advertisement

Chapter 7 A1 Glencoe Precalculus DATE PERIOD Before you begin Chapter 7 Conic Sections and Parametric Equations Anticipation Guide A D A 7. The transverse axis of a hyperbola is longer than the conjugate axis. 8. A circle is a special type of ellipse. After you complete Chapter 7 D 6. The graph of a degenerate conic is a line, two intersecting lines, or a point. A 4. There are formulas to find the coordinates of a point on the graph of a rotated conic section. 5. A conic section could be a triangle or a square. D D 3. The general form of the equation of a conic section is y = mx + b. A 2. The eccentricity of an ellipse is greater than 1. 3 Chapter Resources 10/24/09 3:59:34 PM Glencoe Precalculus • For those statements that you mark with a D, use a piece of paper to write an example of why you disagree. • Did any of your opinions about the statements change from the first column? Chapter 7 STEP 2 A or D 1. The graph of y 2 = 4x is a parabola. Statement • Reread each statement and complete the last column by entering an A or a D. Step 2 STEP 1 A, D, or NS • Write A or D in the first column OR if you are not sure whether you agree or disagree, write NS (Not Sure). • Decide whether you Agree (A) or Disagree (D) with the statement. • Read each statement. Step 1 7 NAME 0ii_004_PCCRMC07_893808.indd Sec1:3 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. DATE PERIOD (3, -4) (3, -1) (h, k) (h, k + p) directrix: axis of symmetry: y = -7 x=3 y=k-p x=h 1 -3 − 4 11 -3 − 12 11 -3 − 12 1 -3 − 4 y 0 y x 2 ) Answers Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 005_020_PCCRMC07_893808.indd 5 Chapter 7 2 1 y = 3; x = 5 − (6, 3); 6 −1 , 3 ; ( 3. (y − 3)2 = 2(x - 6) y = -1; x = 1 (3, -1); (5, -1); 0 4 8 12 y 0 4 8 12 x x 5 x = 3; y = -5 (3, -2); (3, 1); 1 (x − 3)2 = y + 2 4. − ) ( 12 x = -2; y = 0 (-2, 1); (-2, 2); 12 −2 2 0 4 y 4 8 x 12 8x Lesson 7-1 11/9/09 10:07:10 PM Glencoe Precalculus −4 4 8 −4 −8 −4 0 4 8 For each equation, identify the vertex, focus, axis of symmetry, and directrix. Then graph the parabola. y y 1. (y + 1)2 = 8(x - 3) 2. (x + 2)2 = 4(y - 1) Exercises 6 4 2 0 x Graph the vertex, focus, axis, and directrix of the parabola. Then make a table of values to graph the general shape of the curve. vertex: focus: The equation is in the form (x - h)2 = 4p(y - k), so h = 3 and k = -4. Use the values of h, k, and p to determine the characteristics of the parabola. The equation is in standard form and the squared term is x, which means that the parabola opens vertically. Because 4p = 12, p = 3 and the graph opens upward. Example For (x - 3)2 = 12(y + 4), identify the vertex, focus, axis of symmetry, and directrix. Then graph the parabola. A parabola is the locus of all points in a plane equidistant from a point called the focus and a line called the directrix. The standard form of the equation of a parabola that opens vertically is (x - h)2 = 4p(y - k). When p is negative, the parabola opens downward. When p is positive, it opens upward. The standard form of the equation of a parabola that opens horizontally is (y - k)2 = 4p(x - h). When p is negative, the parabola opens to the left. When p is positive, it opens to the right. Parabolas Study Guide and Intervention Analyze and Graph Parabolas 7-1 NAME Answers (Anticipation Guide and Lesson 7-1) Parabolas Study Guide and Intervention DATE −4 A2 −4 −4 y 4 8x x −8 Glencoe Precalculus 005_020_PCCRMC07_893808.indd 6 Chapter 7 y 4 8x 6 −8 −4 0 4 8 y 4 8x 2(x - 1) = (y - 1)2 −8 8 0 −4 −8 4 2 4x Glencoe Precalculus 4. focus (1.5, 1); opens right; directrix x = 0.5 −8 8 (x - 1)2 = -4(y - 5) −4 y −12 −4 y 2. focus (1, 4); opens down; contains (-3, 1) −4 −4 0 4 8 (x - 5)2 = -12(y - 3) 3. directrix y = 6; opens down; vertex (5, 3) −8 4 8 12 (y - 5)2 = -12(x - 2) 1. focus (-1, 5) and vertex (2, 5) Write an equation for and graph a parabola with the given characteristics. Exercises Graph the vertex, focus, and parabola. y - (-3)2 = 4(-5)(x - 1) p = -5, h = 1, and k = -3 Simplify. (y + 3)2 = -20(x - 1) The standard form of the equation is (y + 3)2 = -20(x - 1). 4 −2 0 Write an equation for and graph a parabola with focus (-4, -3) and vertex (1, -3). 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 -4 - 1 or -5. Because p is negative, the graph opens to the left. Write the equation for the parabola in standard form using the values of h, p, and k. Standard form (y - k)2 = 4p(x - h) Example Specific characteristics can be used to determine the equation of a parabola. (continued) PERIOD 10/24/09 4:05:41 PM Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Chapter 7 Equations of Parabolas 7-1 NAME Parabolas Practice DATE PERIOD x 0 y x 0 y (y - 1)2 = 8x x 4. focus (2, 1); opens right; contains (8, -7) Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 005_020_PCCRMC07_893808.indd 7 Chapter 7 x 2 = 8y 7 Lesson 7-1 3/13/09 4:23:17 PM Glencoe Precalculus 6. SATELLITE DISH Suppose the receiver in a parabolic dish antenna is 2 feet from the vertex and is located at the focus. Assume that the vertex is at the origin and that the dish is pointed upward. Find an equation that models a cross section of the dish. (x + 4)2 = -4(y - 2); (-4, 2); (-4, 1); x = -4; y = 3 5. Write x2 + 8x = -4y - 8 in standard form. Identify the vertex, focus, axis of symmetry, and directrix. 0 y (x + 2)2 = -4(y - 4) 3. vertex (-2, 4), focus (-2, 3) x vertex: (1, -3); focus: (-2, -3); axis of symmetry: y = -3; directrix: x = 4 2. y2 + 6y + 9 = 12 - 12x Write an equation for and graph a parabola with the given characteristics. 0 y vertex: (1, 2); focus: (1, 4); axis of symmetry: x = 1; directrix: y = 0 1. (x - 1)2 = 8(y - 2) Identify the vertex, focus, axis of symmetry, and directrix for each equation. Then graph the parabola. 7-1 NAME Answers (Lesson 7-1) Chapter 7 Parabolas A3 8 40 ft 10 ft T −30 −20 −10 0 y 20x 60 ft 22.5 ft Glencoe Precalculus b. Find the length of a supporting wire that is 100 feet from the center. x2 = 800 y a. Write an equation for the parabolic cable. 400 ft 10 ft 5. BRIDGES The cable for a suspension bridge is in the shape of a parabola. The vertical supports are shown in the figure. −20 5 b. Graph the equation. x2 = -40 y 3/13/09 4:23:22 PM Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Glencoe Precalculus 005_020_PCCRMC07_893808.indd 8 Chapter 7 1 2 y=− x 2 3. FLASHLIGHT A flashlight contains a parabolic mirror with a bulb in the center as a light source and focus. If the width of the mirror is 4 inches at the top and the height to the focus is 0.5 inch, find an equation of the parabolic cross section. 21 ft b. What is the maximum height that a T-shirt reaches? 1 (x - 1)2 = - − (y - 21) 16 a. Write the equation in standard form. 2. T-SHIRTS The cheerleaders at the high school basketball game launch T-shirts into the stands after a victory. The launching device propels the shirts into the air at an initial velocity of 32 feet per second. A shirt’s distance y in feet above the ground after x seconds can be modeled by y = -16x2 + 32x + 5. I a. Write an equation that models the parabola. N VERSI Y 4 in. x = 16y, where the values of x and y are measured in inches. Find the distance from the vertex to the focus of this mirror. 2 PERIOD 4. ARCHWAYS The entrance to a college campus has a parabolic arch above two columns as shown in the figure. Word Problem Practice 1. REFLECTOR The figure shows a parabolic reflecting mirror. A cross section of the mirror can be modeled by 7-1 DATE U NAME Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Enrichment DATE d y-x-2 2 ) D(a, b) P(x, y) Chapter 7 9 y = -x - 2 F(1, 1) m x Lesson 7-1 3/13/09 4:23:28 PM Glencoe Precalculus (0, 0), because (0, 0) is midway between point F and line d. b. Use your answer from part a to find the coordinates of the vertex of the parabola. Justify your answer. Answers Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 005_020_PCCRMC07_893808.indd 9 0 y = x, because y = x contains F(1, 1) and is perpendicular to d. 4. a. Every parabola has an axis of symmetry. Find an equation for the axis of symmetry of the parabola described above. Justify your answer. x2 - 2xy + y2 - 8x - 8y = 0 y PERIOD 3. Use the coordinates of F, P, and D, along with (PD)2 = (PF )2 to find an equation of the parabola with focus F and directrix d. y-x-2 Add to get b = − . 2 a + b = -2. Subtract to get a = − . x-y-2 2 From the equation for line m, -a + b = -x + y. From the equation for d, of point D are D(a, b) = − , − . ( x - y2 - 2 2. Use the equations for lines m and d to show that the coordinates x - y + (b - a) = 0 1. Find an equation for line m. Refer to the discussion above. The diagram at the right shows a fixed point F (1, 1) and a line d that has an equation y = -x - 2. If P (x, y) satisfies the condition that PD = PF, then P is on a parabola. Our objective is to find an equation for the tilted parabola, which is the locus of all points that are the same distance from (1, 1) and the line y = -x - 2. First find an equation for the line m through P(x, y) and perpendicular to line d at D(a, b). Using this equation and the equation for line d, find the coordinates (a, b) of point D in terms of x and y. Then use (PD) 2 = (PF ) 2 to find an equation for the parabola. Tilted Parabolas 7-1 NAME Answers (Lesson 7-1) DATE Ellipses and Circles Study Guide and Intervention y PERIOD A01_A14_PCCRMC07_893808.indd 4 b b a a A4 vertical (-2, 1) (-2, 6), (-2, -4) (-5, 1), (1, 1) x = -2 y=1 y=k x=h 25 −8 −8 −4 −4 0 4 8 y 25 x 4 8 Pdf Pass y 005_020_PCCRMC07_893808.indd 10 −8 −4 −4 0 Chapter 7 −8 16 4 9 8x (y - 1)2 (x + 3)2 3. − + − = 1 64 (y + 2)2 (x + 5)2 1. − + − = 1 25 64 −8 −4 0 y 9 4 25 8x 10 −8 −4 −4 0 y 4 8x (y + 3)2 (x - 2)2 4. − + − = 1 −8 4 8 (y + 1)2 (x + 2)2 2. − + − = 1 (h ± b, k) (h, k ± a) (h, k) Graph the ellipse given by each equation. Exercises orientation: center: vertices: co-vertices: major axis: minor axis: 0 7 9 ' $ −8 y 4 8x ' 7 x Glencoe Precalculus −8 0 4 8 (y - 1)2 (x + 2)2 Graph the ellipse given by the equation − + − = 1. The equation is in standard form. Use the values of h, k, a, and b to determine the vertices and axes of the ellipse. Since a2 > b2, a2 = 25 and b2 = 9, or a = 5 and b = 3. Since a2 is the denominator of the y-term, the major axis is parallel to the y-axis. Example a2 is in the denominator of the y-term. In both cases, c2 = a2 - b2. case, a2 is in the denominator of the x-term. The standard form is (y - k)2 (x - h)2 − +− = 1 when the major axis is vertical. In this case, 2 2 11/15/09 4:12:41 PM Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Chapter 7 The standard form of the equation of an ellipse is (y - k)2 (x - h)2 + − = 1 when the major axis is horizontal. In this − 2 2 An ellipse is the locus of points in a plane such that the sum of the distances from two fixed points, called foci, is constant. Analyze and Graph Ellipses and Circles 7-2 NAME DATE Ellipses and Circles Study Guide and Intervention (continued) PERIOD 4 2 2 Divide each side by 36. ( 2) 4 = 9, - − =4 Factor. (−62 ) 2 = 64 Standard form Factor. (−162 ) Complete the square. Original equation 5 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 005_020_PCCRMC07_893808.indd 11 Chapter 7 15 12 (y - 3) (x + 1)2 − + − = 1; ellipse 2 5. 4x2 + 8x + 5y2 - 30y - 11 = 0 (y + 2)2 = -4(x - 3); parabola 3. 4x - 8 + y2 + 4y = 0 30 (y + 1)2 (x - 2)2 − + − = 1; ellipse 1. y2 + 2y + 6x2 - 24x = 5 11 3 Lesson 7-2 10/24/09 4:28:19 PM Glencoe Precalculus 1 (x + 2)2 = - − (y - 17); parabola 6. 6x2 + 24x + 2y - 10 = 0 (x + 2)2 + (y - 1)2 = 54; circle 4. x2 + 4x + y2 - 2y - 49 = 0 (x - 12)2 + (y + 1)2 = 169; circle 2. y2 + 2y + x2 - 24x = 14 Write each equation in standard form. Identify the related conic. Exercises Because only one term is squared, the graph is a parabola with vertex (8, 2). (x − 8)2 = 8(y − 2) (x − 8)2 − 8(y − 2) = 0 (x2 − 16x + 64) − 8y + 80 - 64 = 0 (x2 − 16x + ?) − 8y + 80 = 0 b. x2 - 16x - 8y + 80 = 0 x2 - 16x - 8y + 80 = 0 an ellipse with center (−3, 2). (y – k)2 (x - h)2 Because the equation is of the form − +− = 1, the graph is b2 a2 9 2 4(x + 3) + 9(y - 2) = 36 (y - 2)2 (x + 3)2 −+− =1 2 4(x2 + 6x + 9) + 9(y2 - 4y + 4) = -36 + 36 + 36 Example Write each equation in standard form. Identify the related conic. a. 4x2 + 9y2 + 24x - 36y + 36 = 0 4x2 + 9y2 + 24x - 36y + 36 = 0 Original equation 4(x2 + 6x + ?) + 9(y2 - 4y + ?) = -36 + ? + ? Complete the square. If you are given the equation for a conic section, you can determine what type of conic is represented using the characteristics of the equation. The standard form of an equation for a circle with center (h, k) and radius r is (x - h)2 + (y - k)2 = r2. Determine Types of Conic Sections 7-2 NAME Answers (Lesson 7-2) Glencoe Precalculus 3/15/10 10:03:09 PM Chapter 7 Ellipses and Circles Practice x PERIOD −4 0 4 y 4 x 2. 25x2 + 9y2 - 50x - 90y + 25 = 0 DATE 28 2 A5 16 12 2 25 16 64 √ 55 − 8 9 (y + 2)2 (x + 1)2 10. − + − = 1 12 10/24/09 4:28:30 PM Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Glencoe Precalculus 005_020_PCCRMC07_893808.indd 12 Chapter 7 1.5 ft to the sides of the center 11. CONSTRUCTION A semi-elliptical arch is used to design a headboard for a bed frame. The headboard will have a height of 2 feet at the center and a width of 5 feet at the base. Where should the craftsman place the foci in order to sketch the arch? 3 − 5 (y + 1)2 (x + 1)2 9. − + − = 1 Glencoe Precalculus (x + 10)2 + (y - 5)2 = 121; circle 8. x2 + y2 + 20x - 10y + 4 = 0 9 2 (y - 3) (x - 4) − + − = 1; ellipse 6. 4x2 - 32x + 3y2 - 18y = -55 Determine the eccentricity of the ellipse given by each equation. (x - 4)2 + (y - 12)2 = 169; circle 7. x2 + y2 - 8x - 24y = 9 (y - 2)2 + 4(x + 5); parabola 5. y2 - 4y = 4x + 16 Write each equation in standard form. Identify the related conic. 25 (y - 4)2 (x + 2)2 − + − =1 4. foci (-2, 1), (-2, 7); length of major axis 10 units 64 2 (y - 6) (x + 4) − + − =1 3. vertices (-12, 6), (4, 6); foci (-10, 6), (2, 6) Write an equation for the ellipse with each set of characteristics. 0 y 1. 4x2 + 9y2 - 8x - 36y + 4 = 0 Graph the ellipse given by each equation. 7-2 NAME Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Ellipses and Circles y2 368.64 y2 225 Answers Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 005_020_PCCRMC07_893808.indd 13 Chapter 7 ≈13 ft b. Find the height of the tunnel 10 feet from the center. 400 x − +−=1 2 a. Write an equation that models the ellipse. 40 ft 15 ft 3. TUNNEL The entrance to a tunnel is in the shape of half an ellipse as shown in the figure. 38.4 in. the sign? b. What is the maximum height of 576 x − +− =1 2 a. Write an equation for the ellipse if the center of the sign is at the origin and the major axis is horizontal. 2. SIGNS A sign is in the shape of an ellipse. The eccentricity is 0.60 and the length is 48 inches. ≈35 ft on either side of the center on the major axis b. Find the location of the foci. y2 x2 − +−=1 1764 529 a. Write an equation modeling the shape of the room. Assume that it is centered at the origin and that the major axis is horizontal. 13 DATE PERIOD 0y ds 70 Ho us e 100 100 x x2 + y2 = 40,000; x2 + y2 = 810,000 Lesson 7-2 10/24/09 4:06:33 PM Glencoe Precalculus c. Write an equation for the circle of water at the current time and an equation for the circle when the water reaches the house. 7 days b. If the pond continues to overflow at the same rate, how many days will it take for the water to reach the house? ; 900 yd −100 −100 0 y a. Graph the circle that represents the water, and find the distance from the center of the pond to the house. ds 0y 20 increases the radius 100 yards per day, as shown below. 4. RETENTION POND A circular retention pond is getting larger by overflowing and flooding the nearby land at a rate that Word Problem Practice 1. WHISPERING GALLERY A whispering gallery at a museum is in the shape of an ellipse. The room is 84 feet long and 46 feet wide. 7-2 NAME Answers (Lesson 7-2) ⎪ by ⎥ ellipse y 2 x 2 b. − + − =1 2 circle A6 8 y ⎪−2x ⎥ + ⎪−3 ⎥ 6 ⎪−2x ⎥ ⎪ ⎥ 8 =1 Glencoe Precalculus 005_020_PCCRMC07_893808.indd 14 Chapter 7 14 a rectangle that is 6 units long and 4 units wide, centered at the origin approximate for greater and greater even, whole-number values of n? ⎪⎥ ⎪⎥ y n x n + − =1 3. What shape will the graph of − 2 3 c. a = 2, b = 3, n = 8 =1 b. a = 2, b = 3, n = 6 6 y + − 3 y =1 4 ⎪−2x ⎥ + ⎪−3 ⎥ a. a = 2, b = 3, n = 4 4 2. In each of the following cases, you are given values of a, b, and n to use in the general equation. Write the resulting equation. Then graph. Sketch each graph on the grid at the right. ⎪3⎥ ⎪2⎥ 2 ⎪2⎥ ⎪2⎥ x a. − + − =1 y 1. Consider two curves that are not superellipses. Graph each equation on the grid at the right. State the type of curve produced each time. For even values of n greater than 2, the curves are called superellipses. x for the family is ⎪− a ⎥ + − = 1, with a ≠ 0, b ≠ 0, and n > 0. n The circle and the ellipse are members of an interesting family of curves that were first studied by the French physicist and mathematician Gabriel Lamé (1795-1870). The general equation n Enrichment DATE 1 0 1 0 1 1 x x Glencoe Precalculus y y PERIOD 3/13/09 4:23:56 PM Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Chapter 7 Superellipses 7-2 NAME DATE (y + 2)2 32 2 2 18 Take the square root of each side. Divide each side by 18. Subtract 32(x - 3)2 from each side. Multiply each side by 576. Original equation PERIOD 3 ) 2nd ) 1 ) (–) ÷ 18 [{]1 , — 2nd 2 ENTER [}] 2nd ZOOM 6. [ √ ] ( 576 — 32 ( 32 1–4. See students’ work. 18 32 (y + 6)2 (x - 3)2 4. − + − = 1 (y + 2)2 (x + 1)2 2. − + − = 1 18 32 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 005_020_PCCRMC07_893808.indd 15 Chapter 7 15 Sample answer: the graph will move up or down. Lesson 7-2 3/13/09 4:24:01 PM Glencoe Precalculus 6. Describe the effects of replacing y + 2 in Exercises 1 and 2 with (y ± c) for c > 0. Sample answer: the graph will move right or left. 5. Describe the effects of replacing x - 3 in Exercises 3 and 4 with (x ± c) for c > 0. 18 32 (y - 2)2 (x - 3)2 3. − + − = 1 18 (x - 7)2 (y + 2)2 1. − + − = 1 Graph each equation on a graphing calculator. Exercises Like other graphs, there are families of ellipses. Changing certain values in the equation of an ellipse creates a new member of that family. — Y= The result is two equations. To graph both equations in Y1, replace ± with {1, -1}. Be careful to include the proper parentheses or you will get an error message. √ 576 - 32(x - 3)2 y=± − -2 576 - 32(x - 3)2 (y + 2)2 = − 18 18(y + 2) = 576 - 32(x - 3) 2 32(x - 3) + 18(y + 2) = 576 2 − + − =1 (x - 3)2 18 graphing calculator, you must first solve for y. (x - 3)2 18 To graph an ellipse, such as − + − = 1, on a (y + 2)2 32 Graphing Calculator Activity Translations of Ellipses 7-2 NAME Answers (Lesson 7-2) Chapter 7 Hyperbolas Study Guide and Intervention DATE PERIOD b b a y2 16 2 4 Simplify. a = 4 and b = 2 A7 (0, 0) vertices: (0, 4), (0, -4) -5.65, 5.65 2 −8 4 4 16 4 8x −8 4 −4 0 8x 3/18/09 11:56:28 PM Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Glencoe Precalculus 005_020_PCCRMC07_893808.indd 16 4 Glencoe Precalculus −8 −4 0 −8 −8 −8 8x 8 −4 4 36 y 16 y (x - 1)2 −4 y 8x (y + 2)2 3. − - − = 1 4 b a y - k = ±− (x - h) (y - 3)2 (x + 2)2 2. − - − = 1 25 9 −8 −4 0 −4 −4 0 Chapter 7 −8 8 y2 x2 1. − -−=1 25 36 Graph the hyperbola given by each equation. Exercises -4.5, 4.5 1 -4.5, 4.5 -4, 4 -1 0 y -5.65, 5.65 x -2 y asymptotes: y = 2x, y = -2x (h, k ± a) 8 foci: (h, k) Make a table of values to sketch the hyperbola. center: (0, √ 20 ), (0, - √ 20 ) Equation relating a, b, and c Determine the characteristics of the hyperbola. c2 = a2 + b2 c2 = 42 + 22 or about 4.47 c = √20 Because a2 = 16 and b2 = 4, a = 4 and b = 2. Use the values of a and b to find the value of c. (h, k ± c) x Graph the hyperbola given by the equation − - − = 1. The equation is in standard form. Both h and k are 0, so the center is at the origin. Because the x-term is subtracted, the transverse axis is vertical. Use the values of a, b, and c to determine the vertices and foci of the hyperbola. Example cases, a2 + b2 = c2. (y - k)2 (x - h2) − -− = 1 when the transverse axis is vertical. In both 2 2 a Analyze and Graph Hyperbolas A hyperbola is the locus of all points in a plane such that the difference of their distances from two foci is constant. The standard form of the equation of a hyperbola is (y - k)2 (x - h)2 − - − = 1 when the transverse axis is horizontal, and 2 2 7-3 NAME Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. DATE Hyperbolas Study Guide and Intervention (continued) PERIOD hyperbola greater than 0 Use the discriminant to identify each conic section. parabola equal to 0 2 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Answers 005_020_PCCRMC07_893808.indd 17 Chapter 7 parabola 7. 25x2 + 100x − 54y = −200 ellipse 5. 5x2 + 2xy + 4y2 + x + 2y + 17 = 0 hyperbola 3. -2x + 6xy + y - 4x - 5y + 2 = 0 2 circle 1. 4x2 + 4y2 - 2x - 9y + 1 = 0 17 hyperbola Lesson 7-3 10/24/09 4:28:42 PM Glencoe Precalculus 8. 16x2 + 100x − 54y2 = −100 parabola 6. x2 + 2xy + y2 + x + 10 = 0 hyperbola 4. x2 + 6xy + y2 - 2x + 1 = 0 ellipse 2. 10x2 + 6y2 - x + 8y + 1 = 0 Use the discriminant to identify each conic section. Exercises A is 12, B is 12, and C is 3. Find the discriminant. B2 - 4AC = 122 - 4(12)(3) or 0 The discriminant is 0, so the conic is a parabola. c. 12x2 + 12xy + 3y2 - 7x + 2y - 6 = 0 A is 5, B is 8, and C is -2. Find the discriminant. B2 - 4AC = 82 - 4(5)(-2) or 104. The discriminant is greater than 0, so the conic is a hyperbola. b. 5x2 + 8xy - 2y2 + 4x - 3y + 10 = 0 A is 2, B is 0, and C is 6. Find the discriminant. B2 - 4AC = 02 - 4(2)(6) or -48 The discriminant is less than 0, so the conic must be either a circle or an ellipse. Because A ≠ C, the conic is an ellipse. a. 2x2 + 6y2 - 8x + 12y - 2 = 0 Example circle ellipse less than 0; B ≠ 0 or A ≠ C Conic Section less than 0; B = 0 and A = C Discriminant You can determine the type of conic when the equation for the conic is in general form, Ax2 + Bxy + Cy2 + Dx + Ey + F = 0. 2 The discriminant, or B - 4AC, can be used to identify a conic when the equation is in general form. Identify Conic Sections 7-3 NAME Answers (Lesson 7-3) Hyperbolas Practice x y2 16 (x – 1)2 0 4 y 2. − – − = 1 x PERIOD A8 16 9 (y - 1)2 x2 − -− =1 4. foci (0, 6), (0, –4); length of transverse axis 8 units 121 y2 5,759,600 Glencoe Precalculus 005_020_PCCRMC07_893808.indd 18 Chapter 7 parabola 9. 4x2 + 8xy + 4y2 + x + 11y + 10 = 0 ellipse 7. 5x2 + xy + 2y2 - 5x + 8y + 9 = 0 18 ellipse Glencoe Precalculus 10. 2x2 + 4y2 - 3x - 6y + 2 = 0 hyperbola 8. 16x2 − 4y2 - 8x - 8y + 1 = 0 Use the discriminant to identify each conic section. 1,210,000 x − - − =1 2 6. ENVIRONMENTAL NOISE Two neighbors who live one mile apart hear an explosion while they are talking on the telephone. One neighbor hears the explosion two seconds before the other. If sound travels at 1100 feet per second, determine the equation of the hyperbola on which the explosion was located. √ 157 − 6 36 5. Determine the eccentricity of the hyperbola given by the equation (y + 10)2 (x - 7)2 − - − = 1. (y - 6)2 (x + 3)2 − - − =1 49 32 3. vertices (–10, 6), (4, 6); foci (–12, 6), (6, 6) Write an equation for the hyperbola with the given characteristics. 0 y 1. x2 - 4y2 - 4x + 24y - 36 = 0 DATE 3/13/09 4:24:16 PM Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Chapter 7 Graph the hyperbola given by each equation. 7-3 NAME Hyperbolas (x – 50)2 (y – 35)2 y 40 80 x 2 y 8 x 16 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 005_020_PCCRMC07_893808.indd 19 Chapter 7 − √ 317 14 c. Find the eccentricity. 11 14 y = ±− x b. Write the equations of the asymptotes. −8 −16 −8 0 8 16 a. Graph the hyperbola. 121 x by − - − = 1. y 196 2 2. SHADOWS A lamp projects light onto a wall in the shape of a hyperbola. The edge of the light can be modeled (-14, 35), (114, 35) b. Find the locations of the seismographs. 0 20 40 80 a. Graph the hyperbola. − - − = 1, where the 1600 2500 seismographs are located at the foci. DATE PERIOD (3, 4) x 4 9 19 2 y2 52 y2 144 −8 −4 −16 −8 0 4 8 y Sample answer: 8 b. Graph one hyperbola. 144 2 Lesson 7-3 10/24/09 4:06:46 PM Glencoe Precalculus 16 x 52 x x − - − = 1; − - − =1 a. Write two equations of the hyperbola in standard form if the center is at the origin, given that the path contains a transverse axis of 24 millimeters with one focus 14 millimeters from the center. 4. SHADOWS The path of the shadow cast by the tip of a sundial is usually a hyperbola. 12 ft b. If each unit on the coordinate plane represents 3 feet, what is the narrowest vertical width of the play area? (y - 4)2 (x - 3)2 − -− =1 a. Write an equation that models the curved sides of the play area. 0 y 3. PARKS A grassy play area is in the shape of a hyperbola, as shown. Word Problem Practice 1. EARTHQUAKES The epicenter of an earthquake lies on a branch of the hyperbola represented by 7-3 NAME Answers (Lesson 7-3) Chapter 7 Enrichment y2 b DATE -4 focus A9 -2 -4 -2 0 -2 -4 y 20 4 focus x 3/13/09 4:24:30 PM Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Glencoe Precalculus 005_020_PCCRMC07_893808.indd 20 Chapter 7 2 Glencoe Precalculus PERIOD The vertices recede to infinity and the branches become flatter and farther from the center. As c approaches infinity, the graphs tend to disappear. 4. Suppose b is held fixed and c approaches infinity. How does the graph of the hyperbola change? The vertices remain at (±a, 0), but the branches become more vertical. The graphs approach the vertical lines x = -a and x = a. 3. Suppose a is held fixed and c approaches infinity. How does the graph of the hyperbola change? The asymptotes remain the same, but the branches become sharper near the vertices. The graphs approach the lines y = x and y = -x. 2. Use a graphing calculator or computer to graph x2 - y2 = 1, x 2 - y 2 = 0.1, and x 2 - y 2 = 0.01. (Such hyperbolas correspond to smaller and smaller values of c.) Describe the changes in the graphs. What shape do the graphs approach as c approaches 0? Because 0 < a < c, as c approaches 0, a approaches 0. So the x-coordinates of the foci and vertices approach 0, which is the x-coordinate of the center. Because the y-coordinates are equal, the points become the same. 1. Write a convincing argument to show that as c approaches 0, the foci, the vertices, and the center of the hyperbola become the same point. Refer to the hyperbola described above. What happens to the shape of the graph as c grows very large or very small? b y=±− a x. Such a hyperbola is shown at the right. (c, 0), where c 2 = a 2 + b 2, the vertices are at (-a, 0) and (a, 0), and the asymptotes have equations a x - −2 = 1. The foci are at (-c, 0) and equation − 2 2 Recall that the equation of a hyperbola with center at the origin and horizontal transverse axis has the Moving Foci 7-3 NAME Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. DATE PERIOD 2 √2 2 4 √2 2 2 √2 x - √2 y 2 √2 x + √2 y 2 − 2xy √2 2 √2 x + √2 y 2 − 4y2 2 2 1 2 2 5 1 1 -− (x)2 - 5 x y - − (y)2 + − =0 2 2 2 2 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Answers 021_032_PCCRMC07_893808.indd 21 21 Simplify. binomials. Expand the Lesson 7-4 10/24/09 4:10:01 PM Glencoe Precalculus 7(y)2 = 160; hyperbola y - 2 = 0; circle 2 √2 Chapter 7 19(x)2 - 26 √ 3 xy - 2. 8x2 - 5y2 = 40, θ = 30° x + (x)2 + (y)2 - 2 √2 4 π 1. x2 - 4x + y2 = 2, θ = − Write each equation in the xy-plane for the given value of θ. Then identify the conic. Exercises The equation of the hyperbola after the 45° rotation is 5(x)2 + 10xy+ (y)2 = 1. 2 1 1 1 (x)2 - xy + − (y)2 - (x)2 + (y)2 - 2 (x)2 - 4xy - 2 (y)2 + − =0 − √2 x - √2 y 2 Replace x and y. 2 2 1 +− =0 √ 2 √ 2 = − x + − y y = x sin θ + y cos θ ( − ) − 2( − )( − ) − 4( − ) + − = 0 x2 4 π π sin − = − and cos − =− Rotation equations for x and y Substitute into the original equation. √2 2 = − x − − y x = x cos θ − y sin θ The conic is a hyperbola because B2- 4AC > 0. Find the equations for x and y. Example Use θ = 45° to write the standard form of 1 x2 - 2xy - 4y2 + − = 0 in the xy-plane. Then identify the conic. An equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 in the xy-plane can be rewritten as A(x)2 + C(y)2 + Dx + Ey + F = 0 in the xy-plane by rotating the coordinate axes through an angle θ. The equation in the xy-plane can be found using x = x cos θ - y sin θ and y = x sin θ + y cos θ. Rotations of Conic Sections Study Guide and Intervention Rotation of Conic Sections 7-4 NAME Answers (Lesson 7-3 and Lesson 7-4) Rotations of Conic Sections Study Guide and Intervention DATE 2 (x) (y) Graph − + − = 1 if it has been rotated 30° from 36 64 2 2 √3 2 1 sin 30° = − and cos 30° = − Rotation equations for x and y A10 Multiply. x = 0 and y = 0 √3 2 =0 Glencoe Precalculus −8 −4 y 021_032_PCCRMC07_893808.indd 22 Chapter 7 −8 −4 y' 4 −8 8x 22 y' −8 −4 0 4 y 4 x' 8x (y)2 (x)2 2. − - − = 1; 60° 4 9 x' 2 π 1. (x + 6)2 = 12(y + 2); − Graph each equation at the indicated angle. Exercises The new center, vertices, and co-vertices can be used to sketch the graph of the ellipse in the xy-plane. 4 Glencoe Precalculus −8 −4 −8 −4 0 4 8 y 8x √3 1 y=− (0) + − (0) 2 2 2 1 =− x + − y y = x sin θ + y cos θ , 3) (-6, 0) → (-3 √ (0, 8) → (−4, 4 √ 3 ) (0, -8) → (4, -4 √ 3 ) (6, 0) → (3 √3 3 , −3) Likewise, convert the coordinates of the vertices and co-vertices into xy-coordinates. =0 √3 1 x = − (0) - − (0) 2 2 Use the equations to convert the xy-coordinates of the center into xy-coordinates. 1 = − x - − y √3 2 x = x cos θ - y sin θ Find the equations for x and y for θ = 30°. The equation represents an ellipse, and it is in standard form. Use the center (0, 0); the vertices (0, 8), (0, -8); and the co-vertices (6, 0), (-6, 0) to determine the center and vertices for the ellipse in the xy-plane. its position in the xy-plane. Example 2 When the equations of rotated conics are given for the xy-plane, they can be graphed by finding points on the graph of the conic and then converting these points to the xy-plane. (continued) PERIOD 10/24/09 4:29:05 PM Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Chapter 7 Graph Rotated Conics 7-4 NAME Rotations of Conic Sections Practice DATE PERIOD (x)2 25 (y)2 4 -71x + 58 √ 3 xy - 13y2 = 400 2 π 4. − - − = 1; θ = − 3(x) 4 (y) 4 2 (y)2 3 (x)2 9 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 021_032_PCCRMC07_893808.indd 23 Chapter 7 y' y x' x 23 10. Graph (x + 1) 2 = -16(y + 3) if it has been rotated 45° from its position in the xy-plane. 45° (y)2 22 33.7°; − + − = 1 Lesson 7-4 10/24/09 4:10:10 PM Glencoe Precalculus 8. 18x2 + 12xy + 13y2 - 198 = 0 (x)2 9 63°; − + − = 1 6. 13x2 − 8xy + 7y2 - 45 = 0 9. COMMUNICATIONS Suppose the orientation of a satellite dish that monitors radio waves is modeled by the equation 4x 2 + 2xy + 4y2 + √ 2 x - √ 2 y = 0. What is the angle of rotation of the satellite dish about the origin? 53°; (y)2 = 2x 7. 16x2 − 24xy + 9y2 - 30x - 40y = 0 45°; − + − = 1 2 5. x2 + xy + y2 = 2 Determine the angle θ through which the conic with each equation should be rotated. Then write the equation in standard form. y x + 16 √ 2 x + 2xy - 16 √2 + y2 = 0 2 3. (x)2 = 16(y); θ = 45° 3 120 = 0; ellipse hyperbola Write an equation for each conic in the xy-plane for the given equation in xy form and the given value of θ. xy + 23(y)2 21(x)2 + 2 √3 2 2. 5x2 + 6y2 = 30; θ = 30° 1 1 − (x)2 - − (y)2 - 1 = 0; 2 4 π 1. xy = 1; θ = − Write each equation in the xy-plane for the given value of θ. Then identify the conic. 7-4 NAME Answers (Lesson 7-4) Chapter 7 A11 (x)2 16 y (y)2 20 60˚ x' x x 24 ) 45˚ x x' elliptical Glencoe Precalculus 5. SHAPES The shape of a reflecting mirror in a telescope can be modeled by 25x2 + 13xy + 2y2 = 100. Determine whether the reflector is elliptical, parabolic, or hyperbolic. 40° counterclockwise 4. LOGIC A hyperbola has been rotated 40° clockwise. Through what angle must it be rotated to return it to its original position? y' y c. Draw the graph in the xy-plane. -x y = 3 √2 b. Find the equation for the axis of symmetry in the xy-plane. ( √ 2 7 √2 - −, − 2 2 a. Find the vertex in the xy-plane. 3. ART Mimi is drawing a picture with rotated parabolas. She wants to graph (x - 3)2 = 12(y - 4) if it has been rotated 45° from the xy-plane. PERIOD 3/13/09 4:22:02 PM Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Glencoe Precalculus 021_032_PCCRMC07_893808.indd 24 0 y b. Draw the graph. 17x2 + 2 √ 3 xy + 19y2 - 320 = 0 a. Write an equation for the ellipse formed by the gear in the xy-plane. xy-plane is − + − = 1. 2. GEARS Suppose the equation of an elliptical gear rotated 60° in the y' b. Draw the graph. (x)2 √ 3 3 2 − - − xy + − y 32 16 32 √ 3 1 − x - − y = 0 2 2 a. Write an equation that models the new orientation of the satellite dish. when it is directly overhead. Later in the day, the dish has rotated 60°. 8 1 2 x modeled by the equation y = − Chapter 7 DATE Rotations of Conic Sections Word Problem Practice 1. COMMUNICATIONS A satellite dish is 7-4 NAME Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Enrichment DATE 1 0 y=x+6 y 1 y= ± √4x - x 2 y=6-x y = 6 - x ± √4x - x 2 x 2. y = x ± 0 1 √ x y Answers Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 021_032_PCCRMC07_893808.indd 25 Chapter 7 See students’ graphs. 7 + 6x - x2 ellipse; 3. y = 2x ± √ 1 # # y = ± √x 1 y = x ± √x 25 See students’ graphs. x Lesson 7-4 3/13/09 4:22:08 PM Glencoe Precalculus x y=x 4. y = - 2x ± √ -2x parabola; " y = ± √4x - x 2 y = x + 6 ± √4x - x 2 " PERIOD parabola Use a separate sheet of graph paper to graph these equations. State the type of curve for each graph. 0 1 y 1. y = 6 - x ± √ 4 - x2 ellipse Graph each equation. State the type of curve for each graph. Often, however, you can use the graphs of two simpler equations to graph a more complicated equation. For example, the graph of the ellipse in the diagram at the right is obtained by adding the y-coordinate of each point on the circle and the y-coordinate of the corresponding point of the line. Equations of parabolas, ellipses, and hyperbolas that are “tipped” with respect to the x- and y-axes are more difficult to graph than the equations you have been studying. Graphing with Addition of y-Coordinates 7-4 NAME Answers (Lesson 7-4) DATE Parametric Equations Study Guide and Intervention PERIOD A01_A14_PCCRMC07_893808.indd 12 -15 -11 -7 -3 -3 -2 -1 0 1 2 3 12 7 4 A12 x+2 4 13 9 5 1 19 12 7 4 3 y −8 −16 −8 0 8 16 24 y 8 x+2 4 4 4 Pdf Pass 4 021_032_PCCRMC07_893808.indd 26 Chapter 7 −12 −8 −4 −4 0 y 8 12 16 6 20 x 1. x = t2 + 4 and y = −t - 3; -4 ≤ t ≤ 4 26 0 3 y x Glencoe Precalculus 2. x = −t and y = √t + 2; 0 ≤ t ≤ 8 Sketch the curve given by each pair of parametric equations over the given interval. Exercises x2 x 5 The rectangular equation is y = − +− +− . 16 4 4 4 Simplify. 2 x x 5 =− +− +− x+2 4 Substitute − for t in the equation for y. Solve for t. Square − . 16 16x Parametric equation for x = − +1 x2 + 4x + 4 16 ⎛ x + 2 ⎞2 y=⎪−⎥ +1 ⎝ 4 ⎠ −=t x -3 Plot the (x, y) coordinates for each t-value and connect the points to form a smooth curve. Write x = 4t - 2 and y = t2 + 1 in rectangular form. 4 0 3 t y 19 x = 4t - 2 Example 2 x -19 t -4 Make a table of values for -4 ≤ t ≤ 4. 10/24/09 4:10:41 PM Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Chapter 7 Example 1 Sketch the curve given by the parametric equations x = -3 + 4t and y = t2 + 3 over the interval -4 ≤ t ≤ 4. Parametric equations are used to describe the horizontal and vertical components of an equation. Parameters are arbitrary values, usually time or angle measurement. Graph Parametric Equations 7-5 NAME DATE (continued) PERIOD 48˚ v0 = 35, θ = 48°, g = 32, and h0 = 2 Parametric equation for vertical position Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 021_032_PCCRMC07_893808.indd 27 Chapter 7 27 Lesson 7-5 11/15/09 4:17:20 PM Glencoe Precalculus 2. Jerome hits a tennis ball at an initial velocity of 38 feet per second and an angle of 42° with the horizontal from a height of 1.5 feet. How far away will the ball land if it is not hit by his opponent? about 46.6 ft 1. Julie is throwing a ball at an initial velocity of 28 feet per second and an angle of 56° with the horizontal from a height of 4 feet. How far away will the ball land? about 25 ft Exercises Step 4 Determine the horizontal position of the ball at 1.7 seconds. x = tv0 cos θ Parametric equation for horizontal position = 1.7(35) cos 48 v0 = 35, θ = 48°, and t = 1.7 ≈ 39.8 Use a calculator. The ball will travel about 39.8 feet before it hits the ground. Step 3 Graph the equation for the vertical position and the line y = 0. Use 5: INTERSECT on the CALC menu of a calculator to find the point of intersection of the curve with y = 0. The value is about 1.7 seconds. You could also use 2: ZERO and not graph y = 0. 2 1 = t(35) sin (48) - − (32)t2 + 2 2 1 2 gt + h0 y = tv0 sin θ - − Step 2 Write a parametric equation for the vertical position of the ball. 2 ft 35 ft/sec Step 1 Make a diagram of the situation. Example Luigi is kicking a soccer ball. He kicks the ball with an initial velocity of 35 feet per second at an angle of 48° with the horizontal. The ball is 2 feet above the ground when he kicks it. How far will the ball travel horizontally before it hits the ground? 2 1 2 and the vertical position y by y = tv0 sin θ - − gt + h0. Parametric equations are often used to simulate projectile motion. For an object launched at an angle θ with the horizontal at an initial velocity v0, where g is the gravitational constant, t is time, and h0 is the initial height, the horizontal distance x can be found by x = tv0 cos θ Parametric Equations Study Guide and Intervention Projectile Motion 7-5 NAME Answers (Lesson 7-5) Glencoe Precalculus 3/16/10 9:37:55 PM Chapter 7 Parametric Equations Practice DATE PERIOD y 6 10 14x A13 2 25 x −+− =1 y2 16 6. y = 4 sin θ, x = 5 cos θ y = -3(x - 5) 2 4. x = t + 5, y = -3t2 −12 −8 −4 y t = −1 4 8x t=0 2 t = −3 t = −5 0 t=3 t=1 t = −1 y x 8. t = 4x - 1 for y = x2 + 2 28 Glencoe Precalculus 3/13/09 4:22:25 PM Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Glencoe Precalculus 021_032_PCCRMC07_893808.indd 28 Chapter 7 0.72 s; 48.43 ft b. How long will it take the rocket to travel 10 feet horizontally from its starting point? What will be its vertical distance at that point? x = 80t cos 80°; y = 80t sin 80° - 16t2 a. Write parametric equations to represent the path of the rocket. 9. MODEL ROCKETRY Manuel launches a toy rocket from ground level with an initial velocity of 80 feet per second at an angle of 80° with the horizontal. t=2 −8 t=1 −4 0 3 2-x 3-x 7. t = − for y = − 2 Use each parameter to write the parametric equations that can represent each equation. Then graph the equation, indicating the speed and orientation. y2 x2 − +−=1 9 4 5. x = 3 sin θ, y = 2 cos θ 1 11 y=− x-− 2 2 3. x = 2t + 3, y = t - 4 Write each pair of parametric equations in rectangular form. −8 −4 0 2 −12 x −24 24 −16 16 2 −4 8 2 2 t 2. x = 2t + 6 and y = - − ; -5 ≤ t ≤ 5 −8 −8 0 y 2 1. x = t2 + 1 and y = −t - 6; -5 ≤ t ≤ 5 Sketch the curve given by each pair of parametric equations over the given interval. 7-5 NAME Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Parametric Equations Answers Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 021_032_PCCRMC07_893808.indd 29 Chapter 7 vertical: y = t(28) sin (60) - 16t2 + 5; horizontal: x = t(28) cos (60) 4. BASKETBALL Mandy throws a basketball with an initial velocity of 28 feet per second at an angle of 60° with the horizontal. If Mandy releases the ball from a height of 5 feet, write a pair of equations to determine the vertical and horizontal positions of the ball. Yes, at 20 feet the ball is about 3.8 feet above the ground. 3. TENNIS Melinda hits a tennis ball with an initial velocity of 42 feet per second at an angle of 16° with the horizontal from a height of 2 feet. She is 20 feet from the net and the net is 3 feet high. Will the ball go over the net? 8.97 ft b. What is the maximum height of the ball? 2.53 ft a. How high above the ground will the ball be when it gets to Sarah? 2. PLAYING CATCH Tom and Sarah are playing catch. Tom tosses a ball to Sarah at an initial velocity of 38 feet per second at an angle of 28° from a height of 4 feet. Sarah is 40 feet away from Tom. 1.98 m 29 DATE PERIOD 34˚ Lesson 7-5 3/13/09 4:22:33 PM Glencoe Precalculus Sample answer: Yes; at 400 ft the ball is 15.4 feet above the ground so it will clear the fence. c. If the fence is 8 feet tall and 400 feet from home plate, will the ball clear the fence to be a home run? Explain. 73.36 ft b. What is the maximum height the ball will reach? 421.6 ft a. How far will the ball travel horizontally before it hits the ground? 3 ft 120 ft/sec 6. BASEBALL Micah hit a baseball at an initial velocity of 120 feet per second from a height of 3 feet at an angle of 34°. about 61.9 ft b. Find the maximum height the ball reaches. x = 100t cos 39°, y = 100t sin 39° - 16t2 a. Write parametric equations for the flight of the ball. 5. GOLF Julio hit a golf ball with an initial velocity of 100 feet per second at an angle of 39° with the horizontal. Word Problem Practice 1. PHYSICS A rock is thrown at an initial velocity of 5 meters per second at an angle of 8° with the ground. After 0.4 second, how far has the rock traveled horizontally? 7-5 NAME Answers (Lesson 7-5) Enrichment DATE Time x A14 g cos α ) Glencoe Precalculus 021_032_PCCRMC07_893808.indd 30 Chapter 7 y = -0.013x 2 + 5.67x 1. a projectile fired at 80° to the horizontal with an initial velocity of 200 ft/s 30 Glencoe Precalculus y = -0.00037x 2 + 0.84x 2. a projectile fired at 40° to the horizontal with an initial velocity of 150 m/s Find the equation of the path of each projectile. y = -0.00035x 2 + 0.18x 9.8 y=- − x 2 + (tan 10°)x 2 2 ( 2(120) cos 10° ) Example Write the equation of a projectile fired at an angle of 10° to the horizontal with an initial velocity of 120 m/s. where g is the acceleration due to gravity, 9.8 m/s 2 or 32 ft/s 2, v 0 is the initial velocity, and α is the angle at which the projectile is fired. 0 x 2+ (tan α)x, y=- − 2 2 ( 2v The equation of the path of a projectile on the coordinate plane is given by The path assumes that gravity is the only force acting on the projectile. 0 Height y The path of a projectile after it is launched is a parabola when graphed on a coordinate plane. PERIOD 3/13/09 4:22:37 PM Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Chapter 7 Coordinate Equations of Projectiles 7-5 NAME Spreadsheet Activity DATE 2 4 6 8 10 2 3 4 5 6 B -4.75 -0.99302 2.935661 6.356741 8.676882 x C 8.209921 9.428068 9.016016 7.045013 3.855863 y y 15x Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 021_032_PCCRMC07_893808.indd 31 Chapter 7 = 6*SIN(A2*3) and = -4*COS(A2) 31 3. Given the parametric equations x = 6 sin 3t and y = -4 cos t, write the formulas to put into your spreadsheet if time is found in cell A2. (0.97, -9.43) Lesson 7-5 3/13/09 4:22:41 PM Glencoe Precalculus In the spreadsheet, the values in column A represent t, and the values in columns B and C represent the calculated x and y values of Saturn’s position at time t. 2. Find the coordinates representing Saturn’s position after 23 years. −15 0 5 5 15 PERIOD Use the parametric equations for Saturn’s position in its orbit in a spreadsheet. To calculate the x position, enter the following formula in B2: = 9.5*COS(PI()15*A2). To calculate the y position, enter the formula = 9.48*SIN(PI()15*A2) in C2. Use the fill handle to paste these formulas into the remaining cells in columns B and C. 1. Complete the spreadsheet to find the position of Saturn every 2 years from t = 2 to t = 32. Use the spreadsheet to make an X-Y graph of Saturn’s position. Exercises t 1 A You have learned that the motion of orbiting planets can be modeled using parametric equations. A spreadsheet can be used to evaluate parametric equations and to provide a graph of a planet’s orbit. Parametric Equations 7-5 NAME Answers (Lesson 7-5) Chapter 7 Assessment Answer Key (Lessons 7-1 and 7-2) Page 33 2 (y + 1) (x - 3) 2 − + − = 1; 16 4 1. ellipse y 2. (6, 9) 6 (14, -1) 0 −6 (-2, -1) −12 (6, -1) 12 18 x Quiz 3 (Lesson 7-4) Page 34 ellipse; 3 xy + 31(x) 2- 10 √ 2 21(y) - 144 = 0 (x + 2) (y - 5) 2 −+− =1 11 D 5. 4. Quiz 4 (Lesson 7-5) Page 33 Page 34 y 8 1. 1. −4 −8 2 (y + 1) (x + 1) − - − =1 3. 6 √2 − 11 4. hyperbola B 4. H 0 x x 0 7 3. y 4 −12 −8 −4 9 F D Quiz 2 (Lesson 7-3) 2. 2. 11x 2- 10 √ 3 xy + 2 y + 144 = 0 5. 2 B 14˚ 3. x = 2; y = 1 36 1. 2. (6, -11) 2 4. Page 35 hyperbola; x 2 2 1.4xy - y - 12 = 0 (2, -1); (2,-3); 3. Mid-Chapter Test 2 2. 3. (y - 1) 2 (x - 2) 2 − - − =1 5. D 5. Chapter 7 16 25 y2 49 x − +−=1 25 D about 2.33 s A15 x 2 = -32y 6. y x = 65t cos 35˚, y = 65t sin 35˚- 16t 2; 4. yes 5. Answers Quiz 1 12 (-1, 6) 7. −8 −4 8 (-1, 9) 4 0 (-1, 3) 8x Glencoe Precalculus Chapter 7 Assessment Answer Key Vocabulary Test Page 36 Form 1 Page 37 1. Page 38 10. H 11. B 12. G 13. B 14. J 15. C 16. F 17. C 18. F B 2. F 3. C true false; greater than 4. 5.false; transverse axis 4. G 6. false; parametric 5. C true 1. 2 2 2. false; a - b = c 2 3. true 7. false; perpendicular 8. H true 9. 10. 6. a false; ± − 7. b 11. the set of all points in a plane such that the difference of their distances from two foci is constant 8. 12. a conic formed by the intersection of a double-napped right cone and a plane through the vertex of the cone 9. Chapter 7 A F D B: A16 y = 4x - 4 Glencoe Precalculus Chapter 7 Assessment Answer Key Page 40 10. 1. 4. 5. 6. 7. 13. D 14. J 15. B 16. F G G Chapter 7 B 4. H 5. C 6. G 8. B: J 11. D 12. F 13. C 14. H 15. C 16. J 17. D 18. J G A B G B 10. D 3. 7. A 18. 9. J A A Page 42 D 2. 17. 8. 1. H 12. 3. J B 11. 2. Form 2B Page 41 J F 1 y = -− x-2 2 9. A17 Answers Form 2A Page 39 A B: y=x-3 Glencoe Precalculus Chapter 7 Assessment Answer Key Form 2C Page 43 Page 44 (-4, 1) 1. 12. (y + 3)2 (x + 2)2 3 9 −-− =1 2. (-4, 3), (-4, -1) 3. (-4 ± 2 √ 2 , 1), 13. (-6, 1), y = 1 3) (-4, 1 ± 2 √ 27° 4. 2 7. (x + 2)2 = -20(y - 1) y2 4 x − +−=1 64 5. 6. 14. t+3 x = −, 2 2 t + 6t + 13 y= − 4 2 (x') − + (y')2 = 1 3 hyperbola 15. y 16. 16 12 x = 100t cos 35°, y = 100t sin 35° + 4 - 16t2 8. 8 4 −4 Sample answer: x2 = -48y 9. 11. 8 12 17. about 0.7 s 18. about 32.1 ft 19. hyperbola 20. 10. 0 x (x - 2)2 + (y - 7)2 = 225 145 ) (9, 6 ± √ 8 y - 6 = ±− (x - 9) Chapter 7 9 B: A18 (x - 2)2 (y + 5)2 9 4 −+−=1 Glencoe Precalculus Chapter 7 Assessment Answer Key Form 2D Page 45 Page 46 (y + 2)2 1. (3, -1) 2. (3 ± √ 5 , -1) (6, -1), (0, -1), (3, 1), (3, -3) (2, 1), y = 1 13. 63° 4. y2 x2 −+−=1 100 9 5. (x - 2)2 = -16(y + 5) 14. t-4 , x=− 2 t - 8t + 24 y= − 4 2 6. 7. (x')2 2 (y')2 26 hyperbola 15. −+−=1 16 16. Answers 3. 12. (x + 1)2 -− =1 − 4 16 y 12 x = 110t cos 32°, y = 110t sin 32° + 3 - 16t2 8. 8 4 −4 4 0 8 12 16x −4 x2 = -60y 9. 10. 11. 17. about 0.62 s 18. about 34.6 ft 19. ellipse 20. (x + 6)2 + (y + 9)2 = 121 (-3 ± √ 5 , 2) y - 2 = ± 2(x + 3) Chapter 7 B: A19 (x + 4)2 (y - 1)2 25 4 −+−=1 Glencoe Precalculus Chapter 7 Assessment Answer Key Form 3 Page 47 Page 48 (-3, 2) 1. 2. 12. (x + 2)2 (y - 4)2 25 16 −-− =1 (-3, 2 ± √ 5) (-3, 5), (-3, -1), 3. (-1, 2), (-5, 2) 4. hyperbola; 2° 5. y2 x − +−=1 4 9 2 t+2 x = −, 3 6. -t2 - 4t + 32 y= − 9 2 7(x' ) 40 7. 13. 14. 15. (2, -2), y = -2 (y + 5)2 = -40(x - 2) hyperbola y 16. 2 11(y' ) 40 −+−=1 0 x x = 38t cos 29°, 8. y = 38t sin 29° + 4 - 16t2; Sample answer: Yes, because at 38 feet the ball is only 4.1 ft above the ground 17. 18. about 0.68 ft Sample answer: Yes, when the ball is 20 feet in front of her it is at a height of 0.94 feet. √2 2 − Sample answer: 125 y x2 = - − 3 9. 10. 11. 19. (x - 13)2 + (y - 12)2 = 325 20. (-1 ± 2 √ 5 , -2) y + 2 = ±2(x + 1) Chapter 7 (x + 1)2 (y - 3)2 625 400 −+−=1 B: A20 Glencoe Precalculus Chapter 7 Assessment Answer Key Page 49, Extended-Response Test Sample Answers 1e. The graph is an ellipse since 1a. The equation is a circle if A = B. Sample answer: (-1) 2- 4(3)(2) < 0. 2 2 (x - 1) + (y - 2) = 4 3-2 = - 1, 2θ = 135˚, cot 2θ = − -1 θ ≈ 68˚ y (1, 2) x 0 2a. The graph of x 2 - 4y 2 = 0 is two lines 1 1 of slope − and slope - − that intersect 2 2 at the origin. x 2 - 4y 2 = 0 x 2 = 4y 2 ⎪x⎥ = 2⎪y⎥ 1b. The equation is an ellipse if A ≠ B and A and B have the same sign. Sample answer: x -x − = y or − =y 2 2 2b. y 2 y y = 2x x x 0 y = - 2x 2c. degenerate hyperbola 1c. The equation is a parabola when A or B is zero, but not both. Sample answer: y - 2 = 4(x + 1) 2 3. Sample answer: Most lamps with circular shades shine a cone of light. When this light cone strikes a nearby wall, the resulting shape is a hyperbola. The hyperbola is formed by the cone of light intersecting the plane of the wall. y (reﬂected light) (-1, 2) 0 x 1d. The equation is a hyperbola if A and B have opposite signs. Sample answer: (reﬂected light) 2 y x2 −-− =1 4 Chapter 7 1 A21 Glencoe Precalculus Answers (y - 1) 2 (x + 2) − + − =1 4 1 ⎪ x⎥ − = ⎪y⎥ 2 Chapter 7 Assessment Answer Key Standardized Test Practice Page 50 1. A B C D 2. F G H J 3. A B C D 4. F G H A B C D 6. F G H J 8. A F B G Chapter 7 C H 9. A B C D 10. F G H J 11. A B C D 12. F G H J 13. A B C D 14. F G H J 15. A B C D 16. F G H J 17. A B C D 18. F G H J 19. A B C D J 5. 7. Page 51 D J A22 Glencoe Precalculus Chapter 7 Assessment Answer Key Standardized Test Practice (continued) Page 52 2 y2 16 x − +− = 1 100 20. y 21. x 0 (x - 4)2 = 8(y - 9) Answers 22. 26.57°, 5(x') - √ 5 y' = 13 2 23. 24. (1, -4) 25. 25.4 mi log 6 4 − log4 e 26. 27. (-4, 1), (3, ∞) x = 85t cos 34°, y = 85t sin 34° + 3 - 16t2 28a. about 38.3 ft 28b. 28c. about 213.5 ft Chapter 7 A23 Glencoe Precalculus