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nsvalue for the w idth of the box w ould be negative or zero.3. height = 2.94 in., length « 14.12 in., w idth ~ 10.12 in., m axim um volum e ~ 420.11 in 3 5. radius = 1.55 in., height = 3.09 in. m inim um surface area = 45.19 in 2 7. (6 .67 ,1 .83 ), Sam ple answ er: A m inim um tim e o f 1.83 hours w ill be obtained if the racer takes a path that w ill have h im /her at the sidew alk 3.33 m iles before the pier.9. m = 0; Sam ple answ er: The slope of the tangent line is 0. The tangent line is a horizontal line.

CHAPTER 7Conic Sections and Parametric Equations

Page 421 Chapter 7 Get Ready

1. x = 1; (0, - 1 2 ) ; (1, - 1 3 ) 3. x = - 1 ; (0, - 8 ) ; ( - 1 , - 1 0 )5. x = 2; (0, - 4 ) ; (2, - 1 6 ) 7 .x = 25; (0, 550); (25 ,543 .75 ) 9 .1 0 8 1 1 .1 0 0 1 3 .—172 15. vertical asym ptote at x = —5 17. vertical asym ptotes at x = —5 and x = 1; horizontal asym ptote at y = 0 19. vertical asym ptote at x = 4.

Pages 428-431 Lesson 7-1

1. vertex: (3, 7); focus: (3 ,1 0 ); directrix: y = 4; axis of sym m etry: x = 3

3. vertex: (—2, 4); focus: (3, 4); directrix: x = — 7; axis of sym m etry: y = 4

-12

( y - 4)2 = 20(x+ 2)

5. vertex: (—8 ,3 ); focus: (—8 ,5 ); directrix: y = 1; axis of sym m etry: x ■

7 . vertex: (1, —5); focus: (7, —5); directrix: x = —5; axis o f sym m etry: y = —5

R90 ; Selected Answers

9. vertex: ( - 8 , - 2 ) ; focus: (—8, —3); directrix: y = — 1; axis o f sym m etry: x = — 8

11. Sam ple answ er: The focus is 4 feet above the ground.

(13) a. y 2 - 180x + lOy + 565 = 0

y 2 + 10y = 180x - 565

y2 + lOy + 25 = 180x - 565 + 25

y 2 + lOy + 25 = 180x - 540

(y + 5 )2 = I8 0 (x - 3)

b. The equation in standard form has y as the squaredterm , w hich m eans that the parabola opens horizontally. Because 4p = 180, p = 45 and the graph opens to the right. The equation is in the form (y — fc)2 = 4p(x — h), so h = 3 and fc = —5. Since the stern is located at the vertex o f the parabola form ed, it is at the p oin t (/;, fc) or (3, —5). The sw im m er is at the focus, located a t (h + p, fc), w hich is (3 + 45, —5) or (48, —5). The d istance the sw im m er is from the stem o f the boat represents the length of rope needed to attach the sw im m er to the stern. U sing the distance form ula, the d istance betw een these tw o points is

V (4 8 - 3 )2 + ( - 5 - (—5))2 or 45. Therefore, the length o f rope attach ing the sw im m er to the stern o f the boat is 45 feet.

15. x 2 = 8(y + 7); vertex: (0, —7); focus: (0, —5); directrix: y = —9; axis o f sym m etry: x = 0

17. x2 = —24(y + 1); vertex: (0, —1); focus: (0, —7); directrix: y = 5; axis o f sym m etry: x = 0

19. y2 = 20(x - 3); vertex: (3, 0); focus: (8, 0); directrix: x = —2; axis o f sym m etry: y = 0

21. (y — 4 )2 = 10(x — 2); vertex: (2 ,4 ); focus: (4 .5 ,4 ); directrix: x = —0.5; axis of sym m etry: y = 4

23. (x + 6)2 = 18(y + 9); vertex: (—6, —9); focus: (—6, —4.5); directrix: y = —13.5; axis of sym m etry: x = — 6

25. 9 in.27. (y + l ) 2 = 24(x + 4)

f 2 — 1 8 y + 1 2 x = 1 2 6

\ 1 J\ 1 /1 /

1- 1 0 1 o n o X

\1

Y

, k f y

—6- ■o — ■* w

\

( y + l ) 2 = 2 4 (x + 4 )

29. (x + 3 )2 = 8(y - 2)i y

\ i8 /

v /\\ 1-4\V /

- 8 )\ i O X(x + 3) 2 _ OO 2)

T 1

31. (y - 4 )2 = —32(x - 7)

ys

I \

r“-20 -10 0 10 20x

1/

I><C\JCOIIIC\JI

33. (x - l ) 2 = — 12(y - 6) 35. (x - 8)2 = 16(y + 7)I *

1►1

11 S* \1

—L 0 1 ix\ 1

I ' T( x - 1 )2 = —12 (y — 6)

37. (x - l ) 2 = —12(y - 5) 39. (x + 4 )2 = - 4 ( y - 1)

•8:i,

11

- 4 O 1 8X

O r-1)z = 12(y-- 5 ) -t1t

k y

-kx + 4 2 = -4 y-- 1 )-A K

-►

-1 2 - 8 / X\

/ \\

For Homework Help, go to H o tm a th .co m

41. (y + 9 )2 = 4 (x + 6)

45. y = —8x — 45

(4!

47. y = 4x + 14

The graph opens vertically. D eterm ine the vertex and focus.—0.25(x - 6 )2 = y - 9

(x - 6 )2 = —4(y - 9)Because 4p = —4, p = — 1. The vertex is (6 ,9 ) and the focus is (6 ,8 ) . We need to determ ine d, the distance betw een the focus and the point of tangency, C.This is one leg of the isosceles triangle.

d = \j{x2 - x l )2 + (y2 - y l )2

= \/( 10 - 6 )2 + (5 - 8 )2

= 5U se J to find A , the end poin t o f the other leg o f the isosceles triangle. S ince p is negative, the parabola opens dow n and A w ill be above the focus.

A = (6, 8 + 5) or (6 ,1 3 )Points A and C b o th lie on the line tangent to the parabola. To find an equation for this line, first use tw o points to find the slope m.

Then use a p oin t on the line and the slope to w rite an equation for the line.

Point-slope formula

(x „ y t ) = (1 0 ,5 ) and m = - 2 Simplify.Solve for y.

51 . opens dow nw ard 53. opens upw ard 55. (x — 3 )2 =—4 (y — 5) 57. (y — l ) 2 = —16(x + 5) 59a. Sam ple answ er:

x 2 = —180.27(y + 20) 59b. about 30.35 m63. (y + 5 )2 = 8(x + 3)

y - y i = m ( x - * i )y — 5 = —2(x — 10) y — 5 = —2x + 20

y = —2x + 25

■y

-6 "(5,3 h 5, 7)

0 6 12 18 X(--1, - 1)

12

65. (2, 8) 67. (0, - 1 ) 69a. y 2

71a. (y — 2 )2 = 8(x + 3) or (y -3

2 )2 :

69b. 1.53 ft

—8(x + 3)

R91

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Sam ple answ er: To prove that the endpoints o f the latus rectum X and W and the point o f intersection o f the axis and directrix D are the vertices of a right isosceles triangle A X D W , w e need to show that ZXDW is a right angle and that X D = WD. Since X F = FW , FD = FD, and Z X F D = ZWFD, A X F D = A FW D by SA S. Thus, X D = WD. To prove that ZXDW is a right angle, w e can first p rove that Z X D F and ZFDW are b oth 45° angles. Since XF = FD and ZXFD is a right angle, A X F D is an isosceles right triangle. Therefore, ZXDF is 45°. This also m eans that ZFDW is 45°. L ikew ise, ZXDW is a right angle. Thus, AXDW is an isosceles right triangle. 73. Sam ple answ er: x 2 = 4 (y — 2) 75. Sam ple answ er: y 2 = 12x 77a. i. 1 unit; ii. 2 units; iii . 4 units

77c. A s the focus is m oved farther aw ay from the vertex, the parabolas becom e wider. 77d. Sam ple answ er:(x + l ) 2 = 4 (y + 7)

77e. Sam ple answ er: The parabolas all have a vertex of ( 0 , - 1 ) and open dow nw ard. T he first equation produces the narrow est parabola and the second equation p roduces the w idest parabola.

(79) The w idth of the sector is 3 units, so y = 1.5. A = —xy

and the area o f the sector is 2.4 square units. Substitute the values into the equation and solve for x.2-4 = 4 x (1.5)

2.4 = 2x 1.2 = xTherefore, the vertex o f the parabola is (0, 0), and tw o other points on the parabola are (1 .2 ,1 .5 ) and (1.2, —1.5). Set up and solve a system o f equations using y as the independent variable. ay2 + by + c = x

a( 0 )2 + b{ 0 ) + c = 0 a(1.5)2 + b(1.5) + c = 1.2

a (—1.5)2 + b(—1.5) + c = 1.2 c = 0

2.25 a + 1.5b = 1.2 2.25a - 1.5b = 1.2

4.5a = 2.4

2.25 • ^ + 1.5b = 1.2

1.2 + 1.5b = 1.2 1.5b = 0

b = 0The equation of the parabola is -j|-y2

81. Q uadrants I and IV; the vertex is (—2, 5) and p = —2. Since the vertex is to the left of the y-axis, and the parabola opens to the left, no points w ill be to the right o f the y-axis, or in Q uadrants I and IV.

- 1

2 15: x or yL = — x.O

85. m ax at (7 ,8 .5 ) = 81.5, m in at (2 ,0 ) = 16 8 7 . — +

89. cot 6 = 7- , csc 6> =6 6

y + 2 y + 1

97. y

/\\

0 X

-8

f i x ) = x 3 - x ‘ - 4 x + 4

101. G 103. G

Pages 438-441 Lesson 7-2

1.

3. 5 .

x2 + 9y 2 - 1 4 x + 3 6 y + 49 = 0

— 9

A

9x2 + y 2 + 126x + 2y+ 433 = 0

7 (x — 3)2 | (y + 3)2 l a ( x - 2 ) 2 | (y — 2)2 1100 36 25

„ (* + <>)2 , ( y - 3)2 , « ( * - 2)2 , ( y - s ) 2 ,64 100 225 81

1 5 .0 .8 3 7 1 7 .0 .4 2 6 19 .0 .511 2 1 .0 .4 4 7 23a. 28.8 in. 2 i'2

324 ' 2072 3 b . J ^ + ^ = l 2 5 . (X 1} + = 1 ; ellipse

27. (y + 9) = 12(x — 6); parabola 29. ellipse

2x 2 + 7y2 + 24x + 84y + 310 = 0

2x2 + 24x + 7y2 + 84y + 310 = 0

( x - 7 ) 2 (y + 2)2 1 n

R92 I Selected Answers

2{x2 + 12x + 36) + 7(y2 + 12y + 36) + 310 = 0 + 72 + 252

2(x + 6 )2 + 7(y + 6 )2 = 14 (x + 6)2 (y + 6)2

■ + ■ = 17 2

The related conic is an ellipse becau se a =/= b and the graph is

of the form + (y 7 ^ = 1 .a

33. (x - 3 )2 + y2 = 4b2

35. (x + 4 )2 + (y + 3) = 36

- H 3, 0) - ^ —

K/ A

S" n

\ ,- 1 2 t i

r i

¥ r

O \ 4 *

v -

37.

P F i + PF2 = 2a

\J(x - 0 )2 + (y - c)2 + ^/(x - 0 )2 + (y - ( - c ) ) 2 = 2a

\jx2 + (y - c)2 + \Jx2 + (y + c)2 = 2a

\jx2 + (y - c)2 = 2a - \Jx2 + (y + c)2

x2 + y2 — 2cy + c2 = 4a2 - 4 a ^ x2 + (y + c)2 +

x2 + y 2 + 2cy + c2

4a\jx2 ^Jy7\-~tf = 4a2 + 4cy

a\Jx2 + (y + c)2 = a 2 + cy

a2(x2 + y2 + 2cy + c2) = a4 + 2a2cy + c2!/2

a2x 2 + a2y2 + 2fl2cy + a2c2 = a 4 + 2a2cy + c ^ 2

— <P~y2 + a2x 2 = a4 — a2c2

y 2(a2 — c2) + a2x 2 = fl2(a2 — c2)

y 2b2 + a2x 2 = a2b2

y- + ^ = \ a2 b2

3 9 . i i ^ + (y - 3) = 1 4 1 . ^ + ^ = 116 1 16 49

43) a. The length o f the m ajor axis is 43.4 + 28.6 + 0 .87 or

72.87. The value o f a is or 3 6 .435 . The d istance0 87from the focus (the sun) to the vertex is 28.6 + — or

29.035. Therefore, the value o f c, the focus to the center, is 36.435 — 29.035 or 7.4. In an ellipse,

c2 = a2 — b2. The value o f b is V « 2 — c2 =

V 3 6 .4352 - 7 .42 ~ 35.676. T he valu e o f 2b, the length o f the m inor axis, is about 35.676 • 2 or about 71.35 m illion mi.

c 7 4b. The eccentricity e is equal to —. e = ^ ^ : 0.203

45. center: (0, —6); foci: (+ 5 V 3 , —6); vertices: (± 1 0 , —6 )

47. center: (—1, 0); foci: (—1, ± 7 ) ; vertices: ( —1, ± V 6 5 )

49. — + — = 1 51.100 64

(y + 4 ) , ( x — 2 )36 + '

■2 - 78416

53a. x 2 + y2 = 64, x 2 + y 2

For Homework Help, go to H o tm a th .c o m

53b. I y** >

*1 5

>/ V

\f f N 111 - 1 5 V 15 1 X\

15/

\ ✓

JU

53c. 5 days

55. ( 7 , - 1 ) , ( 3 , - 5 ) 57. ( - 2 ,3 ) , ( 6 , - 5 ) 59. ^ ^ = i

61. (x - 6.5)2 + (y - 4 .5 )2 = 32.5 63. (x - l )2 + (y + 7)2 = 16; center: (1, —7), radius: 4 65. (x + l )2 + (y — 4 )2 = 64; center: (—1 ,4 ), radius: 8 67. Sam ple answ er: N o; if a2 = p and b2 = p + r, then c = V —r. If a 2 = p + r and b2 = p, then c = \fr and the foci are (0 , ± V r ) .

The area o f the ellipse A = 24tt. Substitu te into the area form ula to obtain an equation w ith a and b.24tt = -nab

24 = abSubstitu te a = b + 5 into the equation and solve for b.

24 = (b + 5 )b 24 = b2 + 5b

0 = b2 + 5b — 24 0 = (b + 8 )(b - 3)

Since b cannot b e negative, b = 3 and a = 3 + 5 or 8 . Therefore, an equation for the e llip se is

x 2 v2 x 2 v 2

8 3 64 971 . Yes; sam ple answ er: If (x, y) is a point on the ellipse, then(—x, —y) m u st also b e on the ellipse.

x2 y2 (- — = 1

2 b2a( - x ) 2 , ( - y )

+ - b2= 1

X2 y 2 h — = 1a2 b2

Thus, (—x, —y) is also a p oin t on the ellipse and the ellipse is sym m etric w ith respect to the origin.

73. Sam ple answ er: W hen a is m u ch greater than c, j is close to 0. S ince e = the valu e o f e is close to zero and the foci are near the center o f the ellipse. So, the ellipse is nearly circular.

75. vertex: —y - j ; focus: —7^; d irectrix: y =

av,'s o f sym m etry: y — —

27 4 '

7 7 .160 M y R eal Babies and 320 M y First Babies

79. sin (» + f ) - cos (0 + f j

= sin 0 cos ^ + cos 0 s in ~ — cos 0 cos ^ + sin 0 sin -3 3 6

= i sin 0 + cos 0 — cos 0 + j sin 0

$connectE ^Ticgrav^iil^orJ R 9 3

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and Solutions

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= j sin 8 + j sin 9

= sin 6

81.21 J>2L 83. ^ 85. [ - 7 ,5 ] 87. 4 real zeros and4 4 2 6 6

3 turning points; 0 , —4, and —2 89. 5 real zeros and4 turning points; 0 ,5 , and —2 91 . 3 + 4 i 93. B 95. C

Pages 449 -452 Lesson 7-3

1.

(8.89, 0)

t- y

/\S\ /■

- 5 48 ,0\\ Me ,c )? 5 , 18, 0)

- 8i 0 \s 8x

( - 3 0)-,r/v £ - 4 - 1

// \\- —o1 -1K

7.

-1 6 - 8 On

-(o, —9.431:

rfp T 9.43)- Y(0. 9)

9.

if. _ - 181 8

y

\ ✓

31 m <72,\i3)rk

tn 3 .1 5,0

0 < /\ X/ (2* 0)V

/ \\/ \S)3x2 2 y _ 12

11. V V

7 ?

(

- ^ J U - - t — /--------

-2 C - 1 (> f

10 20 it

-/ ------

J 20 -

-a I y5

^ = 1 B1

(x -5 )2 _ (y-1)249 17

19.- x z + 3yz - 4x + 6y = 28

k j - -2,

" 8 V✓O

- U s--------

- ■ ( - 2I l

i - 8

21. M -7 . 7.92)* ys JV /\ /

vA' - 1 2 / \ 0 X

-7, -3 .9 2 ^

»Ss

VL

- 5 x 2 + 2y2 - 7 0 x - 8y= 287

23.

27.

(3?

(y - l ) 2 (x + l ) 215

(y + 8)216

49

(x + 3)2

33

= 1 25.

= 1 29.

(x - 3)2 (y + 1)2

27 9

(x + 7)2

25(y - 2)2

49 = 1

The vertices are 4 units apart, so 2a = 4, a = 2, and a2 = 4. The center is the m idpoint of the vertices, or (—2, —3). Therefore, h = —2 and k = —3. T he conjugate axis is of length 12, so 2b = 12, b = 6, and b2 = 36. The x-coordinates of the vertices are different, so the transverse axis is horizontal and the a2-term goes w ith the x 2-term . The equation for the hyperbola is [x + 2)2 (y + 3)2

33a.

39.49.

4

(y - 4)2

36

(x - 5)2

225 7

= 1.

1 33b. 90 ft 3 5 .1 .2 7 3 7 .1 .3 3

1.68 4 1 .1 .3 2 4 3 .2 .8 3 45. hyperbola 47. parabola ellipse 51 . hyperbola 53. circle

) a. The com m on d ifference is 18 kilom eters and the absolu te value of the difference of the distances from any point on a hyperbola to the foci is 2a, so 2a = 18, a = 9 , and a 2 = 81. T he tw o airports are the foci o f the hyperbola and are 72 kilom eters apart, so 2c = 72, c = 36, and c2 = 1296. c2 = a2 + bz, so b2 = 1296 — 81 or 1215. A irport B is due south of airport A, so the transverse axis is vertical and n2-term goes w ith the y2-term . The equation for the hyperbola is

t . '

81 1215= 1.

b. I y C(0 36)-

20Airpo

1r t /be:f e

,

- 4 0 - 2 0 0 20 4 Ox-*-

201 1 -

-irport 0, - 3

bj1 ( B)

I

T he p lane is on the top branch because it is closer to airport A.

C. Because the h ighw ay is the transverse axis and the p lane is 40 kilom eters from the highw ay, x = 40. Substitute 40 for x in the equation.

R94 ! Selected Answers

V

y2 4o2 = 181 1215y2j - - 1.317 = 1ol 2

f r = 2 3 1 7V2 = 187.67y ~ 13.7

The coordinates of the plane are (40 ,13 .7 ).

|P F i - P F 2| = 2a

| -\/(x - 0 )2 + (y - c)2 - \/(x - 0 )2 + [y - (—c)]2 | = 2a

\jx2 + (y - c)2 - ^/x2 + (y + c)2 = 2«

^/x2 + (y - c)2 = 2a + \Jx2 + (y + c)2

x2 + (y - c)2 = 4a2 + 4a\Jx 2 + (y + c)2 + x 2 + (y + c)2

—4cy — 4n2 = A a \ jx 2 + (y + c) 2

—cy — a 2 = a\j x2 + (y + c)2

c2y2 + 2fl2cy + a4 = «2x 2 + 2«2cy + a2c2 + a 2y 2 c2y 2 — a2y 2 — a2x 2 = a2c2 — a4

(c2 — a2)y2 — a2x 2 = a 2(c2 — a 2)

By the Pythagorean Theorem , c2 — a 2 = b2.59. ( - 1 .3 , - 5 .7 ) , (7, - 1 .5 ) 61. ( - 5 , - 1 0 ) , (6 .9 ,13 .7 )

r 2 y263. (0 ,6 ) , (0, —6) 65a. —— = 1 65b. on the right

2 550 7901 b

branch 67. ^ = 1 69. (641 .2 ,3178 .3 ), (1212.8, 640.6),2 (

(331 .6 ,4454 .4), (2351 .0 ,100 .8) 71. (* * g4) ■ - (?/ ~g } = 1

(73) The foci are 14 units apart, so c = 7 and c2 = 49.The center is the m idpoint of the foci, (6, —2), so h = 6

c 7 oand k = —2. The eccentricity — is —, so a = 6 and « = 36.

b2 = c2 — a2 = 49 — 36 or 13.The x-coordinates of the vertices are different, so the transverse axis is horizontal and the a2-term goes w ith the x2-term . The equation for the hyperbola is

(x - 6)2 (y + 2)2 36 13

9y2 2t/2 r 2 I/275‘ 121 " 121 = 1 ?7‘ Sample answer: 16 _ 48 = 179. Sam ple answ er: If the transverse is horizontal, then a is ahorizontal distance and b is a vertical distance. So, the slopes

of the asym ptotes are ±-|-. If the transverse is vertical, then a

is a vertical distance and b is a horizontal distance. So, the slopes of the asym ptotes are ±~.

For Homework Help, go to H o tm a th .c o m

y x281. —— — = 1 83. Sam ple answ er: First, determ ine

w hether the orientation of the hyperbola is vertical or horizontal. T hen use the foci to locate the center of the hyperbola and determ ine the values o f h and k. U se the transverse axis length to find a 2. F ind c, the distance from the center to a focus. U se the equation b2 = c2 — a2 to find b2. Finally, use the correct standard form to w rite the equation, depending on w hether the transverse axis is parallel to the x-axis or to the y-axis.

85.

. 8 *

87a. 105 ft 87b. 5 seconds

— h64 +(y + 5)2

49= 1

1 - 7 8 ■xi - 3 '89. 6 5 - 2 • x2 = 2

.3 - 4 9. X3. . 2 6 .

; (—5 ,1 0 , 9) 9 1 .2 « 7 r ; n e .

93. no solution 95. sin 9 = cos 9 = tan 9 = ■ ,I d 2d 7

csc 9 = sec 9 = cot 9 = 97. —|, —1 (multiplicity: 2),

4 + 9 1, 4 - 91; (2x + 3)(x + l ) 2(x - 4 + 9i)(x - 4 - 9i)99. H 101. G

Pages 4 5 9 -4 6 1 Lesson 7 -4

1 . hyperbola; ( x 7) 2 + 2 \[3 x 'y ' — (y ' ) 2 + 18 = 0 3. parabola;

(yO2 — 8 x ' = 0 5. parabola; ( x ^ 2 + 2\[?>x'y' + 3(y')2 +

1 6 V 3 x ' - 16y' = 0 7. ellipse; 2 (x0 2 + 2(y02 - 5\ flx +

5\ fly — 6 = 0 9. hyperbola; 25(x ')2 — 4(y ')2 + 64 = 0

1 3 . 4 + ^ = 1 1 5 . ^ - ^ - ^ = !x'2 _ 1 1 o x'2 j - l i e0 8~ “ T T _ 15- ‘

17_ ( f2! _ ( 3 ^ i= l 19. x 2 — xy + y 2 0

21. x 2 + 10V 3xy + 111/2 - 144 = 0 2 3 .13x2 + 6y/3.xy +

7y2 - 112 = 0 25. 3 x 2 - lOxy + 3y2 + 128 = 0

27. 23x2 + 2 6 V 3 x y - 3y2 - 144 = 0

a. Find the equations for x and y for 9 = 30°. x' = x cos 9 + y sin 0

_ V3 , 1 - —x + 2 yy' = y cos 9 — x sin 9

V3 1= ^ - V ~ 2 X

Substitu te into 144(x')2 + 64(y^2 = 576

144 I f x + | y )2 + 64 ( ^ y - | x )2= 576

144 x2 + } y 2 + ^ - x y ■64 ( I y z + } * 2 - ^ f x y )■■576

108x2 + 36y2 + 72\/3xy + 48y2 + 16x2 - 32V 3xy = 576

124x2 + 40\/3xy + 84y2 - 576 = 0

31x2 + 10\/3xy + 21y2 - 144 = 0

b. G raph the equation by solv ing for y.

21y2 + ( l0 V 3 x )y + (31x2 - 144) = 0 U se the quadratic form ula.

connectED.mcgraw-hill.com I R95

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and

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y = -

—10\/3x ± \/(10V3x)2 - 4(21)(31x2 - 144) 2(21)

-10\/3x ± V12096 - 2304x242

Graph the conic.

33. i \y\■ 8(x')2 4- 6 (y-)2 =

y"s.V

( \v 0 ) X\ /

39. Sx2 + bxy - 4 y 2 = - 2

\ J YAK

[ -6 .6 1 ,1 4 .6 ] scl: 1 by [ - 2 , 1 2 ] scl: 1

—7.58, 7 .58] scl: 1 by [ - 5 , 5] scl: 1

41.

2 x 2 + 4xy 4- 2 y 2 4- 2\/2x - 2\fiy = - 1 2

[ -1 0 .5 8 ,4 .5 8 ] scl: 1 by [ - 2 , 8] scl: 1

[ - 1 0 , 1 0 ] scl: 1 by [ - 1 0 ,1 0 ] scl: 1

R 96 S e le c t e d A n s w e r s

47. intersecting lines y = 4x and y = —4x 49. point at (0, 0) 51 . b 53. c

[ - 1 0 , 1 0 ] scl: 1 by [ - 1 0 , 1 0 ] scl: 1

55a. A x2 + Bxy 4- Cy2 + Dx + Ey + F = 0 is equivalent to A'(x')2 + B'x'y' + C '(y')2 + D'x' + E'y' 4- F' = 0 by rotating the

A — Cconic section through 9 such that cot 29 = — - — . F is unaffected b y this rotation, so F = F'.

55b. Ax2 4- Bxy 4- Cy2 + Dx + Ey + F = A '(x')2 + B 'x'y ' 4- C'(y')2 + D 'x' + E'y' + F'; by subtracting out the xy, x, y, and F term s, the rem aining statem ent is true. A x2 4- Cy2 = A '(x ')2 4- C'(y')2; A m ust equal A ’ and B m ust equal B' in order for the statem ent to be true. 55c. A x2 + Bxy 4- Cy2 4- Dx + Ey +F = A '(x ')2 + B'x'y' 4- C (y r)2 + D 'x ' + E'y' 4- F'; b y subtracting out the x, y, and F term s, the rem aining statem ent is true.A x2 + Bxy + Cy2 = A ’ix1)2 + B'x'y' 4- C '(y')2; A m ust equal A', B m ust equal B', and C m ust equal C' in order for the statem ent to be true. Therefore, B 2 — 4AC = (S ')2 — 4 (A'C); A(x cos 9 + y sin 9)2 + B(x cos 9 4- y sin 9)(y cos 9 — x sin 9).

(57) For each equation, solve for y using the quadratic form ula. T hen graph using a graphing calculator.

9x2 4- 4xy 4- 5y2 - 40 = 0

5y2 4- 4x(y) 4- (9x2 — 40) = 0

—4x ± \J(4x)2 — 4(5)(9x2 — 40)

—4x + V800 - 164x2 10

x2 - xy - 2y2 - x - y 4- 2 = 0 —2y 2 4- (—x — l)y 4- (x2 — x 4- 2) = 0

x + 1 ± y (—x — l)2 — 4(—2)(x2 — x 4 2)

x + 1 + V?x2 — 6x + 17

[ - 1 0 , 1 0 ] scl: 1 by [ - 1 0 , 1 0 ] scl: 1

The graphs in tersect at four points: (—1.9, 2.2), (—1.5, —1.5), (1.9, - 2 .2 ) , (1.9, 0.8).

Equation GraphM inim um Angle of

Rotation

x2 - 5 x 4 - 3 — y = 0 parabola 360°

6x2 4 10y2 = 15 ellipse OO 0 02 x y = 9 hyperbola 180°

59b. Sam ple answ er: A parabola has 1 line o f sym m etry and the m inim u m angle of rotation is a com plete circle. A n ellipse and a hyperbola have 2 lines of sym m etry and the m inim um angle of rotation is a half circle. 59c. 130°

61 . Let x = x' cos 9 + y' sin 9 and y = —x' sin 9 4- y' cos 9. r 2 = x 2 + y 2

= (x ' cos 9 + y' sin 9)2 + (—x' sin 9 + y' cos 9)2 = (x ^ c o s 2 9 4- Ix'y'cos 9 sin 9 4- (y ^ s in 2 9 + (x^sin2 9

- Ix'y'cos 9 sin 9 4- (y ^ c o s 2 9 = [(xr)2 4- (yO ^cos2 9 4- [(xO2 4- (yO ^sin2 9 = [ (*0 2 + (y ^ K c o s2 9 4- s in 2 9)= [(*0 2 + (y02K 1)= M 2 + (yO2

63. cos 9 (x = x' cos 9 — y' sin 9)sin 9 (y = x' sin 9 4- y' cos 9) x cos 9 = x' co s2 9 — y' sin 9 cos 9

4- V sin 9 = x' s in 2 9 4- v ' sin 9 cos 9 x cos 9 4- y sin 9 = x' co s2 9 4- x' s in 2 9 x cos 9 4- y sin 9 = x' (cos2 9 4- s in 2 9) x cos 9 4- y sin 9 = x'

sin 9 (x = x' cos 9 — y' sin 9) cos 9 (y = x' sin 9 4- y' cos 9)

x sin 9 = x' cos 9 sin 9 — y' s in 2 9 — y cos 9 = x' cos 9 sin 9 4- y ' cos2 9

x sin 9 — y cos 9 = — y' s in2 9 — y' c o s 2 9 x sin 9 — y cos 9 = —y' (sin2 9 4- cos2 9) y cos 9 — x sin 9 = y'

65. Sam ple answ er: The d iscrim inant is defined as B 2 — 4 4 C, or in this instance, (B ')2 — AA'C'. S ince a conic that is rotated has no B' term , the d iscrim inant reduces to —AA'C'. Thus, only the A' and C term s determ ine the type o f conic. Therefore, —AA'C' < 0 w ould be an ellipse or a circle,—AA'C' = 0 w ould be a parabola, and —AA'C' > 0 w ould be a hyperbola. For a circle or an ellipse, A' and C need to share the sam e sign. For a parabola, either A! or C has to be equal to 0. For a hyperbola, A' and C' need to have opposite signs.

67. y iI

- f - 4 O 4 8x

- 4 i/24 - - - S 7 = 1

2 — o r

69. > i y\ s

\ 8J (

A - i O 8 AM

- (x — 3)2 64

( y—7 25

) 2 = r

7 1 .0 .4 4 7 73a. s 4- d = 5000, 0.035s 4- 0 .05d = 227.50 73b. savings account: $1500; certificate of deposit: $3500 75. 2 77. yes, yes; (x — 4)(x - 2)(x 4- 2)(x 4- 3) 79. D 81. A

Pages 469-471 Lesson 7-5

1.

For Homework Help, go to H o tm a th .c o m

3.

x = - f 4 -4r7 ^ 4 4OO1CvJII <r>

_ II-X

!

i-T6

t= ) ^ t = a ~ t = -1

7. “8 y

-60

2y = - V F 4 - 5

-2t p

}h = 7

4 = b t--= 8

O 3 4 X

9. y = 0 .25x2 4- 2 .5x 4- 10.25 11 . y = 5\/x + 2

M 2 - 8

12

- 4

> f- f

t

A -r-

— i\ - A

1 5 .y = ^ + 9

1 6y

0

- 1 6 - f O 1 6 x

16

2 217. y = - x 2 + ^ fx -l- 100 19. 4- = 13 4 49 4

y

i = : 0J

= t

f = ^0 i 1 X

2 2 „ x 1/T + T

I I y I 1 I2 J x = 3 c o s e

y = 3 s in 9

0 = K = 00 X

0 dl2

1

r

1

R97

Selected Answers

and Solutions

Selec

ted

Answ

ers

and

Solu

tions

2 7 .x = | a n d y = ^ 9 - | ^

1t

I= 0

- —'-V -

8x= 9 - x2

I- A

\

-t= -2 4 O f== 24JC

29. x = 5t — 20 and y = - 2 5 f 2 + 200f •390

31. x = 1 — 2f andy = - f 2 + t + \

y I It = 5 \

t--'O 16x

= 'i

—

II

ll0

wl*

1 +

>u *

t= 2 oU \ t= 6

-1 C\J

33. yes

(35) Solve the first equation for t. x = log t

10 x = t

Substitute into second equation. y = t + 3 y = 10* + 3

In x = log t, x m ust be > 0, so the dom ain restriction is x > 0.

3 7 .y = 10* + 4 39. y = 2 • 10* — 8, x =/= 0 4 1 .x = f + 4 a n d y = 7 + 3f 43. x = t — 2 and y = — 6 + 4f, 0 < f < 4 45. b 47. a

(49) a. The position equations are x = tv0 cos 9 and

y = fw0 sin 9 — ~gt2 + h0. The initial velocity z>0 is

0.75 and 9 is 45°. The surface of the w ater is at y = 0, so the value o f the initial height h 0 is 0.3. By substitution, the position equations are x = t •0.75 cos 45° and y = t • 0.75 sin 45° — 4 .9f2 + 0.3.

b. Find the value of t for w hich y = 0 in the vertical position equation.0 = t • 0.75 sin 45° - 4 .9 f2 + 0.3

0 = ~ t - 4 .9 f2 + 0.3.

The positive zero in the graph of the equation is about 0.3074, so t = 0.3074 seconds. Substitute this for t in x = t • 0.75 cos 45°, the horizontal position equation.

x = 0.3074(0.75) • ^ : 0.16.

This m eans that w hen the frog reaches the surface of the water, he is only about 0.16 m from w here he jum ped. Therefore, he is about 0.5 — 0.16 or 0.34 m eters from the other bank.

C. The 0.4 m eter d istance is a horizontal distance, so substitute the values into the horizontal position equation and solve for u0.

x = tv0 cos 8

0.4 = 0.38 - v0 -^y 1.49 = v,0

51a. about 27.2 ft/s 51 b. about 1 second 51c. about 1.84 seconds 53. Sam ple answ er: x = 4 + f, y = 10 + 3f 55. Sam ple answ er: The horizontal d istance is m odeled by the cosine function, w hich is 0 at 90°. This w ould im ply that the projectile has no horizontal m ovem ent. The corresponding param etric equation w ould b e x = 0.57. Sam ple answ er: Param etric equations show both horizontal and vertical positions of an object over tim e, w hile rectangular equations can only show one or the other.

59.[

I(x ' 2 - - ( kOs = 1

| C— —** X

\

(y - 2)2 (x - 3)261 ■ .4 5

63. 2 tan2 x65. 3 In 3 + In 2; 3.9967. 3 In 2 - In 3; 0.97

= 1

69. horizontal asym p tote at y = vertical asym p tote at x = —6; D = (x | x ~ 6 , x € R)

1; IL16 y

J u

- 8 \ k > 1

71 . vertical asym ptotes at x = —5; oblique asym ptote at y = x + 3; x-intercepts 0 and —8; y-intercept 0; D = (—00, —5) U (—5, 00)

73. 7 75. 23 77. E 79a. 21x2 + 31y2 - 10V 3xy = 4

79b. ellipse 79c. 4(x ')2 + 9 (y ')2 = 1 79d. ~ 0.745

Pages 4 7 3 -4 7 6 Chapter 7 Study Guide and Review

I . conic section 3. directrix 5. foci 7. center 9. param etric

I I . vertex: (—3, —2), focus: (—3 ,1 ) ; axis of sym m etry: x = —3; directrix: y = —5

8 - - 44

+ - y —

*( - 3 , - 2 ) - 4 -

— 8 -

R98 Selected Answers

25- x + 49 = 1

13. vertex: (2, —1), focus: (2, —2); axis of symmetry: x = 2; directrix: y = 0

y

(2, — 11

- f —I 0 ix/ \

/t/ \

r

15. (x - l )2 = —16(i/ - 5)y \(1 5

-1 6 O 16x-4 1

\\3

19. (y + 4)2 = -4 ( x + 3) 21. (x - 3)2 = 4(i/ + 7)y

1 -3,l -- I ' s 0 X

V/

M

y

—8 — 4 X

-8 CO

- 1

23.

- 8 - 4

-1 2

(x - 3 )2 ( ( y + 6 ) 2 _ 116 25

( x - 5 ) 2 , (y - 2)2 ^ ' l 2 5 + 9

27. (x - l ) 2 +(y — 2)2 = 30; circle 29. (x - 2)2 =—4(y + 5); parabola

For Homework Help, go to Hotm ath.com

41. V/

43.

/I

-10,10] scl: 1 by [-1 0 ,1 0 ] scl: 1 [-1 0 ,1 0 ] scl: 1 by [-1 0 ,1 0 ] scl: 1

45.

[ -5 , 5] scl: 1 by [ -5 , 5] scl: 1

47. (xO2 - 2\f3x'y' + 3 (y^2 + (2\/3 - 4 )x ' + (4a/3 + 2)y ' = 20; parabola 49. 9(y ')2 + 4 (x ')2 = 36; ellipse

I yf = 4~ ~-4

12

/f

= - 3 l . = 3,1 1

2t= -iA t =0 2 JX

1-1 " f =II I I t-.= n53.y = i - - 2

8 x

57a. x2 + y2 = 900 57b. 5 seconds 59a. 3x2 + 2\/3xy + y2 + 8x — 8\/3y = 0

59b. *

V\

k /—! — O ix

Pages 478-479 Chapter 7 Connect to AP Calculus

1. 376.99 units3 3.100.53 units3 5. Sample answer: The approximation will get closer to the actual volume of the solid because as the cylinders decrease in height, they will better fill the volume of the solid. 7. 188.5 units3

v 1 s r » i s u.* »

VectorsPage 481 Chapter 8 Get Ready

I .3 ; (—± 4 ) 3. V 29; ( - i - 8 ) 5 .5 .4 7 .4 .0 9.22.8 ft

I I . no solution 13. B « 39°, C as 50°, c a 23.0

connectED.mcgraw-hill.com | R 9 9

Selected Answers and

Solutions