Business Calc Homework w answers_Part_93

Business Calc Homework w answers_Part_93 - Appendix A5.2 3....

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Unformatted text preview: Appendix A5.2 3. 3x 2 c F(0, 4. 6x 2 461 2y 2 a2 6 b2 x2 2 y2 3 1 2 0 1 6 3e c a 3 3 11. e 1e a e c a 3 1 3 4 take c 5 4 and a 5; c 2 9b a2 3; b2 3 1 3 ; 3 16 25 x 25 2 b2 b2 y 9 y 4 2 2 therefore 1 1); directrices are y 9y 2 2 x2 54 9 y2 6 c 1 3 a ; F( a e b 2 9 3, 0); directrices are 3 1 3 4 2 2 4 2 4 x x 0 3 3 3 and a y 36 2 5. Foci: (0, b2 3), e 36 9 0.5 c x 27 27 2 c e 3 0.5 6 12. The eccentricity e for Pluto is 0.25, e take c 40 b2 1 and a 15 b 4; c2 a2 b2 1 x 16 2 1 c 8 e 0.2 y2 1 1536 c a 0.25 b2 1 4 16 y 15 2 6. Foci: ( 8, 0), e b 2 0.2 c 64 8 and a 1600 x2 1536 1600 15; therefore, 1 is a model of Pluto's orbit. y 7. Vertices: ( 10, 0), e c ae x2 100 0.24 a 2.4 b 2 10 and 100 5.76 94.24 5 10(0.24) y2 94.24 1 0.1 a 7 b2 70 4900 49 4851 5 x 8. Vertices: (0, and c x 4851 2 70), e 70(0.1) ae y 4900 2 1 9 5 9 5 9. Focus: ( 5, 0), Directrix: x a e 9 5 c e2 13. One axis is from (1, 1) to (1, 7) and is 6 units long; the ae 5 and 5 e 9 5 3 Then PF (x x2 x2 4 9 ae 9 5 2 e2 e 5 5 PD 3 other axis is from (3, 4) to ( 1, 4) and is 4 units long. . Therefore, a 3, b 2 and the major axis is vertical. The center is the point C(1, 4) and the ellipse is given by 5 3 9 5 2 (x 5)2 5)2 2 y2 5x y2 5 4 (y 0)2 5 x 9 x 9 5 (x 4 1)2 (y 9 4)2 1; c2 a2 b2 32 22 5 c 18 5 5; therefore the foci are F(1, 4 c a 3 5 3 5), the y2 x 9 2 5 2 x 9 y 4 2 x 81 5 eccentricity is e y 4 a e 5 3 , and the directrices are 4 9 5 . 5 1 16 c 16 e 2 (y 0)2 ae 4 and 4 10. Focus: ( 4, 0), Directrix: x a e 16 ae e2 16 (x y2 16 4 e2 PF (x x2 1 PD 2 4)2 1 (x 4 1 1 e . Then 4 2 1 x 16 2 y 8 4)2 8x y2 16)2 32x 1 5 2 5 x2 3 4 1 2 (x 4 2 x y2 48 64 48 y2 256) x 462 Appendix A5.2 4)2 y2 2 x 3 1 c 2 14. Using PF (x x2 x2 15. 9x 2 c 5 9 e PD, we have 2 (x 9 22. Focus ( 2, 0) and directrix x a e 1 ae 2 2 e 1 2 2 2 e 1 e2 2 ae 2. 2x 2 and 4) y 2 4 (x 9 9) 2 4e (y 0)2 8x y2 16 y2 4 2 (x 9 18x 81) x2 36 y2 20 Then PF 1. (x x2 c a 5 ; 4 2PD y2 4 y2 (y 1 (x 4 x 2)2 1 2 2 1 2 20 5x 2 144 x2 16 9y 2 y2 9 180 or 1 5e 2)2 4x 16y 2 a 2 y2 4 x2 y 3 x 2 1 4 b 2 16 9 23. (x 3x 2 1)2 2x 3 x2 3)2 y2 8x 1) (x 1) 45 3 y 2 1 2 asymptotes are y x 16. y 2 c 0 x2 a e 16 . 5 3 x; F( 5, 0); directrices are 4 x2 x2 8 6y 60y 5(y 2 9 4 12y 9 2 (y 4 4y 4) 8 a2 y2 8 1 8 8 4e c a 4 8 4x 2 2; 4(x 2 24. c 2 1 c2 thus, 10 2x; 2 5 2 10 x a2 2 5y 2 2x 2 0 36) 4 4 180 b2 (y asymptotes are y y 17. 8x 2 c e F( 18. 8y 2 c e F(0, c a c a x; F(0, 8 2 x2 2 4); directrices are 6) 36 2 1 c2 e2a 2 x a2 2 0 a e 2 y2 8 a2 b2 b2 e2a 2 b2 y b2 2 a 2; e a2 y a 2(e2 2 c c a ea 1); 2y 2 a2 16 b2 10 2 a 2(e2 1) 2 8 1 1; the asymptotes of this hyperbola are y 5; asymptotes are y 0 1 8 10 x ; 2 2 5 c a a e e2 1x. As e increases, the slopes of the asymptotes increase and the hyperbola approaches a single straight line. 25. The ellipse must pass through (0, 0) c ( 1, 2) lies on the ellipse a 2b 0; the point 8. The ellipse is 0 10, 0); directrices are x 2x 2 a2 16 b2 10 2 y 2 2 x 8 2 2 5; asymptotes are y 0 a e tangent to the x-axis its center is on the y-axis, so a 2 10 10); directrices are y 1) and e 3 b2 1 3c a2 9 3a c2 a2 and b 4x 2 x2 y2 (y 4 the equation is 4x 2 4y 2) 4 2 y2 2)2 4y 4 0. Next, 19. Vertices (0, c y2 3a x2 8 1 and e 9 1 8 3 4 4 4x 2 1a (y 2 and b a2 3 1 (now using the 4 1 3 standard symbols) c2 c 3 and e 1 c a b2 3 2 3e 20. Foci ( 3, 0) and e a 1 b2 c2 3c y 8 2 c a 2 . 3a 1 8 x2 ae 2. (y 0)2 21. Focus (4, 0) and directrix x ae e2 2c 2e 4)2 2 4 and a e 2 2 4 e2 2 e2 (x 2) Then PF (x x2 x2 4) 2 2PD y 2 2x 2 2(x y2 8 x 8 2 8x y2 16 2(x 2 y 8 2 4x 1 4) Appendix A5.2 26. We first prove a result which we will use: let m1 and m2 be the slopes of two nonparallel, nonperpendicular lines. Let be the acute angle between the lines. Then tan y 463 Similarly, tan b2 . Since tan cy0 tan , and . and are both less than 90 , we have m1 1 m2 m1m2 . (To see this result for positive-slope lines, P(x0, y0) x let and 1 be the angle of inclination of the line with slope m1, 2 be the angle of inclination of the line with slope m2. m2. Then tan ( tan 1 1 1 2 F1(c, 0) F2(c, 0) Assume m1 Then tan since m1 and we have 1 1 2 . , 2 ) tan 1 tan 1 2 2 m1 1 m2 m1 m2 tan 2 tan 27. To prove the reflective property for hyperbolas: b 2x 2 2b 2x 2 and m2 tan .) a 2y 2 2a 2yy b x a 2y a 2b 2 0 Now we prove the reflective property of ellipses (see the accompanying figure): 2b x 2 y b 2x . a2y Let P(x0, y0) be a point of tangency (see the accompanying figure). The slope from P to F( c, 0) is and from P to F2(c, 0) it is y0 x0 c y0 x0 c 2a yy 2 0 y Let P(x0, y0) be any point on the ellipse y (x0) bx0 a a 2 . Let the tangent through b x0 x02 a 2y0 1 2 P meet the x-axis in point A, . Let F1(c, 0) and y0 x0 c and define the angles will show that tan y F1PA tan . and F2 PA . We F2( c, 0) be the foci. Then mPF mPF y0 x0 c 2 and . Let and be the angles between the tangent line and PF1 and PF2 respectively. Then b 2x0 y0 x0 b x0y0 a 2y0(x0 c) 2 tan 1 a 2y0 c F1(c, 0) P(x0, y0) A x F2(c, 0) b 2x02 a y0x0 b 2x0c a y0 c b 2x0c a y0c 2 2 2 b 2x0c a y0c (b 2x02 (a 2 2 a 2y02 b 2x0y0 a 2y02) b 2)x0y0 b2 . cy0 a 2b 2 c x0y0 2 464 Appendix A5.3 10. 3x 2 6xy 3y 2 4x 5y 12 B 2 4AC 62 4(3)(3) 0 Parabola 11. 3x 2 5xy 2y 2 7x B 2 4AC ( 5)2 12. 2x 2 4.9xy 3y 2 4x B 2 4AC ( 4.9)2 Hyperbola c 27. continued From the preliminary result in Exercise 26, x0b 2 y0a 2 x0 x0 y0 c y0 14y 1 4(3)(2) 1 7 4(2)(3) 0 Hyperbola 0.01 0 tan 1 x02b 2 x0y0a 2 a 2b 2 x0y0c 2 x0b 2 y0a 2 x0 b 2 c y0 a 2 c x0b 2c y0a 2c y0 x0 c y0 x0 c y02a 2 x0y0b 2 b2 . In a similar manner, y0c x0b 2 y0a 2 x0b 2 13. x 2 3xy 3y 2 6y 7 B 2 4AC ( 3)2 4(1)(3) 21xy 14. 25x B 2 4AC 2 2 3 41 0 Ellipse 0 Hyperbola 0 Ellipse 4y 212 2 2 350x 0 4(25)(4) 3xy 2y 17y 2 0 15. 6x B 2 4AC 32 4(6)(2) 39 16. 3x 2 12xy 12y 2 B 2 4AC 122 17. cot 2 tan , and therefore x x 2 2 1 2 2 2 2 A B C 0 1 435x 9y 72 0 4(3)(12) 0 Parabola 02 2 tan 1 y0a 2 b2 . Since tan y0c x sin 2 2 4 ; y cos x cos x 2 2 1 2 y 2 2 2 y sin , y y ,y 2 2 2 2 and are acute angles, we have . APC, and the APB. Since APC x 2 y 28. The tangent to the ellipse of P bisects tangent to the hyperbola at P bisects and APB are a linear pair, so that m APB m LAPB 2 x y x 2 2 y 2 4 Hyperbola ; x 18. cot 2 2x 1 1 1 y 2 A B C 02 y sin, y m APC m APC 2 180 and therefore x 90 , the tangents to the ellipse and x 2 2 2 2 2 4 x cos x 2 2 2 2 2 x sin 2 2 2 2 y cos hyperbola are perpendicular. y ,y 2 2 2 2 2 2 x y y 2 2 2 x 2 y 2 x 1 x 2 2 y s Appendix A5.3 (pp. 612618) 1. x 3xy y B 2 4AC 2 2 2 x y x x 0 ( 3)2 2 4(1)(1) 5 0 Hyperbola 1 2 1 2 1 2 1 2 1 2 xy y x y x xy 2 2 2 2 2 1 2 3 1 2 y 1 x2 y 1 3x 2 y 2 2 2 2 2 18xy 27y 5x 7y 4 2. 3x B 2 4AC ( 18)2 4(3)(27) 0 Parabola 3. 3x 2 7xy B 2 4AC Ellipse 4. 2x 2 B2 15xy 4AC 17y 2 1 ( 7)2 4(3) 17 2y 2 x y 0 ( 15)2 4(2)(2) 0.477 0 Ellipse A B C 3 1 1 3 19. cot 2 2 2 3 3 6 ; therefore x 1 0 Ellipse x 3 2 3 2 3 x cos x 1 y ,y 2 1 y 2 2 y sin , y 1 x 2 3 2 x sin y y cos 5. x 2 2xy y 2 2x y 2 B 2 4AC 22 4(1)(1) 0 0 Parabola x 3 2 6. 2x 2 y 2 4xy 2x 3y 6 B 2 4AC 42 4(2)( 1) 24 0 Hyperbola 7. x 2 4xy 4y 2 3x 6 B 2 4AC 42 4(1)(4) 8. x 2 y 2 3x B 2 4AC 9. xy y 2 3x Hyperbola 2y 02 10 4(1)(1) 4AC 0 Parabola 4 12 0 Ellipse (circle) 4(0)(1) 1 0 4x 2 1 x 2 3 x 3 1 y 2 2 1 x 2 3 2 3 y 2 y 3 2 8 y 2 x 1 y 2 8 2 3 x 16y 1 2 0 5 B2 0 Parabola Appendix A5.3 A B C 1 2 3 1 3 465 20. cot 2 2 3 6 ; 24. cot 2 A B C 0 1 0 02 y sin , y 2 4 ; therefore x x 3 2 3 2 x cos x 1 y ,y 2 1 2 y 2 1 2 3 2 y sin a, y 1 x 2 3 2 3 2 x sin y 1 x 2 y cos therefore x x 2 2 2 2 x cos x 2 2 2 2 2 x sin 2 2 y cos y ,y 2 2 2 2 2 x 2 y 2 2 x 3 2 x 1 y 2 3 2 y x 2 y 2 x 1 2x 2 2 y x 2 2 y 2 x x 21. cot 2 1 2 2 5 2 y 2 A B C y 2 1 5y 2 x y 1x 1 2 1 2 ellipse ; 0 x x 25. cot 2 2 1 2 2 1 2 y 2 1 0 02 y sin , y y A 2 2 C 2x 3 2 3 0 hyperbola 02 ; 2 4 therefore x x 2 2 2 2 x cos x 2 2 2 2 2 2 2 x sin 2 2 y cos B 2 4 y ,y 2 2 x y therefore x y x sin 2 2 2 x cos y cos 2 2 2 2 2 2 2 y sin , x 2 2 2 x 2 2 y 2 2 2 2 2 x 2 2 y, x y 2 2 2 x 2 2 y y 3 x x y 2 x y 2 2 2 y 2 2 2 2 2y 2 2y A B C 1 Parallel horizontal lines 3 2 1 3 1 3 2 3 3 x x y 2 2 2 x 2 2 y 22. cot 2 2 x sin 2 3 y 19 ; 4x 2 + 2y A 2 19 Ellipse 3 ( 1) 4 3 1 3 therefore x x 1 2 1 x 2 x cos 3 2 3 2 3 y sin , y 3 2 y cos 26. cot 2 C B y ,y 2 x 1 y 2 3 2 2 x sin y 1 y 2 3 6 ; 3 x 2 y 2 1 2 y 2 3 x 1 1 2 y 3 2 x 1 y 2 therefore x x 3 2 3 2 3 x cos x 1 y ,y 2 1 2 y 2 3 2 2 y sin , y 1 x 2 3 2 3 2 y cos x 4y 2 1 Parallel horizontal lines A B C 2 2 2 2 x 1 x 2 4 2 3 7 x 1 x 2 3 2 y 23. cot 2 02 2 y 4 ; 5x 27. cot 2 2 3y A B 7 Hyperbola 14 2 16 3 cos 2 4 3 (if we 5 therefore x x 2 2 2 2 x cos x x 2 2 2 2 2 2 2 2 2 2 y sin , y y ,y 2 x sin 2 2 y cos C 2 2 x y 2 2 choose 2 in Quadrant I); 2 2 1 1 cos 2 2 1 1 (3/5) 2 2 5 1 5 y 2 2 2 2 2 2 2 2 8 2 2x 2 x y 2 2 2 x 2 2 y 2 2 thus sin y cos or sin and x y 0 8 x cos 2 2 2 5 (3/5) 2 1 5 x 8 2y y and cos 0x 2 4y 0 Parabola ...
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