COLLEGE ALGEBRA II (MATH 010)
SPRING 2015
HOMEWORK-1A (SOLUTIONS)
1. Graph each equation. Identify the focus and directrix. (a) y 2 = 4x
(b) x2 = −4y
Solution:
(a) 4p = 4, p = 1, F= (1, 0) and directrix: x = −1
5
x=-1
4
3
2
1
(1,0)
0
−5 −4 −3 −2 −1 0
−1
1
2
3
4
5
−2
−3
−4
−5
(b) 4p = −4, p = −1, F= (0, −1) and y = 1
5
4
3
2
y=1
1
0
−5 −4 −3 −2 −1 0
−1
−2
1
2
3
4
5
(0,-1)
−3
−4
−5
1
2. Find the equation of the parabola with vertex at the
origin and the following properties.
(a) Directrix y = −3
(b) Focus (2, 0)
Solution:
(a) Since the directrix is y = −3, a = 3 and 4a = 12.
The parabola opens up. The equation is x2 = 12y.
(b) Since the focus= (2, 0), a = 2 and the parabola
opens right. Since 4a = 8, the equation is y 2 = 8x.
3. Find the equation of the parabola with vertex at
the origin and the indicated axis of symmetry that
contains the given point.
(a) y-axis; (4, 2)
(b) x-axis; (−3, 6)
Solution:
(a) Opens Up
(b) Opens Left
2
x = 4ay
y 2 = 4ax
(x, y) = (4, 2)
(x, y) = (−3, 6)
2
4 = (4)(a)(2)
62 = (4)(a)(−3)
16 = 8a
36 = −12a
a=2
a = −3
2
x = 8y
y 2 = −12x
2
4. Graph each ellipse. Identify the foci and lengths of
the major and minor axes.
(a) 2x2 + y 2 = 12
(b) 4x2 + 7y 2 = 28
Solution:
(a)
y2
x2
2
+
12, b2 = 6
6
12 = 1, a = √
√
c2 = a2 − b2 ,√c = 6, foci= (0,√± 6)
|major| = 2 12, |minor| = 2 6
4
3
2
1
0
−5 −4 −3 −2 −1 0
−1
1
2
3
4
5
−2
−3
−4
(b)
y2
x2
2
+
7, b2 = 4
7
4 = 1, a = √
√
c2 = a2 − b2 ,√c = 3, foci= (± 3, 0)
|major| = 2 7, |minor| = 4
4
3
2
1
0
−5 −4 −3 −2 −1 0
−1
1
2
3
4
5
−2
−3
−4
3
5. Find an equation for each ellipse with center at the
origin, and possessing the following properties.
(a) Major axis on x axis
(b) Major axis on y axis
Major axis length = 10
Minor axis length = 20
√
Minor axis length = 6
Distance of foci from center = 70
Solution:
2
2
2
2
(a) xa2 + yb2 = 1 (b) xb2 + ay2 = 1
2a = 10
2b = 20
a=5
b = 10
√
2b = 6
c = 70
b=3
a2 = b2 + c2 = 170
y2
y2
x2
x2
+
=
1
+
25
9
100
170 = 1
6. Sketch the graph of the hyperbola. Identify the foci
and asymptotes.
2
2
2
2
(b) y4 − x9 = 1
(a) x9 − y4 = 1
Solution:
(a)
√
2
2
2
a = 3, b =
2,
c
=
a
+
b
,
c
=
13
√
2
foci= (± 13, 0), asymp: y = ± 3 x
7
6
5
4
3
2
1
0
−7 −6 −5 −4 −3 −2−1
−1 0
1
2
3
4
5
6
7
−2
−3
−4
−5
−6
−7
4
(b)
√
a = 2, b = √
3, c2 = a2 + b2 , c = 13
foci= (0, ± 13), asymp: y = ± 23 x
7
6
5
4
3
2
1
0
−7 −6 −5 −4 −3 −2−1
−1 0
1
2
3
4
5
6
7
−2
−3
−4
−5
−6
−7
7. Complete the square to put each equation in standard form. Identify the type of conic, and its components (vertices, center, foci, 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
2
2
(x−2)
+ (y−2)
=1
9
4
Ellipse: Center=
(2, 2), vertices= (±3, 0),
√
foci= (± 5, 0)
5
(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 = ± 43 (x + 4)
6