COLLEGE ALGEBRA II (MATH 010)
FALL 2016
——PRACTICE FOR EXAM II(SOLUTIONS)——
1. Find the focus and directrix of the parabola
y = − 18 x2 , and sketch the graph.
Solution:
x2 = −8y
4p = −8, p = −2
Focus (0, −2) and Directrix y = 2.
4
3
2
1
0
−5 −4 −3 −2 −1 0
−1
1
2
3
4
5
−2
−3
−4
−5
2. Find the focus and directrix of the parabola. Sketch the
graph. y 2 − 12x = 0
Solution:
y 2 = 12x, 4p = 12, p = 3
Focus (3, 0) and Directrix x = −3
1
7
6
5
4
3
2
1
0
−4
−3
−2
−1 0
−1
1
2
3
4
5
−2
−3
−4
−5
−6
−7
3. Find the foci, the vertices, and the lengths of the major
and minor axes of the ellipse, and sketch the graph.
x2 y 2
+
=1
16
9
Solution:
√
√
2
2
2
a = √ 16 = 4, b =
√ 9 = 3, and c = a − b
c = 7, Foci (± 7, 0) and Vertices (±4, 0).
Major Axis Length: 2a = 8 and Minor Axis Length: 2b = 6
4
3
2
1
0
−5 −4 −3 −2 −1 0
−1
1
2
3
4
5
−2
−3
−4
2
4. The vertices of an ellipse are (0, ±4), and the foci are (0, ±3).
Find its equation, and sketch the graph.
Solution:
a = 4, c = 3, and c2 = a2 − b2 or b2 = a2 − c2 = 7
2
y2
=1
Equation: x7 + 16
5
4
3
2
1
0
−4 −3 −2 −1 0
−1
1
2
3
4
−2
−3
−4
−5
5. Find the vertices, foci, and asymptotes of the hyperbola,
and sketch the graph.
x2 − 4y 2 = −4
Solution:
√
y2
x2
Rewrite
as
−
=
1,
a
=
1,
b
=
4 = 2, and c2 = a2 + b2
1
4
√
√
c = 5, Vertices (0, ±1), Foci (0, ± 5), y = ± ab x = ± 12 x
5
4
3
2
1
0
−5 −4 −3 −2 −1 0
−1
1
2
3
4
5
−2
−3
−4
−5
3
6. Find the equation of the hyperbola with vertices (±2, 0)
and foci (±3, 0). Sketch the graph.
Solution:
a =√
2, c = 3, and c2 = a2 + b2
2
2
b = 5, Equation x4 − y5 = 1
6
5
4
3
2
1
0
−5 −4 −3 −2 −1 0
1
2
3
4
5
−2
−3
−4
−5
−6
7. Complete the square to put each equation in standard form.
Identify the type of conic, and its components (vertices,
center, foci, asymptotes, etc.)
(a) 4x2 + 9y 2 − 16x − 36y + 16 = 0
(b) x2 + 8x + 8y = 0
(c) −9x2 + 16y 2 − 72x − 96y − 144 = 0
Solution:
(a) 4x2 + 9y 2 − 16x − 36y + 16 = 0
(4x2 − 16x) + (9y 2 − 36y) = −16
4(x2 − 4x) + 9(y 2 − 4y) = −16
4(x2 − 4x + 4) + 9(y 2 − 4y + 4) = −16 + 4(4) + 9(4)
4(xh − 2)2 + 9(y − 2)2 = 36
i
1
1
2
2
4(x
−
2)
+
9(y
−
2)
= 36
· 36
36
4
(x−2)2
9
2
+ (y−2)
=1
4
Ellipse: Center=
(2, 2), vertices= (2 ± 3, 2),
√
foci= (2 ± 5, 2)
(b) x2 + 8x + 8y = 0
(x2 + 8x) = −8y
(x2 + 8x + 16) = −8y + 16
(x + 4)2 = −8(y − 2)
Parabola: Vertex= (−4, 2), focus= (−4, 0),
directrix: y = 4
(c) −9x2 + 16y 2 − 72x − 96y − 144 = 0
(−9x2 − 72x) + (16y 2 − 96y) = 144
−9(x2 + 8x) + 16(y 2 − 6y) = 144
−9(x2 + 8x + 16) + 16(y 2 − 6y + 9) = 144 − 9(16) + 16(9)
16(yh − 3)2 − 9(x + 4)2 = 144
i
1
1
2
2
· 144
16(y
−
3)
−
9(x
+
4)
= 144
144
2
2
(y−3)
− (x+4)
=1
9
16
Hyperbola: Center= (−4, 3),
Vertices= (−4, 3 ± 3), foci= (−4, 3 ± 5),
asymptotes: y − 3 = ± 34 (x + 4)
5
8. Use the discriminate to identify the type of each rotated
conic.
(a) 13x2 + 10xy + 13y 2 − 72 = 0
√
(b) x2 − 6 3xy − 5y 2 − 8 = 0
√
√
(c) x2 + 2 3xy + 3y 2 − 8 3x − 8y − 4 = 0
Solution:
(a) 13x2 + 10xy + 13y 2 − 72 = 0
A = 13, B = 10, and C = 13
B 2 − 4AC = 100 − 4(13)(13) = 100 − 676 = −576 < 0
Ellipse
√
2
(b) x2 − 6 3xy − 5y
√ −8=0
A = 1, B = −6 3, and C = −5
B 2 − 4AC = 36(3) − 4(1)(−5) = 108 + 20 = 128 > 0
Hyperbola
√
√
(c) x2 + 2 3xy +√3y 2 − 8 3x − 8y − 4 = 0
A = 1, B = 2 3, and C = 3
B 2 − 4AC = 4(3) − 4(1)(3) = 12 − 12 = 0
Parabola
6