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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

Ask a homework question - tutors are online