Pre-Calc Homework Solutions 466

# Pre-Calc Homework Solutions 466 - 466 Appendix A5.3 A B C 4...

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Unformatted text preview: 466 Appendix A5.3 A B C 4 4 1 3 cos 2 4 3 (if we 5 7 2 9 28. cot 2 34. tan 2 12 0.38, cos 45 0.92; 10.45, D 22.5 choose 2 in Quadrant II); thus sin 1 cos 2 2 1 5 1 cos 2 2 1 (3/5) 2 1 1 5 2 5 1 2 2 (3/5) 2 2 5 sin and cos then A E 0.55, B 0.00, C 86 2 18.48, 7.65, and F 2 or sin 29. tan 2 sin C 1 1 3 and cos 0.55x an ellipse 13.28 0.88, B 3 3 0, an (b) A D 0.00, 35. (a) A C x 1 2 2 10.45y 18.48x 7.65y 86 0, 26.57 0.97; then A cos 45 sin 45 sin 45 cos 45 1 2 y 2 2 2 2 1 ,F 2 2 2 1 ,B 2 0, 0.23, cos 3.12, D 2 1 2 0.74, E 3.12y 2 1.20, and F 0.74x 1.20y 1x y 2 0.88x ellipse 30. tan 2 sin B F 1 ,C 2 E 2 B y 2 2 1 ( 3) 1 2 5 1 (see part (a) above), 2 1 0, F a x 2 2 1 2 y 2 a 11.31 5.65 2.05, 0.30, and x 2a 0 we have 0.10, cos 0.995; then A 2.99, E 0.00, C 7 2 3.05, D 36. Yes, the graph is a hyperbola: with AC 4AC 0 and B 2 4AC 0. 2.05x 3.05y 2 2.99x 0.30y 7 0, a hyperbola 31. tan 2 sin C 4 1 4 4 2 3 53.13 0.89; then A 0, and F 5 26.57 0.00, B 0.00, 0.45, cos 5.00, D 2 0, E 5 0 or y 5.00y 32. tan 2 sin C 1.00, parallel lines 36.87 18.43 0.00, B 49 0.00, 2 12 18 3 2 4 0.32, cos 20.00, D 2 0.95; then A 0, and F 37. The one curve that meets all three of the stated criteria is the ellipse x 2 4xy 5y 2 1 0. The reasoning: The symmetry about the origin means that ( x, y) lies on the graph whenever (x, y) does. Adding Ax 2 Bxy Cy 2 Dx Ey F 0 and A( x)2 B( x)( y) C( y)2 D( x) E( y) F 0 and dividing by the result by 2produces the equivalent equation Ax 2 Bxy Cy 2 F 0. Substituting x 1, y 0 (because the point (1, 0) lies on the curve) shows further that A F. Then Fx 2 Bxy Cy 2 F 0. By implicit differentiation, 2Fx By Bxy 2Cyy 0, so substituting x 2, y 1, and y 0 (from Property 3) gives 4F B 0 B 4F the conic is Fx 2 4Fxy Cy 2 F 0. Now substituting x 2 and y 1 again gives 4F 8F C F 0 C 5F the equation is now Fx 2 4Fxy 5Fy2 F 0. Finally, dividing through by F gives the equation x 2 4xy 5y 2 1 0. 38. If A C, then B B cos 2 (C A) sin 2 0, E 49 20.00y 33. tan 2 sin C 0, parallel lines B cos 2 . 52 78.69 0.77; then A 5.07, E 2 5 3 2 39.35 5.05, B 0.00, 1 0, 39. Then 4 2 2 B B cos 2 0 so the 0.63, cos 0.05, D 2 xy-term is eliminated. 90 x x cos 90 y sin 90 and y x sin 90 y cos 90 x x2 b2 y2 (b) 2 a 6.19, and F 6.19y 1 y 5.05x 0.05y 5.07x a hyperbola (a) y2 a2 x2 b2 1 1 a2 m( y ) y m( y ) 1 x m (c) x (d) y (e) y 2 y 2 mx x mx bx by 1 x m b m ...
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## This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.

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