Practice Test - Chapter 7

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Gervais Howard
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1 Write an equation for an ellipse with each set of characteristics. 1. vertices (7, 4), ( 3, 4); foci (6, 4), ( 2, 4) The distance between the vertices is 2a. 2a = 7 ( 3) a = 5; a 2 = 25 The distance between the foci is 2c. 2c = 6 ( 2) c = 4 = 2 center: (2, 4) + = 1 2. foci ( 2, 1), ( 2, 9); length of major axis is 12 The distance between the foci is 2c. 2c = 1 ( 9) c = 5 The length of the major axis is 2a. 2a = 12 a = 6; a 2 = 36 = 4 center: ( 2, 4) + = 1 esolutions Manual - Powered by Cognero Page 1

2 3. MULTIPLE CHOICE What value must c be so that the graph of 4x 2 + cy 2 + 2x 2y 18 = 0 is a circle? A 8 B 4 C 4 D 8 The graph of a second degree equation of the form Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 is a circle if B 2 4AC < 0 and B = 0 and A = C. Since A = 4, C must also equal 4. The correct answer is C. Write each pair of parametric equations in rectangular form. Then graph the equation. 4. x = t 5 and y = 3t 4 Solve for t. x = t 5 x + 5 = t Substitute for t. y = 3t 4 y = 3(x + 5) 4 y = 3x y = 3x + 11 Make a table of values to graph y. x y Plot the (x, y) coordinates and connect the points to form a smooth curve. esolutions Manual - Powered by Cognero Page 2

3 5. x = t 2 1 and y = 2t + 1 Solve for t. y = 2t + 1 y 1 = 2t = t Substitute for t. x = t 2 1 x = 1 x + 1 = x + 1 = 4(x + 1) = (y 1) 2 Make a table of values to graph y. x y , 3 3 3, 5 8 5, 7 Plot the (x, y) coordinates and connect the points to form a smooth curve. esolutions Manual - Powered by Cognero Page 3

4 6. BRIDGES At 1.7 miles long, San Francisco s Golden Gate Bridge was the longest suspension bridge in the world when it was constructed. equation for the design of the bridge. a. Suppose the design of the bridge can be modeled by a parabola and the lowest point of the cable is 15 feet above the road. Write an equation for the design of the bridge. b. Where is the focus located in relation to the vertex? a. Since the cable resembles a parabola that opens up, the standard form of the equation that can be used to model the cable is (x h) 2 = 4p (y k). If the parabola is centered on the y axis, the vertex is located at the point (0, 15). One point that lies on the parabola is (2100, 500). Use the values of h, k, x and y to solve for p. (x h) 2 = 4p (y k) (2100 0) 2 = 4(p )(500 15) = 1940p p Use the values of h, k and p to write an equation for the design of the bridge. (x h) 2 = 4p (y k) (x 0) 2 = 4(2273.2)(y 15) x 2 = (y 15) b. The distance between the vertex and the focus is p. Since p was found in part a. to be feet, the focus is located feet above the vertex. Write an equation for the hyperbola with the given characteristics. 7. vertices (3, 0), ( 3, 0); asymptotes Because the y coordinates of the vertices are the same, the transverse axis is horizontal, and the standard form of the equation is = 1. The center is the midpoint of the segment between the vertices, or (0, 0). So, h = 0 and k = 0. The slopes of the asymptotes are ±. So, the positive slope is equal to, where b = 2, b 2 = 4, a = 3, and a 2 = 9. Using the values of h, k, a, and b, the equation for the hyperbola is = 1. esolutions Manual - Powered by Cognero Page 4

5 8. foci (8, 0), (8, 8); vertices (8, 2), (8, 6) Because the x coordinates of the vertices are the same, the transverse axis is vertical, and the standard form of the equation is = 1. The center is the midpoint of the segment between the foci, or (8, 4). So, h = 8 and k = 4. You can find c by determining the distance from a focus to the center. One focus is located at (8, 0) which is 4 units from (8, 4). So, c = 4. You can find a by determining the distance from a vertex to the center. One vertex is located at (8, 2) which is 2 units from (8, 4). So, a = 2 and a 2 = 4. Now you can use the values of c and a find b. b 2 = c 2 a 2 b 2 = b 2 = 16 4 b 2 = 12 b = Using the values of h, k, a, and b, the equation for the hyperbola is = 1. esolutions Manual - Powered by Cognero Page 5

6 Write an equation for each conic in the xy plane for the given equation in x y form and the given value of θ. 9. 7(x 3) = (y ) 2, θ = 60º 7(x 3) = (y ) 2, θ = Use the rotation formulas for x and y to find the equation of the rotated conic in the xy plane. x = x cos θ + y sin θ x = x + y y = y cos θ x sin θ y = y x Substitute these values into the original equation. esolutions Manual - Powered by Cognero Page 6

7 10. + = 1, θ = + = 1, θ = Use the rotation formulas for x and y to find the equation of the rotated conic in the xy plane. x = x cos θ + y sin θ x = x + y y = y cos θ x sin θ y = y x Substitute these values into the original equation. esolutions Manual - Powered by Cognero Page 7

8 Graph the hyperbola given by each equation. 11. = 1 The equation is in standard form, with h = 0 and k = 4. Because a 2 = 64 and b 2 = 25, a = 8 and b = 5. The values of a and b can be used to find c. c 2 = a 2 + b 2 c 2 = c = or about 9.43 Use h, k, a, b, and c to determine the characteristics of the hyperbola. orientation: In the standard form of the equation, the y term is being subtracted. Therefore, the orientation of the hyperbola is horizontal. center: (h, k) = (0, 4) vertices: (h ± a, k) = ( 8, 4) and (8, 4) foci: (h ± c, k) = ( 9.43, 4) and (9.43, 4) asymptotes: Graph the center, vertices, foci, and asymptotes. Then make a table of values to sketch the hyperbola. x y , , , , 9.59 esolutions Manual - Powered by Cognero Page 8

9 12. = 1 The equation is in standard form, with h = 6 and k = 3. Because a 2 = 4 and b 2 = 36, a = 2 and b = 6. The values of a and b can be used to find c. c 2 = a 2 + b 2 c 2 = c = or about 6.32 Use h, k, a, b, and c to determine the characteristics of the hyperbola. orientation: In the standard form of the equation, the x term is being subtracted. Therefore, the orientation of the hyperbola is vertical. center: (h, k) = ( 6, 3) vertices: (h, k ± a) = ( 6, 5) and ( 6, 1) foci: (h, k ± c) = ( 6, 9.32) and ( 6, 3.32) asymptotes: Graph the center, vertices, foci, and asymptotes. Then make a table of values to sketch the hyperbola. x y , , , , 0.60 esolutions Manual - Powered by Cognero Page 9

10 13. MULTIPLE CHOICE Which ellipse has the greatest eccentricity? The greater the eccentricity, the more stretched the graph will appear. Graph D is more circular than the other three graphs, so it cannot be the answer. The other three graphs have similar co vertices located at either ( 2, 0) and (2, 0) or (0, 2) and (0, 2) and all have centers at the origin. Therefore, the graph that has vertices farthest from the origin will have the greatest eccentricity and will be the most stretched. Graph A has vertices at ( 8, 0) and (8, 0). Graph B has vertices at ( 6, 0) and (6, 0). Graph C has vertices at (0, 10) and (10, 0). The correct answer is C. esolutions Manual - Powered by Cognero Page 10

11 Write an equation for and graph a parabola with the given focus F and vertex V. 14. F(2, 8), V(2, 10) Because the focus and vertex share the same x coordinate, the graph is vertical. The focus is (h, k + p ), so the value of p is 8 10 or 2. Because p is negative, the graph opens down. Write the equation for the parabola in standard form using the values of h, p, and k. 4p (y k) = (x h) 2 4( 2)(y 10) = (x 2) 2 8(y 10) = (x 2) 2 The standard form of the equation is (x 2) 2 = 8(y 10). Graph the vertex and focus. Then make a table of values to graph the parabola. esolutions Manual - Powered by Cognero Page 11

12 15. F(2, 5), V( 1, 5) 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 2 ( 1) or 3. Because p is positive, the graph opens to the right. Write the equation for the parabola in standard form using the values of h, p, and k. 4p (x h) = (y k) 2 4(3)[x ( 1)] = (y 5) 2 12(x + 1) = (y 5) 2 The standard form of the equation is (y 5) 2 = 12(x + 1). Graph the vertex and focus. Then make a table of values to graph the parabola. Graph the ellipse given by each equation = 1 The ellipse is in standard form. The values of h and k are 5 and 3, so the center is at (5, 3). a = or 7 b = or 3 c = 6.32 orientation: horizontal vertices: ( 2, 3), (12, 3) covertices: (5, 6), (5, 0) esolutions Manual - Powered by Cognero Page 12

13 17. (x + 3) 2 + = 1 The ellipse is in standard form. The values of h and k are 3 and 6, so the center is at ( 3, 6). a = = 9 b = 1 c = 8.94 orientation: vertical vertices: ( 3, 15), ( 3, 3) covertices: ( 4, 6), ( 2, 6) 18. CAMPING In many U.S. parks, campers must secure food and provisions from bears and other animals. One method is to secure food using a bear bag, which is done by tossing a bag tied to a rope over a tall tree branch and securing the rope to the tree. Suppose a tree branch is 30 feet above the ground, a person 20 feet from the branch throws the bag from 5 feet above the ground. a. Will a bag thrown at a speed of 40 feet per second at an angle of 60º go over the branch? b. Will a bag thrown at a speed of 45 feet per second at an angle of 75º go over the branch? a. To determine whether the bag will go over the branch, you need the horizontal distance that the bag has traveled when the height of the bag is 30 feet. First, write a parametric equation for the vertical position of the bag. y = tv 0 sin θ gt 2 + h 0 = t(40) sin 60 (32)t Graph the equation for the vertical position and the line y = 30. The curve does not intersect the line. Therefore, the height of the bag never reaches 30 feet and the bag will not make it over the branch. esolutions Manual - Powered by Cognero Page 13

14 b. To determine whether the bag will go over the branch, you need the horizontal distance that the bag has traveled when the height of the bag is 30 feet. First, write a parametric equation for the vertical position of the bag. y = tv 0 sin θ gt 2 + h 0 = t(40) sin 60 (32)t Graph the equation for the vertical position and the line y = 30. The curve will intersect the line in two places. The second intersection represents the bag as it is moving down toward the branch. Use 5: intersect function on the CALC menu to find the second point of intersection with y = 30. The value is about 1.89 seconds. Determine the horizontal position of the bag at 1.89 seconds. x = tv 0 cos θ = 1.89(45) cos The bag will travel a distance of about feet before the height reaches 30 feet a second time. Since the bag is being thrown from a distance of 20 feet from the tree, the bag will make it over the branch. esolutions Manual - Powered by Cognero Page 14

15 Use a graphing calculator to graph the conic given by each equation. 19. x 2 6xy + y 2 4y 8x = 0 Graph the equation by solving for y. x 2 6xy + y 2 4y 8x = 0 y 2 + ( 6x 4)y + x 2 8x = 0 [ 10, 10] scl: 1 by [ 10, 10] scl: x 2 + 4y 2 2xy + 3y 6x + 5 = 0 Graph the equation by solving for y. x 2 + 4y 2 2xy + 3y 6x + 5 = 0 4y 2 + ( 2x + 3)y + x 2 6x + 5 = 0 [ 10, 10] scl: 1 by [ 10, 10] scl: 1 esolutions Manual - Powered by Cognero Page 15

Mid-Chapter Quiz: Lessons 7-1 through 7-3

Mid-Chapter Quiz: Lessons 7-1 through 7-3 Write an equation for and graph a parabola with the given focus F and vertex V 1. F(1, 5), V(1, 3) Because the focus and vertex share the same x coordinate, the graph is vertical. The focus is (h, k +