College Algebra for Calculus
Exam 4 Key
Instructions
1. Do NOT write your answers on these sheets. Nothing written on the test papers will be graded.
2. Please begin each section of questions on a new sheet of paper.
3. Please do not write answers side by side.
4. Please do not staple your test papers together.
5. Limited credit will be given for incomplete or incorrect justification.
6. All answers must be exact. Numeric approximations will be marked incorrect.
Questions
1. Solve
(a) (8) Find the min and max values of P = 4y−3x over the region inside 8y−7x ≤ 6, x−6y ≤ 4, 4y+5x ≤ 54,
and x ≥ 0, y ≥ 0.
Based on the graph in Figure 1 there are four corners to check.
x
=
0.
8y − 7x
=
6.
8y
= 6.
y
=
3/4.
y
=
0.
x − 6y
=
4.
x
=
4.
8y − 7x
=
6.
4y + 5x
=
54.
−8y − 10x
=
−108.
−17x
=
−102.
x
=
6.
8y − 7(6)
=
6.
8y
=
48.
y
=
6.
4y + 5x
=
54.
x − 6y
=
4.
x
=
4 + 6y.
4y + 5(4 + 6y)
=
54.
1
Exam 4: Systems and Conics
2
x
y
P
34y + 20
=
54.
34y
=
34.
y
=
1.
x
=
4 + 6(1)
=
10.
0
0
0
0
3/4
3
4
0
−12
6 10
6
1
6 −36
The min occurs at (10, 1) and the max occurs at (6, 6).
(b) Find all solutions to the system 5x2 − 6y 2 = 1 and 10x2 + 4y 2 = 66.
5x2 − 6y 2
=
1.
−2(5x2 − 6y 2 )
=
−2(1).
2
=
−2.
10x + 4y
2
=
66.
16y
2
=
64.
y2
=
4.
y
=
±2.
2
5x − 6(2)
5x2
=
=
1.
25.
x2
=
5.
√
± 5.
2
−10x + 12y
2
2
x =
√
√
√
The four points of intersection are ( 5, 2), ( 5, −2), (− 5, 2), and (− 5, −2).
√
Exam 4: Systems and Conics
3
2. Identify Conics (8 each)
For each quadratic below determine the type of conic and list its center or vertex, foci, and widths as appropriate.
(a) 9x2 − 4y 2 − 54x − 40y = 55.
9x2 − 4y 2 − 54x − 40y
=
55.
9(x2 − 6x) − 4(y 2 + 10y)
=
55.
2
2
9(x − 6x + 9 − 9) − 4(y + 10y + 25 − 25)
=
55.
2
2
=
55.
2
2
9[x − 3] − 81 − 4[y + 5] + 100
=
55.
9[x − 3]2 − 4[y + 5]2
[x − 3]2
[y + 5]2
−
4
9
c2
=
36.
=
1.
=
4+9
√
13
9([x − 3] − 9) − 4([y + 5] − 25)
c =
Hyperbola
Center: (3,-5)
Vertices: (5,-5),
(1,-5)
√
√
Foci: (5 + 13, −5), (1 − 13, −5)
Box width: 2, height: 3
(b) x2 + 16y 2 + 4x − 96y + 132 = 0.
x2 + 16y 2 + 4x − 96y + 132
=
0.
=
0.
x + 4x + 4 − 4 + 16(y − 6y + 9 − 9) + 132
=
0.
(x + 2)2 − 4 + 16([y − 3]2 − 9) + 132
=
0.
=
0.
=
16.
=
1.
2
2
x + 4x + 16(y − 6y) + 132
2
2
2
2
(x + 2) − 4 + 16[y − 3] − 144 + 132
2
2
(x + 2) + 16[y − 3]
(y − 3)2
(x + 2)2
+
16
1
Ellipse
Center: (-2,3)
Width: 4, Height:
1
√
√
Foci: (−2 − 15, 3) (−2 + 15, 3)
Exam 4: Systems and Conics
4
(c) 56 + 12x + 4y − y 2 = 0.
56 + 12x + 4y − y 2
=
0.
56 + 12x − (y 2 − 4y)
=
0.
=
0.
=
0.
2
56 + 12x − (y − 4y + 4 − 4)
2
56 + 12x − ([y − 2] − 4)
Parabola
Vertex: (-5,2)
Focus: (-2,2)
Directrix: x = −8.
60 + 12x =
[y − 2]2 .
12(x + 5)
=
[y − 2]2 .
4(3)(x + 5)
=
[y − 2]2 .
Exam 4: Systems and Conics
5
3. Constructing Conics (6 each)
Write an equation for the conics described below. Sketch a graph for each.
(a) Parabola with vertex (1, 3) and focus (1, 5).
4(2)(y − 3) = (x − 1)2 .
6
5
4
3
2
1
0
-1
1
2
3
(b) Ellipse with center (−2, 5) and points (−2, 11), (−2, −1), (2, 5), (−6, 5).
(x + 2)2
(y − 5)2
+
= 1.
42
62
10
8
6
4
2
-6
-4
2
-2
(c) Hyperbola with center (6, 1) and points (9, 7) and (6, 4).
(7 − 1)2
(9 − 6)2
−
2
3
a2
9
4− 2
a
9
a2
a2
=
1.
=
1.
=
3.
=
3.
Exam 4: Systems and Conics
6
(x − 6)2
(y − 1)2
−
= 1.
32
3
5
2
4
6
8
10
12
-5
6
4
2
2
-2
4
6
-2
Figure 1: Region for Min and Max
8
10