Math 143
Practice Test 5B
Solutions
1.
Parabola
directrix: y = 11
The vertex is midway between
the focus and the directrix, so
vertex = (0, 0)
V
p is the distance from the vertex
to the focus, so
p = -11
F
Equation: (x – h)2 = 4p(y – k)
x2 = -44y
2.
x2 + y2 = 1
36 25
Center: (0, 0)
V
F
c2 = 36 - 25
c2 = 11
c = 11 = 3.3
Foci: (11, 0) and (-11, 0)
The vertices are the endpoints of the major axis.
Vertices: (6, 0) and (-6, 0)
F
V
3.
Ellipse
center: (-3, 5)
Horizontal major axis with length of 12
Minor axis length of 4
V
Equation:
(x + 3)2 + (y – 5)2 = 1
36
4
V
4. (x – 4)2 = 4(y + 1)
y = (1/4)(x – 4)2 – 1
Vertical parabola
opens up
Vertex: (4, -1)
x y
0 3
2 0
6 0
8 3
4p = 4
p=1
Focus: (4, 0)
Directrix: y = -2
F
V
5.
4x2 + 9y2 + 24x – 36y + 36 = 0
4 = -36 + __
4(x2 + 6x + __)
9 + 9(y2 – 4y + __)
36 + __
36
4(x + 3)2 + 9(x – 2)2 = 36
(x + 3)2 + (x – 2)2 = 1
9
4
Center: (-3, 2)
c2 = 9 – 4 = 5
c = 5
Foci:
(-3 + 5, 2)
(-3 – 5, 2)
F
F
6.
9x2 – 16y2 = 144
x2 – y2 = 1
16
9
Horizontal hyperbola
center: (0, 0)
c2 = 16 + 9 = 25
c=5
Foci: (5, 0) and (-5, 0)
Asymptotes:
y= 3x
4
y= -3x
4
F
F
7.
Hyperbola:
Foci: (0, -4) and (0, 4)
Vertices: (0, -2) and (-, 2)
F
The hyperbola is vertical with
an equation in the form
V
(y – k)2 – (x – h)2 = 1
b2
a2
V
F
center: (0, 0)
b2 = 4
and
c2 = a2 + b2
16 = a2 + 4
a2 = 12
c2 = 16
Equation:
y2 – x2 = 1
4
12
8.
11
8
=
=
11!
3! 8!
=
11(10)(9)8!
3! 8!
165
Remember that I want to see the work on a problem like this.
You can also find the answer by using your calculator.
11 nCr 8 = 165
9.
5
(x + 2y)5
5
5
5
5
5
= 0 (x)5 + 1 (x)4(2y)1 + 2 (x)3(2y)2 + 3 (x)2(2y)3 + 4 (x)(2y)4 + 5 (2y)5
= 1x5 + 5(x4)(2y) + 10(x3)(4y2) + 10(x2)(8y3) + 5(x)(16y4) + 1(32y5)
= x5 + 10x4y + 40x3y2 + 80x2y3 + 80xy4 + 32y5
10. Find the fifth term of (2x – 3)6
5th
term =
6 (2x)2 (-3)4
4
= 15(4x2)(81)
= 4860x2