Chapter 9 Homework Pre-Cal (9.1 - 9.2)
9.1 Find the standard form of the equation of the ellipse and give the location of its foci. -10-8-6-4-2246810-10-8-6-4-2246810xy A coordinate system has a horizontal x-axis labeled from negative 10 to 10 in increments of 1 and a vertical y-axis labeled from negative 10 to 10 in increments of 1. An ellipse that has a horizontal major axis is centered at (0, 0) and passes through the plotted points (negative 9, 0), (0, 5), (9, 0), and (0, negative 5). ..... F
..First, we use the graph to determine the center of the ellipse. The center of the ellipse is located at the origin. Notice that the points where the ellipse crosses the x-axis are equidistant from the origin in a horizontal direction. Likewise the points where the ellipse crosses the y-axis are equidistant from the origin in a vertical direction. Now we use the graph to determine the orientation of the major axis. The ellipse is elongated in the horizontal direction. The standard form of the equation of an ellipse with center at the origin and horizontal major axis is x2a2+y2b2=1, where a2>b2, and c2=a2−b2. We need to find the values of a and b. The distance from the center to either one of the vertices gives the value of a. The vertices are the endpoints of the major axis. Remember that the major axis for this ellipse is horizontal. The vertices of the ellipse are located at (−9,0) and (9,0). Therefore, a=9. The distance from the center to one of the endpoints of the minor axis gives the value of b. Remember that the minor axis for this ellipse is vertical. The endpoints of the minor axis are located at (0,−5) and (0,5). Therefore, b=5. We need to compute a2 and b2 to use in the standard form equation. a2=92=81 b2=52=25 Since the ellipse has a horizontal major axis and the center is located at (0,0), the standard form of the equation takes the form x2a2+y2b2=1. We substitute the values of a2 and b2 into the equation. x281+y225=1 The foci are on the major axis, c units from the center. Since the major axis is horizontal and the center is at the origin, the foci are located at (c,0) and (−c,0). We can use the formula c2=a2−b2 to determine c2. c2 = 92−52 = 81−25 = 56 If c2=56, c is the positive square root of 56. Thus, c=56=214. We substitute the value of c into the ordered pairs (c,0) and (−c,0). The foci are located at −214,0 and 214,0.
9.1 Graph the ellipse and locate the foci. x249+y29=1 T
..The standard form of the equation of an ellipse with center at the origin, and major and minor axes of length 2a and 2b (where a and b are positive, and a2>b2) takes one of the forms shown below. x2a2+y2b2=1 or x2b2+y2a2=1 Since the equation given in the problem statement has this form, it represents an ellipse centered at the origin. We determine the values of a2 and b2 from the equation, remembering that a2>b2. a2=49 b2=9 We find the values of a and b by taking the positive square root. a=7 b=3 Because the denominator of the x2-term, 49, is greater that the denominator of the y2-term, 9, the major axis of the ellipse is horizontal. An ellipse centered at the origin with a horizontal major axis has vertices a units from the center, located at (−a,0) and (a,0). The vertices are located at (−7,0) and (7,0). Since the major axis is horizontal, the minor axis is vertical. According to the standard form of the equation, the endpoints of the minor axis are b units from the center, located at (0,−b) and (0,b). The endpoints of the minor axis are located at (0,−3) and (0,3). The equation is that of an ellipse centered at the origin, with vertices at (−7,0) and (7,0) and endpoints of the minor axis at (0,−3) and (0,3). The graph of the ellipse is shown to the right. -8-6-4-22468-8-6-4-22468xy x y graph Now we find the location of the foci. Since the vertices are located at (−a,0) and (a,0) for this ellipse, the foci are located at (−c,0) and (c,0). We can determine the value of c2 using the formula c2=a2−b2. c2 = a2−b2 = 49−9 Substitute 49 for a2 and 9 for b2. = 40 Subtract. Since c2=40, c is the positive square root of 40. Thus, c=40=210. We substitute the value of c into the ordered pairs (−c,0) and (c,0). The foci are located at −210,0 and 210,0.
9.1 Will a truck that is 18 feet wide carrying a load that reaches 13 feet above the ground clear the semielliptical arch on the one-way road that passes under the bridge shown in the figure on the right? 14 ft56 ft B
Because the truck's width is 18 feet, to determine the clearance, find the height of the archway of the bridge 9 feet from the center. Using the equation x2a2+y2b2=1, the archway of the bridge can be expressed as x2784+y2196=1. The truck can be expressed as a rectangle that is 18 by 13. As shown in the figure on the right the 18 foot wide truck corresponds to x=9. -30-25-20-15-10-55101520253015xyTruck13 feet(0,14)9(-28,0)(28,0) A coordinate system has a horizontal x-axis labeled from negative 30 to 30 in increments of 5 and a vertical y-axis labeled from 0 to 15 in increments of 5. From left to right, a semiellipse rises from the plotted and labeled point (negative 28, 0) to the plotted and labeled point (0, 14) and then falls to the plotted and labeled point (28, 0). A rectangle labeled Truck has a horizontal bottom side on the x-axis with the center at the origin and a vertical left side labeled 13 feet. A horizontal line segment which extends from a point on the y-axis to the vertical right side of the rectangle is labeled 9. Find the height of the archway 9 feet from the center by substituting 9 for x and solving for y in the equation x2784+y2196=1. 92784+y2196 = 1 Substitute 9 in for x. 81784+y2196 = 1 Square 9. Isolate the y-term. 784•81784+y2196 = 1•784 Clear fractions by multiplying both sides by 784. 81+4y2 = 784 Use the distributive property and simplify. 4y2 = 703 Subtract 81 from both sides. y2 = 7034 Divide both sides by 4. Solve for y. y2 = 7034 Divide both sides by 4. y = 7034 Take only the positive square root. The archway of the bridge is above the x-axis so y is nonnegative. y ≈ 13.26 Use a calculator and round to two decimal places as needed. Thus, the height of the archway 9 feet from the center is approximately 13.26 feet. Because the truck's height is 13 feet, there is enough room for the truck to clear the archway of the bridge.
9.1 Will a truck that is 10 feet wide carrying a load that reaches 9 feet above the ground clear the semielliptical arch on the one-way road that passes under the bridge shown in the figure on the right?
It ___will___ clear the arch because the height of the archway of the bridge 5 feet from the center is approximately ___9.68__ feet. (Round to two decimal places as needed.)
9.1 Find the standard form of the equation of the ellipse satisfying the given conditions. Endpoints of major axis: (9,11) and (9,−3) Endpoints of minor axis: (12,4) and (6,4) S
Standard Forms of Equations of Ellipses Centered at (h,k)EquationCenterMajor AxisFociVertices(x−h)2a2+(y−k)2b2=1, a2>b2 and c2=a2−b2(h, k)Parallel to the x-axis, horizontal(h−c, k)(h+c, k)(h−a, k)(h+a, k)(x−h)2b2+(y−k)2a2=1, a2>b2 and c2=a2−b2(h, k)Parallel to the y-axis, vertical(h, k−c)(h, k+c)(h, k−a)(h, k+a) Since the vertices of the major axis are located at (9,11) and (9,−3), the major axis is vertical. The center of the ellipse is midway between the vertices and the endpoints of the minor axis, located at (12,4) and (6,4). The center of the ellipse is (9,4) Determine the values for a and b. a is the distance from the center, (9,4), to either vertex. a=7 b is the distance from the center, (9,4), to either endpoint of the minor axis. b=3 Since the major axis is vertical, the standard form of the equation takes the form (x−h)2b2+(y−k)2a2=1 Replacing h, k, b2, and a2 with their respective values determined above, the equation for the given ellipse is (x−9)29+(y−4)249=1
9.1 Find the standard form of the equation of the ellipse satisfying the given conditions. Major axis horizontal with length 22; length of minor axis=10; center: (0 ,0 ) S
Standard Forms of the Equations of an Ellipse The standard form of the equation of an ellipse with center at the origin and major and minor axes of length 2a and 2b (where a and b are positive, and a2>b2) is x2a2+y2b2=1 or x2b2+y2a2=1 The vertices are on the major axis, a units from the center. The foci are on the major axis, c units from the center. For both equations, b2 = a2−c2. Equivalently, c2=a2−b2. For the first form, the major axis is horizontal with length 2a and the foci are (c, 0) and (−c, 0). For the second form, the major axis is vertical with length 2a and the foci are (0, c) and (0, −c). Since the major axis is horizontal and the center of the ellipse located at (0,0), the equation willl be of the form x2a2+y2b2=1 Determine the values for a2 and b2. Since the center is located at (0,0), the value of a is half the length of the major axis. a=11, therefore a2=121. Since the center is located at (0,0), the value of b is equal to half the length of the minor axis. b=5, therefore b2=25. Substitute the values of a2 and b2 in x2a2+y2b2=1 to find the standard form of the equation. x2121+y225=1
9.2 Find the standard form of the equation of the hyperbola shown below. -10-8-6-4-2246810-10-8-6-4-2246810xy A coordinate system has a horizontal x-axis labeled from negative 10 to 10 in increments of 1 and a vertical y-axis labeled from negative 10 to 10 in increments of 1. Dashed horizontal line segments pass through (0, 7) and (0, negative 7). Dashed vertical line segments pass through (negative 5, 0) and (5, 0). The line segments form a rectangle with the following vertices: (negative 5, negative 7), (negative 5, 7), (5, 7), (5, negative 7). An asymptote that rises from left to right and an asymptote that falls from left to right pass through the vertices of the rectangle. A hyperbola centered at (0, 0) opens left and right and has vertices at (negative 5, 0) and (5, 0), approaching both asymptotes. ..... T
The equation of a hyperbola with center at the origin can have one of two standard forms, x2a2−y2b2=1 or y2a2−x2b2=1. In both cases, the vertices are a units from the center. To write an equation for the hyperbola, find the values of a2 and b2, and determine whether the x2-term or the y2-term is positive. Use the graph to identify the transverse axis and the values for a and b. Examination of the graph shows that the tranverse axis is on the x-axis. The hyperbola crosses the transverse axis at the vertices. Examination of the graph shows that the vertices are located on the x-axis. The vertices are (5,0) and (−5,0). From the vertices, we see that a=5. Since values of both a2 and b2 are needed to write the standard form of the equation, our next step is to find b. The rectangle crosses the y-axis at the points (0,7) and (0,−7). These points give b=7. The standard form for the equation of a hyperbola with center at the origin and transverse axis on the x-axis is x2a2−y2b2=1. Thus, the requirement is to compute the values of a2 and b2. Recall that a=5 and b=7. Thus, a2=25 and b2=49. Using the formula for the standard form for the equation of a hyperbola, its clear that the equation for the given graph is x225−y249=1.
9.2 Find the standard form of the equation of the hyperbola satisfying the given conditions. Endpoints of transverse axis: (0,−15), (0,15); asymptote: y=5x T
The equation of a hyperbola with center at (0,0), foci at (±c,0), vertices at (±a,0) and transverse axis along the x-axis is as given below. x2a2−y2b2=1, where b2=c2−a2 The equation of a hyperbola with center at (0,0), foci at (0,±c), vertices at (0,±a) and transverse axis along the y-axis is as given below. y2a2−x2b2=1, where b2=c2−a2 The endpoints of the transverse axis are (0,−15) and (0,15). So the foci are located on the y-axis and are symmetric with respect to the x-axis. Determine if the transverse axis is along the x-axis or the y-axis. The transverse axis is the y-axis because this is the axis which joins the vertices. The center of the hyperbola is midway between the endpoints of transverse axis, located at (0,0). The standard form of the equation will be of the form y2a2−x2b2=1. Remember, the vertices are the endpoints of the transverse axis. One vertex is at (0, a)=(0, 15), so a=15. Recall, the hyperbola y2a2−x2b2=1 has two asymptotes y=abx and y=−abx. The given hyperbola has an asymptote y=5x. The slope of this asymptote is 5. The slope of the asymptote is equal to ab, so substitute the known value for a and solve the equation ab=5 to find the value of b. ab = 5 15b = 5 Substitute 15 for a. 15 = 5b Multiply both sides by b. 3 = b Divide both sides by 5. Substitute a = 15 and b = 3 into the equation y2a2−x2b2=1 and simplify. y2a2−x2b2 = 1 y2225−x29 = 1 Therefore, the hyperbola's equation in the standard form is y2225−x29=1.
9.2 Find the standard form of the equation of the hyperbola satisfying the given conditions. Foci at (0,−9) and (0,9); vertices at (0,8) and (0,−8) T
The equation of a hyperbola with center at (0,0), foci at (±c,0), vertices at (±a,0) and transverse axis along the x-axis is as given below. x2a2−y2b2=1, where b2=c2−a2 The equation of a hyperbola with center at (0,0), foci at (0,±c), vertices at (0,±a) and transverse axis along the y-axis is as given below. y2a2−x2b2=1, where b2=c2−a2 The foci are located at (0,−9) and (0,9), on the y-axis and are symmetric with respect to the x-axis. Similarly, the vertices are located at (0,8) and (0,−8), on the y-axis and are symmetric with respect to the x-axis. Determine if the transverse axis is along the x-axis or the y-axis. The transverse axis is the y-axis because this is the axis which joins the vertices. The center of the hyperbola is midway between the foci, located at (0,0). The standard form of the equation will be of the form y2a2−x2b2=1. Determine the values for a2 and b2. The distance from the center, (0,0), to either vertex, (0,8) or (0,−8), is 8, so a=8. To determine b2, first determine c. The distance from the center, (0,0), to either focus, (0,−9) or (0,9), is 9, so c=9. To find the value of b2, substitute c=9 and a=8 in b2=c2−a2 and solve for b2. b2 = c2−a2 = 92−82 = 81−64 Subtract. b2 = 81−64 = 17 Substitute 17 for b2 and 64 for a2 in y2a2−x2b2=1 to find the standard form of the hyperbola's equation. y2a2−x2b2 = 1 y264−x217 = 1 Therefore, the hyperbola's equation in the standard form is y264−x217=1.
9.1 Find the standard form of the equation of the ellipse satisfying the given conditions. Foci: (−8, 0), (8,0); vertices: (−11, 0), (11, 0) T
The foci are located at (−8, 0) and (8,0). Thus, the foci are on the x-axis. This means that the major axis is horizontal. The center of the ellipse is midway between the foci. Thus, the center is located at (0,0). The standard form of the equation of an ellipse with center at the origin and horizontal major axis is x2a2+y2b2=1, where a and b are positive and a2>b2. The foci are on the major axis c units from the center, and b2=a2−c2. We need to determine the values for a2 and b2. To find the value of b2, we will need to use the equation b2=a2−c2. We begin by finding the values of a and c. The distance from the center, (0,0), to either vertex, (−11, 0) or (11, 0), gives the value of a. Thus, a=11, and a2=121. The distance from the center, (0,0), to either focus, (−8, 0) or (8,0), gives the value of c. Thus, c=8, and c2=64. We use the values of a2 and c2 to compute the value of b2. b2 = a2−c2 = 121−64 Substitute 121 for a2 and 64 for c2. = 57 Subtract. The standard form of the equation of an ellipse with center at the origin and horizontal major axis is x2a2+y2b2=1. We substitute the values of a2 and b2 into the equation. x2121+y257=1
9.1 Find the standard form of the equation of the ellipse satisfying the given conditions. Foci: (−9,0), (9,0); y-intercepts: −4 and 4 T
The foci are located at (−9,0) and (9,0). Thus, the foci are on the x-axis. This means that the major axis is horizontal. The center of the ellipse is midway between the foci. Thus, the center is located at (0,0). The standard form of the equation of an ellipse with center at the origin and horizontal major axis is x2a2+y2b2=1, where a and b are positive and a2>b2. The foci are on the major axis c units from the center, and b2=a2−c2. We need to determine the values for a2 and b2. To find the value of a2, we will need to use the equation b2=a2−c2. We begin by finding the values of b and c. Since the major axis of the ellipse is horizontal, the y-intercepts lie on the minor axis. The distance from the center, (0,0), to either of the given y-intercepts, −4 or 4, gives the value of b. Thus, b=4, and b2=16. The distance from the center, (0,0), to either focus, (−9,0) or (9,0), gives the value of c. Thus, c=9, and c2=81. We use the values of b2 and c2 to compute the value of a2. b2 = a2−c2 16 = a2−81 Substitute 16 for b2 and 81 for c2. 97 = a2 Add 81 to both sides. The standard form of the equation of an ellipse with center at the origin and horizontal major axis is x2a2+y2b2=1. We substitute the values of a2 and b2 into the equation. x297+y216=1
9.1 Find the standard form of the equation of the ellipse and give the location of its foci. -66-66xy(2,−2) x y graph ..... F
First, we use the graph to determine the center of the ellipse. The center is located at (2,−2). Now we use the graph to determine the orientation of the major axis. The ellipse is elongated in the horizontal direction. In other words, the major axis is parallel to the x-axis. The standard form of the equation of an ellipse centered at (h,k) with major axis parallel to the x-axis is (x−h)2a2+(y−k)2b2=1, where a2>b2 and c2=a2−b2. The vertices are at (h+a,k) and (h−a,k). We know the center of the ellipse is located at (2,−2). Thus, h=2 and k=−2. We need to find the values of a and b. The distance from the center to either one of the vertices gives the value of a. The vertices are the endpoints of the major axis. Remember that the major axis for this ellipse is horizontal. The vertices of the ellipse are (−1,−2) and (5,−2). We compute the distance between the center, (2,−2), and one of the vertices to determine the value of a. a=3 The distance from the center to one of the endpoints of the minor axis gives the value of b. Remember that the minor axis for this ellipse is vertical. The endpoints of the minor axis of the ellipse are (2,−4) and (2,0). We compute the distance between the center, (2,−2), and one of the endpoints of the minor axis to determine the value of b. b=2 Since the ellipse has a horizontal major axis and the center is (h,k), the standard form of the equation is (x−h)2a2+(y−k)2b2=1. We compute a2 and b2. a2=32=9 b2=22=4 We substitute the values of h, k, a2, and b2 into the equation. (x−2)29+(y+2)24=1 The foci are on the major axis, c units from the center. Since the major axis is horizontal and the center is (h,k), the foci are located at (h−c,k) and (h+c,k). We can use the formula c2=a2−b2 to determine c2. c2 = 32−22 = 9−4 = 5 If c2=5, c is the positive square root of 5. Thus, c=5. We substitute the value of c into the ordered pairs (h−c,k) and (h+c,k). The foci are located at 2−5,−2 and 2+5,−2.
