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Write an equation for each parabola.1. focus (0, 6), directrix y = –6

SOLUTION: Since the directrix is horizontal, the axis of symmetryis vertical.

The equation is of the form . h = 0, k = 0, and p = 6. The equation is

ANSWER:

2. focus (0, –8), directrix y = 8


The equation is of the form . h = 0, k = 0, and p = –8. The equation is

ANSWER:

3. focus (–2, 0), directrix x = 2

SOLUTION: Since the directrix is vertical, the axis of symmetry ishorizontal.


ANSWER:

4. focus (3, 0), directrix x = –3



ANSWER:

eSolutions Manual - Powered by Cognero Page 1

9-8 Equations of Parabolas

5. vertex , focus

SOLUTION: Since the focus is above the vertex, the axis ofsymmetry is vertical.

So the equation is of the form .

h = , k = 1, p = 2 The equation is

ANSWER:

6. vertex , directrix x =

SOLUTION: Since the directix is vertical, the axis of symmetry ishorizontal.


h = , k = –2, p = The equation is

ANSWER:

7. MODELING The parabolic reflector plate of aflashlight has its bulb located at the focus of theparabola. The distance between the vertex and thefocus is 1.8 centimeters. Write an equation torepresent the cross section of the reflector plate.

SOLUTION: Let the vertex be located at a point corresponding to(0, 0). The parabola opens to the right, so has the

form . h = 0, k = 0, p = 1.8 So the equation is

ANSWER:



Graph each equation.

8.

SOLUTION: Write the equation in standard form to identify h, k ,and p.

Thus, h = 4, k = –6, and p = .Identify some key features of the graph and thendraw.

focus: (4, –6 + ) or (4, –5 )vertex: (4, –6)axis of symmetry: x = 4

directrix: y = –6 – or y = direction of opening: up because p > 0

ANSWER:

9.


Thus, h = –5, k = 3, and p = .Identify some key features of the graph and thendraw.focus: vertex: (–5, 3)axis of symmetry: x = –5

directrix: y = 3 – or y = direction of opening: up because p > 0

ANSWER:



10.


Thus, h = 0, k = 2, and p = 2.Identify some key features of the graph and thendraw.focus: (0, 2 + 2) or (0, 4)vertex: (0, 2)axis of symmetry: x = 0directrix: y = 2 - 2 or y = 0direction of opening: up because p > 0

ANSWER:

11.


Thus, h = –2, k = 4, and p = –1. Identify some key features of the graph and thendraw.focus: (–2 + (–1), 4) or (–3, 4)vertex: (–2, 4)axis of symmetry: y = 4directrix: x = –2 - (–1) or x = –1direction of opening: left because p < 0

ANSWER:



Write an equation for each parabola.

12. focus , directrix


The equation is of the form . The vertex is (0, 0) and p = 2.5. So h = 0, k = 0, and p = 2.5 The equation is

ANSWER:

13. focus , directrix y =


The equation is of the form .

h = 0, k = 0, and p = . The equation is

ANSWER:

14. focus (0, 8), directrix y = –8



ANSWER:




ANSWER:






ANSWER:



The equation is of the form . The vertex is (2, 1) and p = –1 – 1. h = 2, k = 1, and p = –2. The equation is

ANSWER:

18. focus (5, 2), directrix x = –1


The equation is of the form . The vertex is (2, 2) and p = 5 – 2. h = 2, k = 2, and p = 3. The equation is

ANSWER:

19. focus (4, –1), directrix x = 8


The equation is of the form . The vertex is (6, –1) and p = 4 – 6. h = 6, k = –1, and p = –2 The equation is

ANSWER:



20. vertex (0, 0), directrix y = –2



ANSWER:

21. vertex (0, 0), directrix x = 3


The equation is of the form . The vertex is (0, 0) and p = 0 – 3. h = 0, k = 0, and p = –3 The equation is

ANSWER:

22. vertex (4, 4), directrix y = 3


The equation is of the form . The vertex is (4, 4) and p = 4 – 3h = 4, k = 4, and p = 1 The equation is

ANSWER:

23. vertex (1, –1), directrix y = 1


The equation is of the form . The vertex is (1, –1) and p = –1 – 1h = 1, k = –1, and p = –2 The equation is

ANSWER:





The equation is of the form . The vertex is (5, 1) and p = 5 – 0. h = 5, k = 1, and p = 5 The equation is

ANSWER:

25. vertex (–3, 1), directrix x = –4


The equation is of the form . The vertex is (–3, 1) and p = –3 – (–4). h = –3, k = 1, and p = 1 The equation is

ANSWER:



The equation is of the form . The vertex is (0, 6) and p = 0 – 2. h = 0, k = 6, and p = –2 The equation is

ANSWER:

27. vertex (–7, 0), directrix x = –2


The equation is of the form . The vertex is (–7, 0) and p = –7 – (–2). h = –7, k = 0, and p = –5 The equation is

ANSWER:

28. CARS The parabolic reflector plate of a carheadlight has its bulb located at the focus of theparabola. The distance between the vertex and thefocus is 6.5 centimeters.



a. Write an equation of the cross section of thereflector plate.b. Graph the parabola.

SOLUTION: a. The parabola opens upward so the axis ofsymmetry is vertical.

The equation is of the form . The vertex is (0, 0) and the focus is 6.5 cm above thevertex. h = 0, k = 0, p = 6.5 The equation is

b.

ANSWER:

a. b.

Graph each equation.

29.

SOLUTION:

The vertex is (0, 0) and p > 0. The parabola opensupward.

ANSWER:



30.

SOLUTION:


ANSWER:

31.

SOLUTION:

The vertex is (0, 0) and p < 0. The parabola opensdownward.

ANSWER:



32.

SOLUTION:

The vertex is (0, 0) and p > 0. The parabola opens tothe right.

ANSWER:

33.

SOLUTION:

The vertex is (1, –4) and p > 0. The parabola opensupward.

ANSWER:



34.

SOLUTION:

The vertex is (–6, 0) and p < 0. The parabola opensto the left.

ANSWER:

35.

SOLUTION:


ANSWER:



36.

SOLUTION:


ANSWER:

37.

SOLUTION:

The vertex is (–1, –5) and p < 0. The parabola opensto the left.

ANSWER:



38.

SOLUTION:


ANSWER:

39.

SOLUTION:

The vertex is (–4, –4) and p > 0. The parabola opensto the right.

ANSWER:



40.

SOLUTION:

The vertex is (6, 4) and p < 0. The parabola opens tothe left.

ANSWER:

Identify the focus and directrix of each parabola.

41.

SOLUTION:

The vertex is (–2, 3); 4p = 8, so p = 2. The parabolaopens upward, so the axis of symmetry is vertical. The focus is (–2, 3 + 2), or (–2, 5).The directrix is y = 3 – 2, or y = 1.

ANSWER: (–2, 5); y = 1

42.

SOLUTION:

The vertex is (4, 0); 4p = –4, so p = –1. The parabolaopens to the left, so the axis of symmetry ishorizontal. The focus is (4 – 1, 0), or (3, 0).The directrix is x = 4 – (–1), or x = 5.

ANSWER: (3, 0); x = 5

43.

SOLUTION:

The vertex is (–3, 1); 4p = 24, so p = 6. The parabolaopens to the right, so the axis of symmetry ishorizontal. The focus is (–3 + 6, 1), or (3, 1).The directrix is x = –3 – 6, or x = –9.

ANSWER: (3, 1); x = –9



44. SOLAR ENERGY Solar energy can beconcentrated using parabolic reflecting plates. Thecollected energy is used to heat homes and produceelectricity. Write an equation to represent the crosssection of the parabolic plate shown.

SOLUTION: The vertex is (0, 0) and p = 8.5.The parabola opens upward, so is in the form

. h = 0, k = 0, p = 8.5

ANSWER:

Write the equation of the parabola with thegiven directrix and with vertex at (0, 0).

45.

SOLUTION: Since the vertex is at (0, 0) and the directrix is

, . Since the directrix is vertical and p > 0, the parabolaopens to the right.

So the equation has the form .

ANSWER:

y2 = 6x

46.


, . Since the directrix is horizontal and p < 0, theparabola opens downward.


ANSWER:



47.


, . Since the directrix is horizontal and p > 0, theparabola opens upward.


ANSWER:

48.


, . Since the directrix is horizontal and p < 0, theparabola opens downward.


ANSWER:

49.


, . Since the directrix is vertical and p > 0, the parabolaopens to the right.


ANSWER:

y2 = 16x

50.

SOLUTION:

Since the vertex is at (0, 0) and the directrix is ,

. Since the directrix is vertical and p < 0, the parabolaopens to the left.


ANSWER:



51. A parabola with vertex (0, 0) has equation y2 – 16x =0 and is shifted 3 units to the right and 2 units up.Write the equation of the new parabola. Then graphboth parabolas on the same coordinate plane.

SOLUTION: Conjecture: The equation of the new parabola is (y –

2)2 = 16 (x – 3). The graph, shown below, verifies

that (y – 2)2 = 16 (x – 3) is a translation of y2 – 16x= 0 by 3 units to the right and 2 units up.

ANSWER:

(y – 2)2 = 16 (x – 3);

Write an equation for each parabola.52. focus (0, –3), directrix y = 3



ANSWER:




ANSWER:



54. vertex , focus

SOLUTION: Since the focus is above the vertex, the axis ofsymmetry is vertical.


h = , k = 1, p = 2 The equation is

ANSWER:

55. vertex , directrix x =

SOLUTION: Since the directix is vertical, the axis of symmetry ishorizontal.


h = , k = –2, p = The equation is

ANSWER:

Identify the coordinates of the vertex and focus,and the equation of the directrix of eachparabola.

56.

SOLUTION:

The vertex is located at (1, –2).4p = 4, so p = 1. The parabola is open upward.The focus is (1, –2 + 1), or (1, –1).The directrix is y = –2 – 1, or y = –3.

ANSWER: The vertex is located at (1, –2).The focus is (1, –1).The directrix is y = –3.



57.

SOLUTION:

The vertex is located at (4, –2).

4p = , so p = . The parabola is open to the right.

The focus is (4 , –2).

The directrix is x = 3 .

ANSWER: The vertex is located at (4, –2).



58.

SOLUTION:

The vertex is located at (2, 0).

4p = , so p = . The parabola is open upward.

The focus is (2, 0 + ), or (2, ).

The directrix is y =0 – , or y = – .

ANSWER: The vertex is located at (2, 0).

The focus is (2, ).

The directrix is y = – .

59.

SOLUTION:

The vertex is located at (4, –6).

4p = 1, so p = . The parabola is open to the right.



ANSWER: The vertex is located at (4, –6).





60.

SOLUTION:

The vertex is located at (–1, 1).

4p = , so p = . The parabola is open to the right.

The focus is ( , 1).

The directrix is x = .

ANSWER: The vertex is located at (–1, 1).

The focus is ( , 1).

The directrix is x = .

61.

SOLUTION:

The vertex is located at (0, 4).

4p = 1, so p = . The parabola opens upward.

The focus is (0, 4 ).

The directrix is y = 3 .

ANSWER: The vertex is located at (0, 4).

The focus is (0, 4 ).

The directrix is y = 3 .



62. CRITIQUE ARGUMENTS Describe the error inthe graph of each parabola.

a. x2 + 4y = 0

b.

SOLUTION: a. The graph should be open down. This is the graph

of x2 – 4y = 0.

b. The directrix should be the line .

ANSWER: a. The graph should be open down. This is the graph

of x2 – 4y = 0.

b. The directrix should be the line .

63. Graph y = –x2 + 4 and (x – 2)2 = y on the samecoordinate plane. a. Estimate the point(s) of intersection of the twoparabolas.

b. Substitute the expression (x – 2)2 for y into y =

–x2 + 4 and solve for x. c. Substitute the value(s) you found in part b into y =

–x2 + 4 and solve for y.

d. Use your answers to parts b and c to write thecoordinates of the point(s) of intersection of the twographs. Compare this to your estimates in part a. e . Verify that the point(s) you found in part d lie onboth parabolas.

SOLUTION: a. Sample answer: (0, 4), (2, 0)b.

x = 0 or x = 2c.

y = 4 or y = 0d. (0, 4) and (2, 0); They are the same intersectionpoints.

e . Check the solution (0, 4): For y = –x2 + 4,: 4 = –02

+ 4, so (0, 4) is a solution; For (x – 2)2 = (0 – 2)2 = 4,so (0, 4) is a solution. Check the

solution (2, 0): For y = –x2 + 4, 0 = –22 + 4, so (2, 0)

is a solution; For (x – 2)2 = (2 – 2)2 = 0, so (2, 0) is asolution. So, the points lie on bothparabolas.

ANSWER: a. Sample answer: (0, 4), (2, 0)b. x = 0 or x = 2c. y = 4 or y = 0d. (0, 4) and (2, 0); They are the same intersectionpoints.

e . Check the solution (0, 4): For y = –x2 + 4,: 4 = –02

+ 4, so (0, 4) is a solution; For (x – 2)2 = (0 – 2)2 = 4,so (0, 4) is a solution. Check the

solution (2, 0): For y = –x2 + 4, 0 = –22 + 4, so (2, 0)

is a solution; For (x – 2)2 = (2 – 2)2 = 0, so (2, 0) is asolution. So, the points lie on bothparabolas.



64. Graph on a graphing calculator. a. This graph is reflected in the line y = –1. Write theequation of the image. b. Write an equation that represents the parabolaformed by the original graph and its image.

SOLUTION:

a. b. Together, the graphs form a parabola that opens tothe right and has vertex (–2, –1). With 4p = 1, the

equation is .

ANSWER:

a.

b.

65. TELEVISION A television station uses parabolicantennas to transmit their signals. An antenna used totransmit the signal has a focus 50 inches from thevertex. a. Make two sketches of the antenna: one openingupwards and one opening to the right. b. Use your sketches to write an equation for each

sketch. Use the forms x2 = 4py and y2 = 4px. c. Does the equation you found in part b affect thedepth of the antenna? Explain.

SOLUTION: a.

b. Since p = 50, the equations are y2 = 200x and x2 =200y.c. No. The depth of the antenna is the same for bothsketches. It is equal to the distance between thefocus and the vertex, 50 inches.

ANSWER: a.

b. y2 = 200x and x2 = 200yc. No. The depth of the antenna is the same for bothsketches. It is equal to the distance between thefocus and the vertex, 50 inches.



66. ASTRONOMY In astronomy, a radio telescope dishis shaped like a parabola and has special antennas todetect signals from space, known as the “radiouniverse.” A radio telescope can be over 100 metersin diameter, but it can vary greatly in design, size, andconfiguration. a. In a typical antenna, the distance between thefocus and the vertex is 25 meters. What is anequation of the parabola? Assume that vertex is (0,0). b. Explain how a single dish antenna may work in aparabola-shaped telescope dish.

SOLUTION:

a. With vertex (0, 0) and p = 25, the equation is x2 =100y.b. Sample answer: As the signals come in, they areessentially parallel and are therefore reflected fromthe parabolic surface to the focus. The signalsbecome magnified because they are added togetherbefore they are transmitted to a receiver.

ANSWER:

a. x2 = 100yb. Sample answer: As the signals come in, they areessentially parallel and are therefore reflected fromthe parabolic surface to the focus. The signalsbecome magnified because they are added togetherbefore they are transmitted to a receiver.

67. WRITING IN MATH Explain how to find thedistance from the focus to the directrix for the

parabola x = 4y2.

SOLUTION: Sample answer: If you rewrite the equation of the

parabola as , it is of the form y2 = 4px.

Therefore, , and . The value of prepresents the distance from the focus to the vertexand from the vertex to the directrix. So, the distance

from the focus to the directrix must be 2p, or in thiscase.

ANSWER: Sample answer: If you rewrite the equation of the

parabola as , it is of the form y2 = 4px.

Therefore, , and . The value of prepresents the distance from the focus to the vertexand from the vertex to the directrix. So, the distance

from the focus to the directrix must be 2p, or in thiscase.

68. WRITING IN MATH The parabolic reflector plateof a bicycle head lamp has its bulb located at thefocus of the parabola. Explain the possibleadvantages to using this design.

SOLUTION: Sample answer: If the bulb is located at the focus,then the light rays will all be deflected to the focus,which will concentrate the light beams to make theoutput brighter.

ANSWER: Sample answer: If the bulb is located at the focus,then the light rays will all be deflected to the focus,which will concentrate the light beams to make theoutput brighter.



69. REASONING Explain how to find the equation of aparabola which has a vertex at (3, –2) and a focus at(6, –2).

SOLUTION: Sample answer: Make a sketch of the parabola bygraphing the vertex and focus. Since they are on thesame horizontal line and the focus is to the right ofthe vertex, the parabola opens to the right and will be

of the form (y – k)2 = 4p(x – h) with vertex (h, k).So, h = 3 and k = –2.The distance between the focusand the vertex is the value of p, or 3. Substitute the

values into (y – k)2 = 4p(x – h) to get the equation,

(y + 2)2 = 4(3)(x – 3), or (y + 2)2 = 12(x – 3).

ANSWER: Sample answer: Make a sketch of the parabola bygraphing the vertex and focus. Since they are on thesame horizontal line and the focus is to the right ofthe vertex, the parabola opens to the right and will be

of the form (y – k)2 = 4p(x – h) with vertex (h, k).So, h = 3 and k = –2.The distance between the focusand the vertex is the value of p, or 3. Substitute the

values into (y – k)2 = 4p(x – h) to get the equation,

(y + 2)2 = 4(3)(x – 3), or (y + 2)2 = 12(x – 3).

70. WRITING IN MATH As the value of p increases,

how does the width of the graph of change? Justify your reasoning.

SOLUTION: Sample answer: As the value of p increases, the

width of the graph increases because becomessmaller and smaller. This is because if the coefficient

of the x2-term is less than 1, then the graph of theparabola becomes wider.

ANSWER: Sample answer: As the value of p increases, the

width of the graph increases because becomessmaller and smaller. This is because if the coefficient

of the x2-term is less than 1, then the graph of theparabola becomes wider.

71. CHALLENGE a. What part of a parabola can be modeled by theequation ?b. What is the domain and range of the function inpart a?c. Explain how the equation is related to a

parabola of the form y2 = 4px.d. State a rule that you could use to help you use a

graphing calculator to show the graph of y2 = 4px.

SOLUTION: a. Top half of a parabola that has vertex (0, 0) and isopen to the rightb. Domain: x ≥ 0; Range: y ≥ 0c. Sample answer: If you square both sides of

, you get . This is related to y2 = 4px

because the coefficient of x in y2 = 4px is 4p. Sothey are part of the same family of parabolas .

If p = , then the equations are identical. d. Sample answer: Make a list of values for p,including values less than 1, and then use thecalculator to graph the family of functions ofthe form and to see thecomplete parabolas.

ANSWER: a. Top half of a parabola that has vertex (0, 0) and isopen to the rightb. Domain: x ≥ 0; Range: y ≥ 0c. Sample answer: If you square both sides of

, you get . This is related to y2 = 4px

because the coefficient of x in y2 = 4px is 4p. Sothey are part of the same family of parabolas .

If p = , then the equations are identical. d. Sample answer: Make a list of values for p,including values less than 1, and then use thecalculator to graph the family of functions ofthe form and to see thecomplete parabolas.



72. REASONING How are a and p related in the same

parabola expressed both as y = ax2 and x2 = 4py?

SOLUTION: Solve for y in

Compare the equations.

ANSWER:

73. REASONING Predict how a change in the value of

p in the equation y2 = 4px will affect the focus,

directrix, and graph of y2 = 4px. Then verify yourprediction by graphing both in same coordinate plane.

SOLUTION: Sample answer: As p increases, the focus gets

further away from the vertex of the graph of y2 =4px, and the distance between the focusand directrix increases. As p decreases, the focus

gets closer to the vertex of the graph of y2 = 4px, andthe distance between the focus and directrixdecreases. This is verified by graphing.

ANSWER: Sample answer: As p increases, the focus gets

further away from the vertex of the graph of y2 =4px, and the distance between the focusand directrix increases. As p decreases, the focus

gets closer to the vertex of the graph of y2 = 4px, andthe distance between the focus and directrixdecreases. This is verified by graphing.

74. WRITING IN MATH Explain the relationshipbetween the Distance Formula, the equation of aparabola with vertex (0, 0), and the equation of acongruent parabola with a vertex not at the origin.

SOLUTION: Sample answer: This Distance Formula can be usedto find the equations of both parabolas with vertex (0,0), and parabolas that aretranslated. The Distance Formula verifies that thedistance from the focus to a point on the parabola isequal to the distance from the pointto the directrix. This does not change if the parabolais translated.

ANSWER: Sample answer: This Distance Formula can be usedto find the equations of both parabolas with vertex (0,0), and parabolas that aretranslated. The Distance Formula verifies that thedistance from the focus to a point on the parabola isequal to the distance from the pointto the directrix. This does not change if the parabolais translated.



75. CHALLENGE Prove or disprove that the point (–8,-4) lies on a parabola with vertex (0, 0) and

containing the point .

SOLUTION: Sample answer: Graph the three points to see thatthey are located on a parabola with vertex (0, 0) thatis opening downward. Because the

vertex is (0, 0), substitute the coordinates into x2 =4py, and solve for p. If p is the same for both points,then the points are solutions tothe same parabola.

Because the value of p is different for each point, thepoints are not on the same parabola.

ANSWER: Sample answer: Graph the three points to see thatthey are located on a parabola with vertex (0, 0) thatis opening downward. Because the

vertex is (0, 0), substitute the coordinates into x2 =4py, and solve for p. If p is the same for both points,then the points are solutions tothe same parabola.

Because the value of p is different for each point, thepoints are not on the same parabola.

76. Use the Distance Formula to derive the equation of aparabola with focus (0, 4) and directrix y = –4.

SOLUTION: Choose a point on the parabola, (x, y). Then thedistance from the point to the focus is equal to thedistance from the point to the directrix.

ANSWER:

77. MULTI-STEP Consider the parabola with vertex (–4, 2) and focus (–4, 5).a. Write the equation of the parabola.b. Which of the following statements is true about thegraph of the parabola?A The graph opens upward.B The graph opens downward.C The graph opens to the right.D The graph opens to the left. c. Which of the following points are on the graph ofthe parabola? Select all that apply.A (2, 5)

B (–3, )C (0, 0)D (8, 14)E (–4, 2)F (6, 3) d. Which of the following is the equation of thedirectrix of the parabola?A y = –1B y = 1C x = 3D x = –1 e . Describe how the vertex of the graph is translatedfrom (0, 0).



SOLUTION: a. Since the focus is 3 units above the vertex, theparabola opens upward and p = 3. With vertex (–4,2), the equation is

b. Since the focus is 3 units above the vertex, theparabola opens upward. Choice A is correct. c. Substitute the coordinates from each ordered pairinto the equation from part a. The points in choices A,D, and E are all on the parabola. d. The directrix is horizontal and 3 units below thevertex. So its equation is y = –1. Choice A is correct. e . The vertex is translated 4 units left and 2 units up.

ANSWER: a.

b. Ac. A, D, and Ed. Ae . The vertex is translated 4 units left and 2 units up.

78. Which of the following is the equation of a parabolawith vertex (3, –5) and directrix x = 7?

A (y – 5)2 = –16(x + 3)

B (x – 3)2 = –16(y + 5)

C (y + 5)2 = –16(x – 3)

D (y + 5)2 = 16(x – 3)

SOLUTION: Since the directrix is vertical, the axis of symmetry is

horizontal, and so the parabola has the form y2 =4px. With vertex (3, –5), the quantities (y + 5) and (x – 3)will be in the equation, with (y + 5) being squared.Since the directrix is 4 units to the right of the vertex,p = –4, so 4p = –16. So the equation in choice C iscorrect.

ANSWER: C

79. Consider the parabola (y – 1)2 = –8(x + 5). a. Identify the vertex. b. Identify the focus. c. Identify the directrix.

SOLUTION: a. The vertex is (–5, 1).b. Since 4p = –8, p = –2. So this parabola opens tothe left. The focus, then, is 2 units left of the vertex,or (–7, 1).c. The directrix is 2 units right from the vertex, and isvertical, so its equation is x = –3.

ANSWER: a. (–5, 1)b. (–7, 1)c. x = –3



80. The graph of x2 = –2y is translated 1 unit down and 4units left. Which of the following is the equation ofthe new parabola?

A (x + 1)2 = 2(y – 4)

B (x – 4)2 = –2(y – 1)

C (x + 4)2 = –2(y – 1)

D (x + 4)2 = –2(y + 1)

SOLUTION: A translation 1 unit down adds 1 to y and a translation4 units left adds 4 to x. The correct equation is shownin choice D.

ANSWER: D

81. The parabolic reflector of a theatrical stage light hasa bulb located at the focus of the reflector. If thefocus of the parabola is 8.5 inches from the vertex,what are two possible equations of the parabola?

SOLUTION: Let the vertex be located at (0, 0), with p = 8.5. Thenthe equation could be of a parabola that opensupward or one that opens to the right.

The equations could be or .

ANSWER:

or

82. Write an equation of the parabola with focus (–4, –1)and directrix y = 13.

SOLUTION: The focus and directrix show that the vertex is (–4,6), p = –7, and that the parabola opens downward. The equation of the parabola is

.

ANSWER:



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