THE PARABOLA section

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1 698 (3 0) Chapter 3 Nonlinear Sstems and the Conic Sections 49. Fencing a rectangle. If 34 ft of fencing are used to enclose a rectangular area of 72 ft 2, then what are the dimensions of the area? 50. Real numbers. Find two numbers that have a sum of 8 and a product of 0. 0 ft 5. Imaginar numbers. Find two comple numbers whose sum is 8 and whose product is Imaginar numbers. Find two comple numbers whose sum is 6 and whose product is 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 0 ft with an area of 72 ft 2. How long should Rico make the side of the square and what should be the height of the triangle? 54. Designing a bo. Angelina is designing a rectangular bo of 20 cubic inches that is to contain new Eaties breakfast cereal. The bo must be 2 inches thick so that it is eas to hold. It must have 84 square inches of surface area to provide enough space for all of the special offers and coupons. What should be the dimensions of the bo? ft FIGURE FOR EXERCISE 53 GRAPHING CALCULATOR EXERCISES 55. Solve each sstem b graphing each pair of equations on a graphing calculator and using the intersect feature to estimate the point of intersection. Find the coordinates of each intersection to the nearest hundredth. a) e 4 b) 3 ln( 3) 2 c) ft 3.2 THE PARABOLA In this section The Geometric Definition Developing the Equation Parabolas in the Form a( h) 2 k Finding the Verte, Focus, and Directri Ais of Smmetr Changing Forms The parabola is one of four different curves that can be obtained b intersecting a cone and a plane as in Fig These curves, called conic sections, are the parabola, circle, ellipse, and hperbola. We graphed parabolas in Sections 0.3 and.2. In this section we learn some new facts about parabolas. Parabola Circle Ellipse FIGURE 3.3 Hperbola

2 The Geometric Definition 3.2 The Parabola (3 ) 699 In Section 0.3 we called the graph of a 2 b c a parabola. This equation is the standard equation of a parabola. In this section ou will see that the following geometric definition describes the same curve as the equation. Parabola Given a line (the directri) 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. In Section 0.3 we defined the verte as the highest point on a parabola that opens downward or the lowest point on a parabola that opens upward. We learned that b (2a) gives the -coordinate of the verte. We can also describe the verte of a parabola as the midpoint of the line segment that joins the focus and directri, perpendicular to the directri. See Fig The focus of a parabola is important in applications. When parallel ras of light travel into a parabolic reflector, the are reflected toward the focus as in Fig This propert 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 propert is also used in camera lenses, satellite dishes, and eavesdropping devices. Parabola Focus Verte Focus Directri FIGURE 3.4 FIGURE 3.5 p > 0 (0, p) (, ) (0, 0) = p (, p) FIGURE 3.6 Developing the Equation To develop an equation for a parabola, given the focus and directri, choose the point (0, p), where p 0 as the focus and the line p as the directri, as shown in Fig The verte of this parabola is (0, 0). For an arbitrar point (, ) on the parabola the distance to the directri is the distance from (, ) to(, p). The distance to the focus is the distance between (, ) and (0, p). We use the fact that these distances are equal to write the equation of the parabola: ( ) 0 2 ( p) 2 ( ) 2 ( )) ( p 2 To simplif the equation, first remove the parentheses inside the radicals: 2 2 2p p 2 p 2 2 p p p 2 2 2p p 2 2 4p 2 4 p Square each side. Subtract 2 and p 2 from each side.

3 700 (3 2) Chapter 3 Nonlinear Sstems and the Conic Sections = p (, p) (0, 0) (, ) (0, p) So the parabola with focus (0, p) and directri p for p 0 has equation 4p 2. This equation has the form a 2 b c, where a 4 p, b 0, and c 0. If the focus is (0, p) with p 0 and the directri is p, then the parabola opens downward as shown in Fig Deriving the equation using the distance formula again ields 4 p 2. p < 0 FIGURE 3.7 Parabolas in the Form a( h) 2 k The simplest parabola, 2, has verte (0, 0).The transformation a( h) 2 k is also a parabola and its verte is (h, k). The focus and directri of the transformation are found as follows: Parabolas in the Form a( h) 2 k The graph of the equation a( h) 2 k (a 0) is a parabola with verte (h, k), focus (h, k p), and directri k p, where a 4 p. If a 0, the parabola opens upward; if a 0, the parabola opens downward. Figure 3.8 shows the location of the focus and directri for parabolas with verte (h, k) and opening either upward or downward. Note that the location of the focus and directri determine the value of a and the shape and opening of the parabola. a > 0 (h, k + p) (h, k) Directri: = k p a 4p = a( h) 2 + k a < 0 Directri: = k p (h, k) = a( h) 2 + k (h, k + p) FIGURE 3.8 CAUTION For a parabola that opens upward, p 0, and the focus (h, k p) is above the verte (h, k). For a parabola that opens downward, p 0, and the focus (h, k p) is below the verte (h, k). In either case the distance from the verte to the focus and the verte to the directri is p. Finding the Verte, Focus, and Directri In Eample we find the verte, focus, and directri from an equation of a parabola. In Eample 2 we find the equation given the focus and directri. E X A M P L E Finding the verte, focus, and directri, given an equation Find the verte, focus, and directri for the parabola 2.

4 ( 3.2 The Parabola (3 3) 70 ( 0, 4 = 2 ( = 4 FIGURE 3.9 Solution Compare 2 to the general formula a( h) 2 k. We see that h 0, k 0, and a. So the verte is (0, 0). Because a, we can use a 4 p to get 4 p, or p 4. Use (h, k p) to get the focus 0, 4. Use the equation k p to get as the equation of the directri. See Fig E X A M P L E 2 Finding an equation, given a focus and directri Find the equation of the parabola with focus (, 4) and directri 3. Solution Because the verte is halfwa between the focus and directri, the verte is, 7 2. See Fig The distance from the verte to the focus is. Because the 2 focus is above the verte, p is positive. So p 2, and a 4 p. The equation is 2 ( 7, 2 (, 4) 2 ( ( )) = 3 Convert to a 2 b c form as follows: 2 ( )2 7 2 FIGURE (2 2 ) = b 2a or = h (h, k) Ais of Smmetr The graph of 2 shown in Fig. 3.9 is smmetric about the -ais because the two halves of the parabola would coincide if the paper were folded on the -ais. In general, the vertical line through the verte is the ais of smmetr for the parabola. See Fig. 3.. In the form a 2 b c the -coordinate of the verte is b (2a) and the equation of the ais of smmetr is b (2a). In the form a( h) 2 k the verte is (h, k) and the equation for the ais of smmetr is h. Ais of smmetr FIGURE 3. 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 a( h) 2 k to the form a 2 b c, we square the binomial and combine like terms, as in Eample 2. To change from a 2 b c to the form a( h) 2 k, we complete the square, as in the net eample.

5 702 (3 4) Chapter 3 Nonlinear Sstems and the Conic Sections E X A M P L E 3 calculator close-up The graphs of and 2 2( ) 2 3 appear to be identical. This supports the conclusion that the equations are equivalent. 0 Converting a 2 b c to a( h) 2 k Write in the form a( h) 2 k and identif the verte, focus, directri, and ais of smmetr of the parabola. Solution Use completing the square to rewrite the equation: 2( 2 2) 5 2( 2 2 ) 5 2( 2 2 ) 2 5 2( ) 2 3 The verte is (, 3). Because a 4, we have p 4p 2, Complete the square. Move 2( ) outside the parentheses. 5 and p 8. Because the parabola opens upward, the focus is 8 unit above the verte 5, and the directri is the horizontal line 8 unit below the verte, at, 3 8, or, or 2 3. The ais of smmetr is CAUTION Be careful when ou complete a square within parentheses as in Eample 3. For another eample, consider the equivalent equations 3( 2 4), 3( ), and 3( 2) 2 2. E X A M P L E 4 A calculator graph can be used to check the verte and opening of a parabola. calculator close-up Finding the features of a parabola from standard form Find the verte, focus, directri, and ais of smmetr of the parabola , and determine whether the parabola opens upward or downward. Solution The -coordinate of the verte is b a 2( 3) 6 2. To find the -coordinate of the verte, let 3 2 in : The verte is 3 2, 7 4. Because a 3, the parabola opens downward. To find the focus, use 3 4 p to get p. The focus is 2 of a unit below the verte at 3 2, or 3 2, 5 3. The directri is the horizontal line of a unit above 2 the verte, or 6. The equation of the ais of smmetr is 3 2.

6 3.2 The Parabola (3) 703 WARM-UPS True or false? Eplain our answer.. There is a parabola with focus (2, 3), directri, and verte (0, 0). 2. The focus for the parabola 4 2 is (0, 2). 3. The graph of 3 5( 4) 2 is a parabola with verte (4, 3). 4. The graph of is a parabola. 5. The graph of is a parabola opening upward. 6. For 2 the verte and -intercept are the same point. 7. A parabola with verte (2, 3) and focus (2, 4) has no -intercepts. 8. The parabola with focus (0, 2) and directri opens upward. 9. The ais of smmetr for a( 2) 2 k is If a and a, then p (4 p) EXERCISES Reading and Writing After reading this section, write out the answers to these questions. Use complete sentences.. What is the definition of a parabola given in this section? 2. What is the location of the verte? ( 3) What are the two forms of the equation of a parabola? 2. 4 ( 2) What is the distance from the focus to the verte in an parabola of the form a 2 b c? 3. ( ) ( 4) 2 5. How do we convert an equation of the form a 2 b c into the form a( h) 2 k? 6. How do we convert an equation of the form a( h) 2 k into the form a 2 b c? Find the equation of the parabola with the given focus and directri. See Eample Focus (0, 2), directri 2 6. Focus (0, 3), directri 3 Find the verte, focus, and directri for each parabola. See Eample Focus 0, 2, directri 2 8. Focus 0, 8, directri 8

7 704 (3 6) Chapter 3 Nonlinear Sstems and the Conic Sections 9. Focus (3, 2), directri Focus ( 4, 5), directri 4 2. Focus (, 2), directri Focus (2, 3), directri 23. Focus ( 3,.25), directri Focus 5, 7 8, directri 5 8 Write each equation in the form a( h) 2 k. Identif the verte, focus, directri, and ais of smmetr of each parabola. See Eample Solve each problem. 43. World s largest telescope. The largest reflecting telescope in the world is the 6-meter (m) reflector on Mount Pastukhov in Russia. The accompaning figure shows a cross section of a parabolic mirror 6 m in diameter with the verte at the origin and the focus at (0, 5). Find the equation of the parabola (0, 5) Find the verte, focus, directri, and ais of smmetr of each parabola (without completing the square), and determine whether the parabola opens upward or downward. See Eample m FIGURE FOR EXERCISE Arecibo Observator. The largest radio telescope in the world uses a 000-ft parabolic dish, suspended in a valle in Arecibo, Puerto Rico. The antenna hangs above

8 3.2 The Parabola (3 7) 705 the verte of the dish on cables stretching from two towers. The accompaning figure shows a cross section of the parabolic dish and the towers. Assuming the verte is at (0, 0), find the equation for the parabola. Find the distance from the verte to the antenna located at the focus Antenna at focus 200 ft 200 ft ft FIGURE FOR EXERCISE 44 Graph both equations of each sstem on the same coordinate aes. Use elimination of variables to find all points of intersection Solve each problem. 53. Find all points of intersection of the parabola and the -ais. 54. Find all points of intersection of the parabola and the -ais. 55. Find all points of intersection of the parabola and the line Find all points of intersection of the parabola and the line. 57. Find all points of intersection of the parabolas 2 and 2.

9 706 (3 8) Chapter 3 Nonlinear Sstems and the Conic Sections 58. Find all points of intersection of the parabolas 2 and ( 3) 2. c) Sketch the graphs 2( 3) 2 and ( ) 2 2. GETTING MORE INVOLVED 59. Eploration. Consider the parabola with focus ( p, 0) and directri p for p 0. Let (, ) be an arbitrar point on the parabola. Write an equation epressing the fact that the distance from (, ) to the focus is equal to the distance from (, ) to the directri. Rewrite the equation in the form a 2, where a. 4 p 60. Eploration. In general, the graph of a( h) 2 k for a 0 is a parabola opening left or right with verte 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 ais of smmetr? GRAPHING CALCULATOR EXERCISES 6. Graph 2 using the viewing window with and 0. Net graph 2 2 using the viewing window 2 2 and 7. Eplain what ou see. 62. Graph 2 and 6 9 in the viewing window 5 5 and Does the line appear to be tangent to the parabola? Solve the sstem 2 and 6 9 to find all points of intersection for the parabola and the line. 3.3 THE CIRCLE In this section Developing the Equation Equations Not in Standard Form Sstems of Equations r (h, k) (, ) In this section we continue the stud of the conic sections with a discussion of the circle. Developing the Equation A circle is obtained b cutting a cone, as was shown in Fig We can also define a circle using points and distance, as we did for the parabola. Circle A circle is the set of all points in a plane that lie a fied distance from a given point in the plane. The fied distance is called the radius, and the given point is called the center. 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 If (, ) is a point on the circle, its distance from the center is r. So FIGURE 3.2 ( ) h 2 ( 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 ( h) 2 ( k) 2 r 2 with r 0, is a circle with center (h, k) and radius r.

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