1. No category

In Exercises 1-4,find the inclination O(in radians and
,f
i degrees) of the line with the given characteristics.
f. 1. Passes through the points (- 1, 2) and (2, 5)
21. Vertices: (-3, 0), (7, 0); loci: (0, 0), (4, 0)
2, Passes through the points (3, 4) and (-2, 7)
3. Equation: y = 2x + 4
22. Vertices: (2, 0), (2, 4); foci: (2, 1), (2, 3)
'
23. Vertices: (0, 1), (4, 1); endpoints of the minor axis:
4. Equation: 6x - 7y - 5 = 0
7
In Exercises 21-24, find the standard form of the
equation of the ellipse with the given characteristics. Then
graph the ellipse.
(2, 0), (2, 2)
In Exercises 5-8, find the angle 0 (in radians and degrees)
;: between the lines.
5. 4x+y-
;!
1
2
6. -5x+ 3y=3
-5x + y = -1
24. Vertices: (-4,-1), (-4, 11); endpoints of the minor
axis: (-6, s), (-2, s)
25. Architecture A semielliptical archway is to be formed
over the entrance to an estate. The arch is to be set on pil-
-2x + 3y = 1
lars that are 10 feet apart and is to have a height (atop the
pillars) of 4 feetl Where should the foci be placed in order
to sketch the arch?
3-
;I
2-
26. Wading Pool You are building a wading pool that is in
-.b.............ÿ_ X
,/3
I
3
-1
I ÿ-x
1 2
-1
the shape of an ellipse. Your plans give an equation for the
elliptical shape of the pool measured in feet as
x2
y2
32---7 + 1-g = 1.
iÿ'ÿ 7, Zv-7y=8
8. 0.02x + 0.07y = 0.18
iiÿi 0,4x+Y =0
0.09x - 0.04y = 0,17
?;=
Find the longest distance across the pool, the shortest
distance, and the distance between the foci.
!i.ÿ: In Exercises 9 and 10, find the distance between the point
In Exercises 27-30, find the center, vertices, foal, and
::' and the line.
eccentricity of the ellipse.
iÿ:: Point
Line
:ÿ 9. (1, 2)
x-y-3=O
27. (x + 2)2 (y - 1)2
+
=1
81
100
! lO, (0, 4)
x+2y-2=O
28. ('ÿ - s)-----2 + (Y + 3)ÿ _ 1
K-
1
J,
In Exercises 11 and 12, state what type of conic
is formed by the intersection of the plane and the
double-napped cone.
11.
12.
36
29. 16x2 + 9y2 - 321: + 72y + 16 = 0
30. 4x2 + 25y2 + 16x -- 150y + 141 = 0
ln Exercises 31-34, find the standard form of the
equation of the hyperbola with the given characteristics.
31. Vertices: (0,+1); foci: (0,+3)
32. Vertices: (2, 2), (-2, 2); loci: (4, 2), (-4, 2)
33. Foci: (0, 0), (8, 0); asymptotes: y = +2(x - 4)
21ÿ In Exercises 13-16, find the standard form of the equation
34. Foci: (3, +2); asymptotes: y =-+2(x- 3)
:: 0f the parabola with the given characteristics. Then graph
i' the parabola.
In Exercises 35-38, find the center, vertices, foci, and the
:ÿi 13, Veÿte:c (0, 0)
equations of the asymptotes of the hyperbola, and sketch
its graph using the asymptotes as an aid.
14. Vertex: (2, 0)
:i:: Focus: (4, O)
Focus: (0, 0)
". 18. Vertex: (0, 2)
16. Vertex: (2, 2)
:7
i.
Directrix: x = - 3
Directrix: y = 0
3S. (X -- 3)2 (y .qt_ 5)2
16
36. (y- 1)ÿ x2= 1
4
19. Architecture A parabolic archway is 12 meters high
at the vertex. At a height of 10 meters, the width of the
archway is 8 meters (see figure). How wide is the archway
at ground level?
Y
4
37. 9x2 - 1@= - 18x - 32y - 151 = 0
38. --4X2 "ÿ 259 -- 8X + 150y + 121 = 0
In Exercises 41-44, classify the graph of the equation as a
circle, a parabola, an ellipse, or a hyperbola.
41, 5x2 -2yz + 10x - 4y + 17 = 0
ÿ . ÿ ":',
42. -4y2 + + 3y + 7 '-- 0
+++
\
+,v+ 9
44. 4X2 + 4y2 -- 4X + 8y -- 11 = 0
.=j,Q. tÿ 2.Gÿ
j
/ ÿ%/'
L]
/
:t
/
x
i
........
s
="
/,
i
0 7 i
/ <.,,,ÿ,,
.,., _,.
0-ÿ- -j.
l/
rÿ
\
.... ÿ,i ÿ ./
/'l
.,.ÿ / ;t'
<
,,7' ] <<t
.... <-, ....
...m
,., [.,
C'J
--<&
ÿ
7 <L..o.
? "
<-ÿ,
•
/ 1.
._ _
tj- €i
<7 5' .Ii, ',,/
x'.ÿ T
©
7<
2!
'
;%
\.1"--ÿ __ÿ__ _ÿ ,;ÿ
,\
$
I\
.,/%
,
,
....
/
(j
/ ,<ÿi,ÿ
\.' ÿ:' ¢--ÿ
;
/ ÿ-Lÿ?..ÿ =.e
.II
J
t
!
11. Hyperbola
/
21. Vertices: (-3, 0), (7, 0) ÿ a = 5
(h, k) = (2, O)
-i
13. veÿteÿ: (o, o) = (h; ÿ)ÿ
Foci: (0, 0), (4, O) ÿ c = 2
--d
Focus: (4, 0) ÿ p = 4
b2 = a2 - c2 = 25 - 4
21
' (Y - k)2 = 4p¢ '-- h)
(x - h)___ÿ2 + (y - k)___ÿ2 = 1'
a2
(Y - 0)2 = 4(4)¢ - 0)
y
b2
10'
8'
6-
y2 1
(x-2)2 ÿ---=
y2 = 16x
25
21
"l I I l
-8 -6-4
(
-8,
-10-
23. Vemces: (0, 1), (4, 1) ÿ a = 2, (h, k) = (2, 1)
Endpoints of minor taxis: (2,0), (2, 2) ÿ b = 1
' 'ÿif':i
(x - h)2 (y - k)2
a2
15.Directrix:Vertex: (0,x2)= =_3(h,ÿl:) P " 3" --
+
b2
=1
4
'3
(x-2)2.+ (y- 1)2= 1
(y - k)2 = 4p(x - h)
2'
4
:,ÿ----
(Y - 2)2 -=- 12z
-2 -1
.-1-
'S
y
7
6.
5-
4
'
-4-3-2-1
-2
-3
4
5'
25. 2a=lO ÿ a=5
-"+"+ÿ x
2345
b'=4
\
c2=a2-b2=25- 16='9 ÿ c=
19. Parabola
3
-3.
/!;
1
-+-+-+++
2
-2
/
3,
2-
1
The foci occur 3 feet from the center of the arch
on a line connecting the tops of the pillars.
Opens downward
i.
Vertex: (0, 12)
ix - h)2 = 4p(y - 10
27. (x + 2)ÿ ÿ (y - 1)___.__ÿ = 1
.ÿ = 4p(y - 12)
Solution points: (-+4, 10)
16 = @(10 - 12)
81
i
100
a = l O, b = 9, c = ./-1-9
Center: (-=2,1).
16 = -Sp
-2=p
xa = -8(y =- 12)
To find the x-intercepts, let y = 0.
Vertices: (-2, il) and (-2, -9)
i ooi: (-2,1 _
./g
Eccentricity: e =. 10
x2 = 96 I
x = -+ ÿ9"6 = +4v46
At the base, the archway is 2(4-v"6) = aÿ'-6 meters wide.
1
LJi
16x2 + 9y2 - 32x + 7.2y + 16 = 0
9,
16(x=-2x+ 1).+ 9(3*2+ By+ 16) =-16+ 16 + 144
37.
16(x - 1)2 + 9(y + 4)2 = 144
9.'c2 - 1@2 - 18x - 323; - 151 = 0
9(xz-Zx:+ 1)- 16(y2+2y+ 1) = 151 +9- 16
(x- 1)2+ (y+4)2 1
9
16
9(x - i)2 - 16O, + l)2 = 144
(.ÿ - l)2 _ (y + 1)ÿ = 1
a=4, b=3, c=v/-7
16
9 '
Cenÿer: (1,-4)
a=4, b=3, c=5
Vertices: (1, 0) and (1, - 8)
Center: (1, - 1)
Foci: (1,-4 + ./7)
Vertices:. (5, -1) and (-3, -.1)
./7
Eccentricity: e = 4
Foci: '(6, -1) ancl (-4, -1)
3
Asymptotes: y =-1 _+ ÿ(x- 1)
31. Vertices: (0, +1) ÿ a = 1, (h, k) = (0, 0)
' .3
7
3- 1
y'= Tx - 7 or y = -'ÿx - -ÿ
. Foci: (0,+3) ÿ c = 3
b2=c2-a2=9- 1 =8
':"
(3, - k)2
(x - h)2
aa
-1
b2
-6
6 8
33. Foci: (0, 0), (8, 0) ÿ c = 4, (h, k) = (4, 0)
'Asymptoies: y= +2(x - 4) ÿ -b:= 2, b = 2a
a
b2=ca-a2'==> 4a2= 16-a2 ==>
16 b2 = 64
a2 = T'
5
39. Foci: (_+100, 0) ÿ c = 100
(x_._ÿ, h)2 (y - k)2
a2
b2
(X -- 4)2
y2
16/5 64/5
Center: (0,'0)
=1
4
ÿ =0.0005 ÿ d-2-dr =93 =2a ÿ a=465
186,000 186,000
+.
\
--1
b2 = c2 - a2 = 1002 - 46.52 = 7837.75
5(x -- 4)2 5y2
16
x2
64
y2
2162.25 7837.75
• 83
\,
• 3s. (x - 3)2 (y + 5)2 -1
16
/ 602'
--- 1)
y2 7 7.75ÿ1ÿ.25
ÿ5211.5736
y
4
.. y ÿ 72 miles
a = 4, b = 2, c = v/ÿ = 2./5
I
-2
Center: (3,-5)
Qÿ
,'y
Vertices: (7, -5) and (-1,-5)
Fool: (3 ±
1
Asymptotes: y = -5 _+ ÿ(x - 3)
1
13
1
7
y=-ÿx-T Or y=-ÿ:v-ÿ
jj./
- 1