TEST-04/MAC1114
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8. (10 pts.)
Find all the complex cube roots of 3
1/2
+ i. Leave
your answer in polar form with the arguments given in degrees.
If z = 3
1/2
+ i, then we can write z in polar form as
z = 2[cos(30°) + sin(30°)i]. The three cube roots are
w
0
= 2
1/3
[cos(10°) + sin(10°)i],
w
1
= 2
1/3
[cos(130°) + sin(130°)i], and
w
2
= 2
1/3
[cos(250°) + sin(250°)i]
9. (5 pts.)
Find the vertex, focus, and directrix of the
parabola that has the equation given below.
y
2
- 4y = x + 4.
By performing the usual algegraic magic you can transform
the equation above into the standard form equation
(y - 2)
2
= 4(1/4)(x - (-8)).
Using this, you can easily see that the vertex is (-8,2), the
focus is (-8 + (1/4), 2) = (-31/4, 2), and the directrix is the
line defined by x = -8 - (1/4) = -33/4.
10. (5 pts.)
Find the center, foci, and vertices of the ellipse
that has the equation given below.
4x
2
+ y
2
+ 4y = 0.
Again, by playing the complete-the-square game carefully, you
should obtain the standard form equation
(x - 0)
2
+ (1/4)(y - (-2))
2
= 1.
From this you should have c = (4 - 1)
1/2
. Clearly the center is
(0,-2), the two vertices are (0,0) and (0,-4), and the two foci
are (0,2 + 3
1/2
) and (0,2 - 3
1/2
).
11. (5 pts.)
Find the center, foci, and vertices of the
hyperbola that has the equation given below.
y
2
- 4x
2
-16x - 2y - 19 = 0
Finally, by being very puntilious in your algebraic
prestidigitation, you can obtain the standard form equation
(1/4)(y - 1)
2
- (x - (-2))
2
= 1.
The center is (-2,1), the two vertices are (-2,3) and (-2,-1),
and since c = (4 + 1)
1/2
, the two foci are (-2, 1 + 5
1/2
) and
(-2, 1 - 5
1/2
).