# If the parallelogram is a rectangle then u v 0 so u v

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Unformatted text preview: 4 = 1 √ c2 = 4 + 64 = 68, c = 2 17 y y + 3 = 4(x –1) y – 1 = 1 (x + 1) y 2 (–3, 1) (–1 + √5, 1) (3, –3) (–1, –3) x x (1, 1) (–1 – √5, 1) y – 1 = – 1 (x + 1) (–1 – 2 √17, –3) (1 + 2 √17, –3) 2 y + 3 = – 4(x –1) 20. (a) (y − 3)2 /4 − (x + 2)2 /9 = 1 √ c2 = 4 + 9 = 13, c = 13 (b) (y + 5)2 /9 − (x + 2)2 /36 = 1 √ c2 = 9 + 36 = 45, c = 3 5 y y (–2, 3 + √ 13) y + 5 = 1 (x + 2) (-2, 5) (–2, –5 + 3 √ 5) y – 3 = 2 (x + 2) y – 3 = – 2 (x + 2) 3 2 (–2, –2) x 3 x (-2, 1) (–2, 3 – √ 13) (–2, –8) (–2, –5 – 3 √ 5) y + 5 = – 1 (x + 2) 2 21. (a) y 2 = 4px, p = 3, y 2 = 12x (b) y 2 = −4px, p = 7, y 2 = −28x 22. (a) x2 = −4py , p = 4, x2 = −16y (b) x2 = −4py , p = 1/2, x2 = −2y 23. (a) x2 = −4py , p = 3, x2 = −12y (b) The vertex is 3 units above the directrix so p = 3, (x − 1)2 = 12(y − 1). 24. (a) y 2 = 4px, p = 6, y 2 = 24x (b) The vertex is half way between the focus and directrix so the vertex is at (2, 4), the focus is 3 units to the left of the vertex so p = 3, (y − 4)2 = −12(x − 2) 25. y 2 = a(x − h), 4 = a(3 − h) and 9 = a(2 − h), solve simultaneously to get h = 19/5, a = −5 so y 2 = −5(x − 19/5) 26. (x − 5)2 = a(y + 3), (9 − 5)2 = a(5 + 3) so a = 2, (x − 5)2 = 2(y + 3) 27. (a) x2 /9 + y 2 /4 = 1 (b) a = 26/2 = 13, c = 5, b2 = a2 − c2 = 169 − 25 = 144; x2 /169 + y 2 /144 = 1 Exercise Set 12.4 428 28. (a) x2 + y 2 /5 = 1 (b) b = 8, c = 6, a2 = b2 + c2 = 64 + 36 = 100; x2 /64 + y 2 /100 = 1 29. (a) c = 1, a2 = b2 + c2 = 2 + 1 = 3; x2 /3 + y 2 /2 = 1 (b) b2 = 16 − 12 = 4; x2 /16 + y 2 /4 = 1 and x2 /4 + y 2 /16 = 1 30. (a) c = 3, b2 = a2 − c2 = 16 − 9 = 7; x2 /16 + y 2 /7 = 1 (b) a2 = 9 + 16 = 25; x2 /25 + y 2 /9 = 1 and x2 /9 + y 2 /25 = 1 31. (a) a = 6, (2, 3) satisﬁes x2 /36 + y 2 /b2 = 1 so 4/36 + 9/b2 = 1, b2 = 81/8; x2 /36 + y 2 /(81/8) = 1 (b) The center is midway between the foci so it is at (1, 3), thus c = 1, b = 1, a2 = 1 + 1 = 2; (x − 1)2 + (y − 3)2 /2 = 1 32. (a) Substitute (3, 2) and (1, 6) into x2 /A + y 2 /B = 1 to get 9/A + 4/B = 1 and 1/A + 36/B = 1 which yields A = 10, B = 40; x2 /10 + y 2 /40 = 1 (b) The center is at (2, −1) thus c = 2, a = 3, b2 = 9 − 4 = 5; (x − 2)2 /5 + (y + 1)2 /9 = 1 33. (a) a = 2, c = 3, b2 = 9 − 4 = 5; x2 /4 − y 2 /5 = 1 (b) a = 1, b/a = 2, b = 2; x2 − y 2 /4 = 1 34. (a) a = 3, c = 5, b2 = 25 − 9 = 16; y 2 /9 − x2 /16 = 1 (b) a = 3, a/b = 1, b = 3; y 2 /9 − x2 /9 = 1 35. (a) vertices along x-axis: b/a = 3/2 so a = 8/3; x2 /(64/9) − y 2 /16 = 1 vertices along y -axis: a/b = 3/2 so a = 6; y 2 /36 − x2 /16 = 1 (b) c = 5, a/b = 2 and a2 + b2 = 25, solve to get a2 = 20, b2 = 5; y 2 /20 − x2 /5 = 1 36. (a) foci along the x-axis: b/a = 3/4 and a2 + b2 = 25, solve to get a2 = 16, b2 = 9; x2 /16 − y 2 /9 = 1 foci along the y -axis: a/b = 3/4 and a2 + b2 = 25 which results in y 2 /9 − x2 /16 = 1 (b) c = 3, b/a = 2 and a2 + b2 = 9 so a2 = 9/5, b2 = 36/5; x2 /(9/5) − y 2 /(36/5) = 1 37. (a) the center is at (6, 4), a = 4, c = 5, b2 = 25 − 16 = 9; (x − 6)2 /16 − (y − 4)2 /9 = 1 (b) The asymptotes intersect at (1/2, 2) which is the center, (y − 2)2 /a2 − (x − 1/2)2 /b2 = 1 is the form of the equation because (0, 0) is below both asymptotes, 4/a2 − (1/4)/b2 = 1 and a/b = 2 which yields a2 = 3, b2 = 3/4; (y − 2)2 /3 − (x − 1/2)2 /(3/4) = 1. 38. (a) the center is at (1, −2); a = 2, c = 10, b2 = 100 − 4 = 96; (y + 2)2 /4 − (x − 1)2 /96 = 1 (b) the center is at (1, −1); 2a = 5 − (−3) = 8, a = 4, (y + 1)2 (x − 1)2 − =1 16 16 39. (a) y = ax2 + b, (20, 0) and (10, 12) are on the curve so 400a + b = 0 and 100a + b = 12. Solve for b to get b = 16 ft = height of arch. (b) y y2 100 144 x2 + 2 = 1, + 2 = 1, 400 = a2 , a = 20; 2 a b 400 b √ b = 8 3 ft = height of arch. (10, 12) x -20 -10 10 20 429 Chapter 12 40. (a) (x − b/2)2 = a(y − h), but (0, 0) is on the parabola so b2 /4 = −ah, a = − (x − b/2)2 = − b2 (y − h) 4h (b) As in part (a), y = − 4h (x − b/2)2 + h, A = b2 b − 0 b2 , 4h 2 4h (x − b/2)2 + h dx = bh b2 3 41. We may assume that the vertex is (0, 0) and the parabola opens to the right. Let P (x0 , y0 ) be a point on the parabola y 2 = 4px, then by the deﬁnition of a parabola, P F = distance from P to directrix x = −p, so P F = x0 + p where x0 ≥ 0 and P F is a minimum when x0 = 0 (the vertex). 42. Let p = distance (in millions of miles) between the vertex (closest point) and the focus F , then P D = P F , 2p + 20 = 40, p = 10 million miles. P D 40 60° 40 cos 60° = 20 p p Directrix 43. Use an xy -coordinate system so that y 2 = 4px is an equation of the parabola, then (1, 1/2) is a point on the curve so (1/2)2 = 4p(1), p = 1/16. The light source should be placed at the focus which is 1/16 ft. from the vertex. 44. (a) Substitute x2 = y/2 into y 2 − 8x2 = 5 to get y 2 − 4y − 5 = 0; y = −1, 5. Use x2 = y/2 to ﬁnd that there is no solution if y =...
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