C u v 6i 14j 2k is perpendicular to both l1 and l2

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Unformatted text preview: = ±2a, (x + c)2 + y 2 = (x − c)2 + y 2 + 4a2 ± 4a (x − c)2 + y 2 , x2 y2 cx − a , and, since c2 − a2 = b2 , 2 − 2 = 1 a a b Exercise Set 12.5 434 EXERCISE SET 12.5 1. (a) r = 3/2 , e = 1, d = 3/2 1 − cos θ (b) r = 3/2 , e = 1/2, d = 3 1 + 1 sin θ 2 p/ 2 p/ 2 2 -2 –2 0 2 0 2 –2 -2 (c) r = 2 3 2 1+ cos θ , e = 3/2, d = 4/3 (d) r = 5/3 , e = 1, d = 5/3 1 + sin θ p/ 2 p/ 2 7 3 -7 -5 10 7 0 -11 -7 2. (a) r = 1− 0 4/3 , e = 2/3, d = 2 2 3 cos θ (b) r = 1 1− 4 3 sin θ , e = 4/3, d = 3/4 p/ 2 p/ 2 0 0 (c) r = 1/3 , e = 1, d = 1/3 1 + sin θ (d) r = 1/2 , e = 3, d = 1/6 1 + 3 sin θ p/ 2 p/ 2 0 0 435 Chapter 12 3. (a) e = 1, d = 8, parabola, opens up 10 4 , e = 3/4, d = 16/3, 1 + sin θ ellipse, directrix 16/3 units above the pole (b) r = 3 4 5 -15 15 -8 8 -10 -20 2 , e = 3/2, d = 4/3, 1 − 3 sin θ 2 hyperbola, directrix 4/3 units below the pole (c) r = 3 , e = 1/4, d = 12, 1 + 1 cos θ 4 ellipse, directrix 12 units to the right of the pole (d) r = 4 4 -6 6 -5 3 -4 -8 4. (a) e = 1, d = 15, parabola, opens left 20 2/3 , e = 1, 1 + cos θ d = 2/3, parabola, opens left (b) r = 10 -20 20 -15 5 -20 -10 64/7 , e = 12/7, d = 16/3, 1 − 12 sin θ 7 hyperbola, directrix 16/3 units below pole (c) r = , e = 2/3, d = 6, 1 − cos θ ellipse, directrix 6 units left of the pole 20 -30 4 (d) r = 2 3 6 30 -3 -40 13 -6 Exercise Set 12.5 436 5. (a) d = 1, r = 2 2/3 ed = = 2 1 + e cos θ 3 + 2 cos θ 1 + 3 cos θ (b) e = 1, d = 1, r = 1 ed = 1 − e cos θ 1 − cos θ (c) e = 3/2, d = 1, r = 3 3/2 ed = = 3 1 + e sin θ 2 + 3 sin θ 1 + 2 sin θ 6. (a) e = 2/3, d = 1, r = 2 ed 2/3 = = 1 − e sin θ 3 − 2 sin θ 1 − 2 sin θ 3 (b) e = 1, d = 1, r = 1 ed = 1 + e sin θ 1 + sin θ (c) e = 4/3, d = 1, r = 7. (a) r = 4 4/3 ed = = 1 − e cos θ 3 − 4 cos θ 1 − 4 cos θ 3 ed ed ed ,θ = 0 : 6 = ,θ = π : 4 = , 6 ± 6e = 4 1 ± e cos θ 1±e 1e sign to get e = 1/5, d = 24, r = 4e, 2 = 10e, use bottom 24 24/5 = 1 − cos θ 5 − 5 cos θ d d 2 , 1 = , d = 2, r = 1 − sin θ 2 1 − sin θ ed ed ed (c) r = , θ = π/2 : 3 = , θ = 3π/2 : −7 = , ed = 3 ± 3e = −7 ± 7e, 10 = ±4e, 1 ± e sin θ 1±e 1e (b) e = 1, r = e = 5/2, d = 21/5, r = 21 21/2 = 1 + (5/2) sin θ 2 + 5 sin θ ed ed ed ,1 = ,4 = ,1 ± e = 4 1 ± e sin θ 1±e 1e 8 8/5 = r= 5 + 3 sin θ 1 + 3 sin θ 5 8. (a) r = 4e, upper sign yields e = 3/5, d = 8/3, d 6 d , 3 = , d = 6, r = 1 − cos θ 2 1 − cos θ √ √ √ 5 2d √ (c) a = b = 5, e = c/a = 50/5 = 2, r = ; r = 5 when θ = 0, so d = 5 + √ , 1 + 2 cos θ 2 √ 5 2+5 √ . r= 1 + 2 cos θ (b) e = 1, r = 9. (a) r = 3 1+ 1 2 sin θ , e = 1/2, d = 6, directrix 6 units above pole; if θ = π/2 : r0 = 2; if θ = 3π/2 : r1 = 6, a = (r0 + r1 )/2 = 4, b = coordinates), √ √ r0 r1 = 2 3, center (0, −2) (rectangular x2 (y + 2)2 + =1 12 16 1/2 1/2 , e = 1/2, d = 1, directrix 1/2 unit left of pole; if θ = π : r0 = = 1/3; 1 3/2 1 − 2 cos θ √ if θ = 0 : r1 = 1, a = 2/3, b = 1/ 3, center = (1/3, 0) (rectangular coordinates), 9 (x − 1/3)2 + 3y 2 = 1 4 (b) r = 437 Chapter 12 6/5 , e = 2/5, d = 3, directrix 3 units right of pole, if θ = 0 : r0 = 6/7, 1 + 2 cos θ 5 √√ if θ = π : r1 = 2, a = 10/7, b = 2 3/ 7, center (−4/7, 0) (rectangular coordinates), 7 49 (x + 4/7)2 + y 2 = 1 100 12 10. (a) r = (b) r = 2 1− 3 4 sin θ , e = 3/4, d = 8/3, directrix 8/3 units below pole, if θ = 3π/2 : r0 = 8/7, √ if θ = π/2; r1 = 8, a = 32/7, b = 8/ 7, center: (0, 24/7) (rectangular coordinates), 49 72 x+ 64 1024 y− 24 7 2 =1 2 , e = 3, d = 2/3, hyperbola, directrix 2/3 units above pole, if θ = π/2 : 1 + 3 sin θ 2 √ 3 r0 = 1/2; θ = 3π/2 : r1 = 1, center (0, 3/4), a = 1/4, b = 1/ 2, −2x2 + 16 y − =1 4 11. (a) r = 5/3 , e = 3/2, d = 10/9, hyperbola, directrix 3/2 units left of pole, if θ = π : 1 − 3 cos θ 2 5/3 9 9 r0 = 2/3; θ = 0 : r1 = = 10/3, center (−2, 0), a = 4/3, b = 20/9, (x+2)2 − y 2 = 1 1/2 16 20 (b) r = 12. (a) r = 4 , e = 2, d = 2, hyperbola, directrix 2 units below pole, if θ = 3π/2 : r0 = 4/3; 1 − 2 sin θ θ = π/2 : r1 = √ 4 9 = 4, center (0, −8/3), a = 4/3, b = 4/ 3, 1−2 16 y+ 8 3 2 − 32 x =1 16 15/2 , e = 4, d = 15/8, hyperbola, directrix 15/8 units right of pole, if θ = 0 : 1 + 4 cos θ √ 5 4 15 , center (2, 0), 4(x − 2)2 − y 2 = 1 r0 = 3/2; θ = π : r1 = − = 5/2, a = 1/2, b = 2 2 15 (b) r = 13. (a) r = 1+ 8=a= (b) r = 1− 4=a= (c) r = 1− 1 2d 1 2 cos θ 1+ 3 5d 3 5 sin θ = 3d 3 3 , if θ = 3π/2 : r0 = d; θ = π/2, r1 = d, 5 − 3 sin θ 8 2 1 64 3(64/15) 64 15 (r1 + r0 ) = d, d = ,r = = 2 16 15 5 − 3 sin θ 25 − 15 sin θ 3 5d 3 5 cos θ 1 5d 1 5 sin θ 5=c= d , if θ = 0 : r0 = d/3; θ = π, r1 = d, 2 + cos θ 12 2 1 (r1 + r0 ) = d, d = 12, r = 2 3 2 + cos θ d = 16/3, r = (d) = 1 d 2 = = 3d 3 3 3 , if θ = π : r0 = d; θ = 0, r1 = d, 4 = b = d, 5 − 3 cos θ 8 2 4 16 5 − 3 cos θ d , if θ = π/2 : r0 = d/6; θ = 3π/2, r1 = d/4, 5 + sin θ 1 120 11 − d, d = 120, r = = 46 24 5 + sin θ Exercise Set 12.5 14. (a) r = 438 1+ 1 2d 1 2 sin θ 1− 6=a= d , if θ = π/2 : r0 = d/3; θ = 3π/2 : r1 = d, 2 + sin θ 15 2 1 (r0 + r1 ) = d, d = 15, r = 2 3 2 + sin θ...
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This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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