THE PARABOLA section

49. Fencing a rectangle. If 34 ft of fencing are used to en- close a rectangular area of 72 ft², then what are the di- mensions of the area?

50. Real numbers. Find two numbers that have a sum of 8 and a product of 10.

51. Imaginary numbers. Find two complex numbers whose sum is 8 and whose product is 20.

52. Imaginary numbers. Find two complex numbers whose sum is6 and whose product is 10.

53. Making a sign. Rico’s Sign Shop has a contract to make a sign in the shape of a square with an isosceles triangle on top of it, as shown in the figure. The contract calls for a total height of 10 ft with an area of 72 ft². How long should Rico make the side of the square and what should be the height of the triangle?

54. Designing a box. Angelina is designing a rectangular box of 120 cubic inches that is to contain new Eaties breakfast cereal. The box must be 2 inches thick so that it is easy to hold. It must have 184 square inches of sur- face area to provide enough space for all of the special offers and coupons. What should be the dimensions of the box?

F I G U R E F O R E X E R C I S E 5 3

G R A P H I N G C A LC U L ATO R E X E RC I S E S

55. Solve each system by graphing each pair of equations on a graphing calculator and using the intersect feature to es- timate the point of intersection. Find the coordinates of each intersection to the nearest hundredth.

a) y e^x 4 b) 3^y1 x

y ln(x  3) y x²

c) x² y² 4 y x³

10 ft

x ft

x ft

T H E P A R A B O L A

The parabola is one of four different curves that can be obtained by intersecting a cone and a plane as in Fig. 13.3. These curves, called conic sections, are the parabola, circle, ellipse, and hyperbola. We graphed parabolas in Sections 10.3 and 11.2. In this section we learn some new facts about parabolas.

F I G U R E 1 3 . 3

Ellipse Hyperbola

Parabola Circle

I n t h i s

s e c t i o n

● The Geometric Definition

● Developing the Equation

● Parabolas in the Form y a(x  h)² k

● Finding the Vertex, Focus, and Directrix

● Axis of Symmetry

● Changing Forms

13.2

The Geometric Definition

In Section 10.3 we called the graph of y ax² bx  c a parabola. This equa- tion is the standard equation of a parabola. In this section you will see that the fol- lowing geometric definition describes the same curve as the equation.

In Section 10.3 we defined the vertex as the highest point on a parabola that opens downward or the lowest point on a parabola that opens upward. We learned that x b(2a) gives the x-coordinate of the vertex. We can also describe the ver- tex of a parabola as the midpoint of the line segment that joins the focus and direc- trix, perpendicular to the directrix. See Fig. 13.4.

The focus of a parabola is important in applications. When parallel rays of light travel into a parabolic reflector, they are reflected toward the focus as in Fig. 13.5.

This property is used in telescopes to see the light from distant stars. If the light source is at the focus, as in a searchlight, the light is reflected off the parabola and projected outward in a narrow beam. This reflecting property is also used in camera lenses, satellite dishes, and eavesdropping devices.

F I G U R E 1 3 . 4 F I G U R E 1 3 . 5

Developing the Equation

To develop an equation for a parabola, given the focus and directrix, choose the point (0, p), where p 0 as the focus and the line y  p as the directrix, as shown in Fig. 13.6. The vertex of this parabola is (0, 0). For an arbitrary point (x, y) on the parabola the distance to the directrix is the distance from (x, y) to (x,p).

The distance to the focus is the distance between (x, y) and (0, p). We use the fact that these distances are equal to write the equation of the parabola:

^(x^{ 0)}^{2 (y}^{ p)2}^(x^{ x)}^{2 (y}^{ (p))}² To simplify the equation, first remove the parentheses inside the radicals:

^x^{2 y2}^{ 2p}^{y p2}^y^{2 2py}^{ p2}

x² y² 2py  p² y² 2py  p² Square each side.

x² 4py ^{Subtract y2and p2}from each side.

y  4 1

px²

Focus Directrix

Focus Vertex Parabola

Parabola

Given a line (the directrix) and a point not on the line (the focus), the set of all points in the plane that are equidistant from the point and the line is called a parabola.

(0, p)

(0, 0)

(x, y)

(x, –p) y = –p

y

x p > 0

F I G U R E 1 3 . 6

So the parabola with focus (0, p) and directrix y p for p 0 has equation y 

4 1

px². This equation has the form y ax² bx  c, where a ^₄¹_p^, b 0, and c  0.

If the focus is (0, p) with p 0 and the directrix is y  p, then the parabola opens downward as shown in Fig. 13.7. Deriving the equation using the distance formula again yields y^₄¹px².

Parabolas in the Form y  a(x  h)

²

 k

The simplest parabola, y x², has vertex (0, 0). The transformation y a(x  h)² k is also a parabola and its vertex is (h, k). The focus and directrix of the transformation are found as follows:

Figure 13.8 shows the location of the focus and directrix for parabolas with ver- tex (h, k) and opening either upward or downward. Note that the location of the focus and directrix determine the value of a and the shape and opening of the parabola.

F I G U R E 1 3 . 8

For a parabola that opens upward, p 0, and the focus (h, k  p) is above the vertex (h, k). For a parabola that opens downward, p 0, and the focus (h, k p) is below the vertex (h, k). In either case the distance from the vertex to the focus and the vertex to the directrix is p.

Finding the Vertex, Focus, and Directrix

In Example 1 we find the vertex, focus, and directrix from an equation of a parabola.

In Example 2 we find the equation given the focus and directrix.

E X A M P L E 1

Finding the vertex, focus, and directrix, given an equation Find the vertex, focus, and directrix for the parabola y x².

C A U T I O N

Directrix: y = k – p (h, k) (h, k + p)

a > 0

Directrix: y = k – p (h, k)

(h, k + p) a < 0

a  _4p¹

y = a(x – h)² + k y = a(x – h)² + k

y y

x x

Parabolas in the Form y a(x  h)² k

The graph of the equation y a(x  h)² k (a  0) is a parabola with vertex (h, k), focus (h, k p), and directrix y  k  p, where a  ^₄¹_p^. If a 0, the parabola opens upward; if a 0, the parabola opens downward.

y = –p (x, –p)

(0, 0) (x, y) (0, p)

p < 0

x y

F I G U R E 1 3 . 7

Solution

Compare y x² to the general formula y a(x  h)² k. We see that h  0, k 0, and a  1. So the vertex is (0, 0). Because a  1, we can use a ^₄¹_p^ to get

1  4 1

p,

or p^1_4. Use (h, k p) to get the focus

^{0, 14}



. Use the equation y k  p to get y ^1_4as the equation of the directrix. See Fig. 13.9.

■

E X A M P L E 2

Finding an equation, given a focus and directrix

Find the equation of the parabola with focus (1, 4) and directrix y  3.

Solution

Because the vertex is halfway between the focus and directrix, the vertex is

^{1, 72}



. See Fig. 13.10. The distance from the vertex to the focus is _¹

2. Because the focus is above the vertex, p is positive. So p^1_2, and a₄^1_p^^1_2. The equation is

y 1

2(x  (1))² 7 2.

Convert to y ax² bx  c form as follows:

y 1

2(x  1)² 7 2 y 1

2(x² 2x  1)  7 2 y 1

2x² x  4

■

Axis of Symmetry

The graph of y x²shown in Fig. 13.9 is symmetric about the y-axis because the two halves of the parabola would coincide if the paper were folded on the y-axis. In general, the vertical line through the vertex is the axis of symmetry for the parabola. See Fig. 13.11. In the form y ax² bx  c the x-coordinate of the ver- tex is b(2a) and the equation of the axis of symmetry is x  b(2a). In the form y a(x  h)² k the vertex is (h, k) and the equation for the axis of symme- try is x h.

Changing Forms

Since there are two forms for the equation of a parabola, it is sometimes useful to change from one form to the other. To change from y a(x  h)² k to the form y ax² bx  c, we square the binomial and combine like terms, as in Exam- ple 2. To change from y  ax² bx  c to the form y  a(x  h)² k, we com- plete the square, as in the next example.

y = 3 y

x

—7

–1, 2

(–1, 4)

( (

F I G U R E 1 3 . 1 0

F I G U R E 1 3 . 1 1

Axis of symmetry

y

x x = —–^–b

2a (h, k) or x = h

y

– 1 1 x

1

– 1

—1

0, 4

y = –^—1₄

y = x²

( (

F I G U R E 1 3 . 9

E X A M P L E 3

Converting y ax² bx  c to y  a(x  h)² k

Write y 2x² 4x  5 in the form y  a(x  h)² k and identify the vertex, focus, directrix, and axis of symmetry of the parabola.

Solution

Use completing the square to rewrite the equation:

y 2(x² 2x)  5

y 2(x² 2x  1  1)  5 Complete the square.

y 2(x² 2x  1)  2 5 Move 2(1) outside the parentheses.

y 2(x  1)² 3

The vertex is (1, 3). Because a^₄¹_p^, we have

4 1

p  2,

and p^1_8. Because the parabola opens upward, the focus is1

8unit above the vertex at

^{1, 318}



^{, or}

^{1, 285}



, and the directrix is the horizontal line1

8unit below the vertex, y 2^7_8or y^2₈^3. The axis of symmetry is x 1.

■

Be careful when you complete a square within parentheses as in Example 3. For another example, consider the equivalent equations

y 3(x² 4x),

y 3(x² 4x  4  4), and

y 3(x  2)² 12.

E X A M P L E 4

Finding the features of a parabola from standard form

Find the vertex, focus, directrix, and axis of symmetry of the parabola y 3x² 9x 5, and determine whether the parabola opens upward or downward.

Solution

The x-coordinate of the vertex is x 

2a

  b 2(



 9

  3) 

 9

  6 3 2.

To find the y-coordinate of the vertex, let x^3_2in y 3x² 9x  5:

y 3

^32



^{2 9}

^32



^{ 5  247  227  5  74}

The vertex is

^{32, 74}



. Because a 3, the parabola opens downward. To find the focus, use 3 ^₄¹_p^ to get p ^₁¹₂^. The focus is 1

1

2 of a unit below the ver- tex at

^{32,74112}



^or

^32,53



. The directrix is the horizontal line 1

1

2 of a unit above the vertex, y^7_4^₁¹₂^or y^1₆^1. The equation of the axis of symmetry is x3

2.

■

C A U T I O N The graphs of

y₁ 2x² 4x  5 and

y₂ 2(x  1)² 3 appear to be identical. This supports the conclusion that the equations are equivalent.

c a l c u l a t o r

c l o s e - u p

5 10

–5 –5

5

–5 –5

10

A calculator graph can be used to check the vertex and opening of a parabola.

c a l c u l a t o r

c l o s e - u p

5

–10 –5

5

True or false? Explain your answer.

1. There is a parabola with focus (2, 3), directrix y 1, and vertex (0, 0).

2. The focus for the parabola y^1_4x² 1 is (0, 2).

3. The graph of y 3  5(x  4)²is a parabola with vertex (4, 3).

4. The graph of y 6x  3x  2 is a parabola.

5. The graph of y 2x  x² 9 is a parabola opening upward.

6. For y  x²the vertex and y-intercept are the same point.

7. A parabola with vertex (2, 3) and focus (2, 4) has no x-intercepts.

8. The parabola with focus (0, 2) and directrix y  1 opens upward.

9. The axis of symmetry for y a(x  2)² k is x  2.

10. If a (4 1

p)and a 1, then p 1 4.

W A R M - U P S

E X E R C I S E S

1 3 . 2

Reading and Writing After reading this section, write out the answers to these questions. Use complete sentences.

1. What is the definition of a parabola given in this section?

2. What is the location of the vertex?

3. What are the two forms of the equation of a parabola?

4. What is the distance from the focus to the vertex in any parabola of the form y ax² bx  c?

5. How do we convert an equation of the form y ax² bx c into the form y  a(x  h)² k?

6. How do we convert an equation of the form y a(x h)² k into the form y  ax² bx  c?

Find the vertex, focus, and directrix for each parabola. See Example 1.

7. y 2x²

8. y 1 2x² 9. y 1 4x² 10. y 

1 1

2x² 11. y 1

2(x  3)² 2

12. y 1

4(x  2)² 5 13. y (x  1)² 6 14. y 3(x  4)² 1

Find the equation of the parabola with the given focus and directrix. See Example 2.

15. Focus (0, 2), directrix y 2 16. Focus (0,3), directrix y  3 17. Focus

^0,12



, directrix y 1

2 18. Focus

^0,18



, directrix y 1 8

19. Focus (3, 2), directrix y 1 20. Focus (4, 5), directrix y  4 21. Focus (1,2), directrix y  2

22. Focus (2,3), directrix y  1 23. Focus (3, 1.25), directrix y  0.75 24. Focus

^5,18

7



, directrix y 1 8

5

Write each equation in the form y a(x  h)² k. Identify the vertex, focus, directrix, and axis of symmetry of each parabola. See Example 3.

25. y x² 6x  1

26. y x² 4x  7

27. y 2x² 12x  5

28. y 3x² 6x  7

29. y 2x² 16x  1

30. y 3x² 6x  7

31. y 5x² 40x

32. y 2x² 10x

Find the vertex, focus, directrix, and axis of symmetry of each parabola (without completing the square), and determine whether the parabola opens upward or downward. See Example 4.

33. y x² 4x  1

34. y x² 6x  7

35. y x² 2x  3

36. y x² 4x  9

37. y 3x² 6x  1

38. y 2x² 4x  3

39. y x² 3x  2

40. y x² 3x  1

41. y 3x² 5

42. y 2x² 6

Solve each problem.

43. World’s largest telescope. The largest reflecting tele- scope in the world is the 6-meter (m) reflector on Mount Pastukhov in Russia. The accompanying figure shows a cross section of a parabolic mirror 6 m in diameter with the vertex at the origin and the focus at (0, 15). Find the equation of the parabola.

F I G U R E F O R E X E R C I S E 4 3

44. Arecibo Observatory. The largest radio telescope in the world uses a 1000-ft parabolic dish, suspended in a val- ley in Arecibo, Puerto Rico. The antenna hangs above

y

x (0, 15)

6 m

the vertex of the dish on cables stretching from two towers. The accompanying figure shows a cross section of the parabolic dish and the towers. Assuming the ver- tex is at (0, 0), find the equation for the parabola. Find the distance from the vertex to the antenna located at the focus.

F I G U R E F O R E X E R C I S E 4 4

Graph both equations of each system on the same coordi- nate axes. Use elimination of variables to find all points of intersection.

45. y x² 3 46. y x² 3

y x² 1 y x² 5

47. y x² 2 48. y x² x  6

y 2x  3 y 7x  15

Antenna at focus

x y

1000 ft

200 ft 200 ft

49. y x² 3x  4 y x² 2x  8

50. y x² 2x  8 y x² x  12

51. y x² 3x  4 52. y x² 5x  6

y 2x  2 y x  11

Solve each problem.

53. Find all points of intersection of the parabola y x² 2x  3 and the x-axis.

54. Find all points of intersection of the parabola y 80x² 33x  255 and the y-axis.

55. Find all points of intersection of the parabola y 0.01x² and the line y 4.

56. Find all points of intersection of the parabola y 0.02x² and the line y x.

57. Find all points of intersection of the parabolas y x² and x y².

13.3

58. Find all points of intersection of the parabolas y x² and y (x  3)².

G E T T I N G M O R E I N VO LV E D

59. Exploration. Consider the parabola with focus ( p, 0) and directrix x p for p 0. Let (x, y) be an arbi- trary point on the parabola. Write an equation express- ing the fact that the distance from (x, y) to the focus is equal to the distance from (x, y) to the directrix. Rewrite the equation in the form x ay², where a4

1 p. 60. Exploration. In general, the graph of x a(y  h)² k

for a 0 is a parabola opening left or right with vertex at (k, h).

a) For which values of a does the parabola open to the right, and for which values of a does it open to the left?

b) What is the equation of its axis of symmetry?

c) Sketch the graphs x 2(y  3)² 1 and x 

(y  1)² 2.

G R A P H I N G C A LC U L ATO R E X E RC I S E S

61. Graph y x²using the viewing window with 1  x  1 and 0 y  1. Next graph y  2x² 1 using the view- ing window2  x  2 and 1  y  7. Explain what you see.

62. Graph y x²and y 6x  9 in the viewing window

5  x  5 and 5  y  20. Does the line appear to be tangent to the parabola? Solve the system y x² and y 6x  9 to find all points of intersection for the parabola and the line.

T H E C I R C L E

In this section we continue the study of the conic sections with a discussion of the circle.

Developing the Equation

A circle is obtained by cutting a cone, as was shown in Fig. 13.3. We can also define a circle using points and distance, as we did for the parabola.

We can use the distance formula of Section 9.5 to write an equation for the circle with center (h, k) and radius r, shown in Fig. 13.12. If (x, y) is a point on the circle, its distance from the center is r. So

^(x h⁾² (^y k)² r.

We square both sides of this equation to get the standard form for the equation of a circle.

Standard Equation for a Circle The graph of the equation

(x h)² (y  k)² r² with r 0, is a circle with center (h, k) and radius r.

Circle

A circle is the set of all points in a plane that lie a fixed distance from a given point in the plane. The fixed distance is called the radius, and the given point is called the center.

I n t h i s

s e c t i o n

● Developing the Equation

● Equations Not in Standard Form

● Systems of Equations

(x, y)

r (h, k) y

x

F I G U R E 1 3 . 1 2