Chapter 10 Analytic Geometry Section 10.1 13. (e); the graph has vertex ()()hk,1,1= and opens to the right. Therefore, the equation of the graph Not applicable has the form ()yax−=142 () − 1. Section 10.2 14. (d); the graph has vertex ()()hk,0,0= and ()()−+−22 1. xx21 yy 21 opens down. Therefore, the equation of the graph has the form xay2 =−4 . The graph passes 2 −4 through the point ()−−2, 1 so we have 2. = 4 2 2 ()−=−−241a () 2 = 3. ()x +=49 44a = x +=±43 a 1 2 xx+=43 or +=− 4 3 Thus, the equation of the graph is xy=−4 . xx=−1 or =− 7 15. (h); the graph has vertex ()(hk,1,1=− − ) and The solution set is {7,1}−−. opens down. Therefore, the equation of the graph 2 4. (2,5)−− has the form ()xay+=−+141 (). 5. 3, up 16. (a); the graph has vertex ()()hk,0,0= and 6. parabola opens to the right. Therefore, the equation of the graph has the form yax2 = 4 . The graph passes 7. c through the point ()1, 2 so we have 8. (3, 2) ()2412 = a () 44= a 9. (3,6) 1 = a 2 10. y =−2 Thus, the equation of the graph is yx= 4 . 11. (b); the graph has a vertex ()()hk,0,0= and 17. (c); the graph has vertex ()()hk,0,0= and opens up. Therefore, the equation of the graph opens to the left. Therefore, the equation of the 2 has the form xay2 = 4 . The graph passes graph has the form yax=−4 . The graph passes through the point ()2,1 so we have through the point ()−−1, 2 so we have 2 ()2412 = a () ()−=−−241a () 44= a 44= a 1 = a 1 = a 2 Thus, the equation of the graph is xy2 = 4 . Thus, the equation of the graph is yx=−4 . 12. (g); the graph has vertex ()()hk,1,1= and opens 18. (f); the graph has vertex ()(hk,1,1=− − ) and to the left. Therefore, the equation of the graph opens up; further, a = 1 . Therefore, the equation 2 has the form ()yax−=−−1412 (). of the graph has the form ()xy+=141 () +. 968 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 10.2: The Parabola 19. The focus is (4, 0) and the vertex is (0, 0). Both 21. The focus is (0, –3) and the vertex is (0, 0). Both lie on the horizontal line y = 0 . a = 4 and since lie on the vertical line x = 0 . a = 3 and since (4, 0) is to the right of (0, 0), the parabola opens (0, –3) is below (0, 0), the parabola opens down. to the right. The equation of the parabola is: The equation of the parabola is: 2 yax2 = 4 xay=−4 2 yx2 =⋅⋅44 xy=−43 ⋅ ⋅ 2 yx2 = 16 xy=−12 2 Letting xyy===±4, we find 2 64 or 8 . Letting yxx=−3, we find =36 or =±6 . The points (4, 8) and (4, –8) define the latus The points ()−−6, 3 and ()6,− 3 define the latus rectum. rectum. 20. The focus is (0, 2) and the vertex is (0, 0). Both 22. The focus is (–4, 0) and the vertex is (0, 0). Both lie on the vertical line x = 0 . a = 2 and since lie on the horizontal line y = 0 . a = 4 and since (0, 2) is above (0, 0), the parabola opens up. The (–4, 0) is to the left of (0, 0), the parabola opens equation of the parabola is: to the left. The equation of the parabola is: 2 xay= 4 yax2 =−4 2 xy=⋅⋅42 yx2 =−44 ⋅ ⋅ 2 xy= 8 yx2 =−16 2 Letting yxx===±2, we find 16 or 4 . Letting xyy=−4, we find 2 =64 or =±8 . The The points (–4, 2) and (4, 2) define the latus points (–4, 8) and (–4, –8) define the latus rectum. rectum. 969 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 10: Analytic Geometry 23. The focus is (–2, 0) and the directrix is x = 2 . equation of the parabola is: The vertex is (0, 0). a = 2 and since (–2, 0) is to xay2 = 4 the left of (0, 0), the parabola opens to the left. 1 The equation of the parabola is: xy2 =⋅⋅4 2 yax2 =−4 xy2 = 2 2 =− ⋅ ⋅ yx42 1 Letting yxx===±, we find 2 1 or 1. yx2 =−8 2 2 1 1 Letting xyy===±– 2, we find 16 or 4 . The The points 1, and −1, define the latus 2 2 points (–2, 4) and (–2, –4) define the latus rectum. rectum. 24. The focus is (0, –1) and the directrix is y = 1. 1 The vertex is (0, 0). a = 1 and since (0, –1) is 26. The directrix is x =− and the vertex is (0, 0). 2 below (0, 0), the parabola opens down. The 1 1 1 equation of the parabola is: The focus is ,0 . a = and since ,0 is 2 2 2 xay2 =−4 to the right of (0, 0), the parabola opens to the 2 =− ⋅ ⋅ xy41 right. The equation of the parabola is: xy2 =−4 yax2 = 4 ===±2 1 Letting yxx–1, we find 4 or 2 . The yx2 =⋅⋅4 points (–2, –1) and (2, –1) define the latus 2 rectum. yx2 = 2 1 Letting xyy===±, we find 2 1 or 1 . The 2 1 1 points ,1− and ,1 define the latus 2 2 rectum. 1 25. The directrix is y =− and the vertex is (0, 0). 2 1 1 1 The focus is 0, . a = and since 0, is 2 2 2 above (0, 0), the parabola opens up. The 970 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 10.2: The Parabola 27. Vertex: (0, 0). Since the axis of symmetry is vertical, the parabola opens up or down. Since (2, 3) is above (0, 0), the parabola opens up. The equation has the form xay2 = 4 . Substitute the coordinates of (2, 3) into the equation to find a : 2432 =⋅a 412= a 1 a = 3 4 The equation of the parabola is: xy2 = . The 3 1 1 focus is 0, . Letting y = , we find 29. The vertex is (2, –3) and the focus is (2, –5). 3 3 Both lie on the vertical line x = 2 . 2 42 21 =− −−() = xx==± or . The points , and a 532 and since (2, –5) is below 93 33 (2, –3), the parabola opens down. The equation 21 − , define the latus rectum. of the parabola is: 33 ()xh−=−−2 4 ayk () ()xy−=−−−24232 ()()() ()xy−=−+2832 () Letting y =−5 , we find ()x −=2162 xxx−=±24 =− 2 or =6 The points (–2, –5) and (6, –5) define the latus rectum. 28. Vertex: (0, 0). Since the axis of symmetry is horizontal, the parabola opens left or right. Since (2, 3) is to the right of (0, 0), the parabola opens to the right. The equation has the form yax2 = 4 . Substitute the coordinates of (2, 3) into the equation to find a : 3422 =⋅a 98= a 9 a = 30. The vertex is (4, –2) and the focus is (6, –2). 8 Both lie on the horizontal line y =−2 . 2 9 The equation of the parabola is: yx= . The a =−= 2 46 2 and since (6, –2) is to the right of 9 9 (4, –2), the parabola opens to the right. The focus is ,0 . Letting x = , we find 8 8 equation of the parabola is: ()yk−=2 4 axh () − 2 81 9 99 yy==± or . The points , and 2 16 4 84 ()yx−−()2424 =()( − ) 99 , − define the latus rectum. +=−2 84 ()()yx284 971 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Chapter 10: Analytic Geometry Letting x = 6 , we find 32. The vertex is (3, 0) and the focus is (3, –2). Both 2 = =− − = ()y +=216 lie on the horizontal line x 3 . a 20 2 yyy+=±24 =− 6 or =2 and since (3, –2) is below of (3, 0), the parabola opens down. The equation of the parabola is: The points (6, –6) and (6, 2) define the latus 2 rectum. ()xh−=−−4 ayk () ()xy−=−−34202 ()() ()xy−=−382 Letting y =−2 , we find ()x −=3162 xxx−=±34 =− 1 or =7 The points (–1, –2) and (7, –2) define the latus rectum. 31. The vertex is (–1, –2) and the focus is (0, –2). Both lie on the horizontal line y =−2 . a =−10 − = 1 and since (0, –2) is to the right of (–1, –2), the parabola opens to the right. The equation of the parabola is: ()yk−=2 4 axh () − 2 ()yx−−()2411 =()() −− ( ) ()()yx+=2412 + 33. The directrix is y = 2 and the focus is (–3, 4). = Letting x 0 , we find This is a vertical case, so the vertex is (–3, 3). ()y +=242 a = 1 and since (–3, 4) is above y = 2 , the yyy+=±22 =− 4 or =0 parabola opens up. The equation of the parabola −=2 − The points (0, –4) and (0, 0) define the latus is: ()4()xh ayk rectum. (xy−− ( 3))2 = 4 ⋅⋅ 1 ( − 3) (3)4(3)xy+=2 − Letting y = 4 , we find (3)4x +=2 or x +=±32. So, xx=−1 or =−5 . The points (–1, 4) and (–5, 4) define the latus rectum. 972 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall Section 10.2: The Parabola 34. The directrix is x =−4 and the focus is (2, 4). 36. The directrix is y =−2 and the focus is (–4, 4).
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