 3 center: (3, ?3)

+

= 1

6.

+

= 1

SOLUTION:

The ellipse is in standard form. The values of h and k are 3 and 6, so the center is at (3, 6).

a =

or 6

b = or 2

c =

5.7

orientation: vertical

vertices: (3, 0), (3, 12)

covertices: (1, 6), (5, 6)

Write an equation for the ellipse with each set eSolutoiofnscMhaanruaacl t- eProiwsetriecds.by Cognero

7. vertices (9, ?3), (?3, ?3); foci (7, ?3), (?1, ?3)

SOLUTION:

8. foci (3, 1), (3, 7); length of minor axis equals 8

SOLUTION: The distance between the foci is 2c. 2c = |1 - 7| c =3 The length of the minor axis is 2b. 2b = 8 b = 4; b2 = 16

a =

= 5

a2 = 25

= 4 center: (3, 4)

+

= 1

9. major axis (1, ?1) to (1, ?13); minor axis (?2, ?7) to (4, ?7)

SOLUTION:

The length of the major axis is 2a. 2a = |?13 - (?1)| a = 6; a2 = 36

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The length of the minor axis is 2b.

center: (3, 4)

= ?2 center: (8, ?2)

Mid-Chapte+r Quiz: Le=ss1ons 7-1 through 7-3

+

= 1

9. major axis (1, ?1) to (1, ?13); minor axis (?2, ?7) to (4, ?7)

SOLUTION: The length of the major axis is 2a. 2a = |?13 - (?1)| a = 6; a2 = 36

The length of the minor axis is 2b. 2b = |4 - (?2)| b = 3; b2 = 9

= ?7 center: (1, ?7)

+

= 1

10. vertices (8, 5), (8, ?9); length of minor axis equals 6

SOLUTION:

The distance between the vertices is 2a. 2a = |5 - (?9)| a = 7; a2 = 49 The length of the minor axis is 2b. 2b = 6 b = 3, b2 = 9

= ?2

center: (8, ?2)

+

= 1

11. SWIMMING The shape of a swimming pool is designed as an ellipse with a length of 30 feet and an eccentricity of 0.68. a. What is the width of the pool? b. Write an equation for the ellipse if the point of origin is the center of the pool.

11. SWIMMING The shape of a swimming pool is designed as an ellipse with a length of 30 feet and an eccentricity of 0.68. a. What is the width of the pool? b. Write an equation for the ellipse if the point of origin is the center of the pool.

SOLUTION:

a. The pool has a horizontal major axis. Find a. 2a = 30 a = 15

Use the eccentricity of the ellipse to solve for c. e = = 0.68

c = 0.68 ? 15 = 10.2

Solve for b.

b =

b =

11

So, 2b = 22.

The width of the pool is 22 feet.

b. The standard form of the equation is

+

= 1. Use the values for h, k, a, and b to

write an equation for the ellipse.

+

= 1

+

= 1

+

= 1

SOLUTION:

a. The pool has a horizontal major axis. Find a. eSolutions2Maanua=l -3P0owered by Cognero

a = 15

12. MULTIPLE CHOICE Which of the following is a possible eccentricity for the graph? A 0

B

C 1

D

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+

= 1

+

= 1

Mid-Chapter Quiz: Lessons 7-1 through 7-3

12. MULTIPLE CHOICE Which of the following is a possible eccentricity for the graph? A 0

B

C 1

D

SOLUTION:

The eccentricity of an ellipse is a value between 0 and 1 that describes how `stretched' an ellipse is. An eccentricity of 0 would represent a circle and an

eccentricity of is not possible since it is greater

than 1. An eccentricity of 1 would require a = c, which would mean that b = 0. The only possible

answer that can represent the eccentricity is .

The correct answer is B.

Graph the hyperbola given by each equation.

13. ?

= 1

SOLUTION:

The equation is in standard form, with h = 0 and k = ?7. Because a2 = 81 and b2 = 81, a = 9 and b = 9. The values of a and b can be used to find c. c2 = a2 + b2 c2 = 81 + 81

c =

or about 12.73

Use h, k, a, b, and c to determine the characteristics of the hyperbola.

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, ?7) eSolutvioenrstiMcaensu:a(lh- P?owae,rked) b=y (C?o9g,ne?r7o) and (9, ?7)

foci: (h ? c, k) = (?12.73, ?7) and (12.73, ?7) asymptotes:

which would mean that b = 0. The only possible

answer that can represent the eccentricity is .

The correct answer is B.

Graph the hyperbola given by each equation.

13. ?

= 1

SOLUTION:

The equation is in standard form, with h = 0 and k = ?7. Because a2 = 81 and b2 = 81, a = 9 and b = 9. The values of a and b can be used to find c. c2 = a2 + b2 c2 = 81 + 81

c =

or about 12.73

Use h, k, a, b, and c to determine the characteristics of the hyperbola.

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, ?7) vertices: (h ? a, k) = (?9, ?7) and (9, ?7) foci: (h ? c, k) = (?12.73, ?7) and (12.73, ?7) asymptotes:

Graph the center, vertices, foci, and asymptotes.

Then make a table of values to sketch the hyperbola.

x

y

?15

?19, 5

?11

?13.3, ?0.68

11

?13.3, ?0.68

15

?19, 5

14.

?

= 1

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