Selected Solutions to Homework # 2
For each quadratic equation below,
(a) identify the curve as a parabola, an ellipse, or a hyperbola;
(b) identify the focus and directrix if it is a parabola, the two focal
points and the semi-major axis if it is an ellipse, or the two focal points and
the asymptotes if it is a hyperbola; and
(c) sketch a graph of the curve.
1. y 2 + x − 1 = 0
This is a parabola. The basic equation that we use to find the focus
and directrix is 4px = y 2 . For this parabola, the focus lies at (p, 0) and
the directrix is the line x = −p. When the equation of the parabola is
not in this exact form, the focus and directrix can be determined by
completing the square (if necessary) and changing coordinate systems.
We will take a similar approach, but rather than changing coordinates,
we will think about how the graph of a curve y = f (x) or x = g(y) is
affected by the addition of contstant terms.
In general, the graph of y − k = f (x − h) is the same as the graph of
y = f (x), except that the former is the latter translated by k units up
and h units to the right.
For functions of y, the change is analagous: the graph of x − h =
g(y − k) is the same as that of x = g(y) except that the former has
been translated k units up and h units to the right.
Now let’s consider the example above: y 2 + x − 1 = 0. Re-write this
as (x − 1) = −y 2 . We can see that this is the same as the parabola
x = −y 2 , except that has been translated by 1 unit to the right.
The focus of x = y 2 is at (1/4, 0) and its directrix is x = −1/4.
Since x = −y 2 is the reflection of x = y 2 in the y-axis, the focus and
directrix are switched: the focus and directrix of x = −y 2 are (−1/4, 0)
and x = 1/4, respectively.
Finally, by considering the effect of translating the graph, we conclude
that the focus and directrix of (x − 1) = −y 2 are (−1/4 + 1, 0) and
x = 1/4 + 1, respectively.
1
2. x2 + 2y 2 − 4 = 0
This is an ellipse. In standard form, it is written as
x2 y 2
+ 2 = 1,
a2
b
√
where in this case a = 2 and b = 2. Therefore, the focal points are
at (−p, 0), and (p, 0), where p2 = a2 − b2 . (Please refer to the power
point slides refered to in homework #2 for more details.)
3. −x2 − 4x + y = 0
This is a parabola. We write it as y = x2 + 4x. Then, complete the
square: y = x2 + 4x + 4 − 4 = (x + 2)2 − 4. Finally, write it as
y − (−4) = (x − (−2))2
This is the parabola y = x2 except that it has been translated 4 units
down and 2 units to the left (or, equivalently, −4 units up and −2
units to the right).
Therefore, the focus lies at (0 − 2, 1/4 − 4) and the directrix is y =
−1/4 − 4.
4. x2 − y 2 − 4 = 0
This is a hyperbola. In standard form, this hyperbola has equation
x2 y 2
− 2 = 1,
a2
b
where a = 2 and b = 2. The focal points lie at (−p, 0) and (p, 0),
where p2 = a2 + b2 . The asymptotes are found by solving for y:
p
y = ± x2 − 4 ≈ ±x for |x| >> 1.
So, the asymptotes are y = x and y = −x. (Please refer to the power
point slides refered to in homework #2 for more details.)
5. x2 + y 2 + 4x + 3 = 0
This is an ellipse. We need to complete the square:
x2 + 4x + 4 + y 2 = −3 + 4.
The above can be factored into
(x − (−2))2 + y 2 = 1.
Thus, the curve is, in fact, a circle of radius 1 centered at (−2, 0).
There is only one focus in the case of a circle: the center is the focus.
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6. −4x2 + y 2 − 1 = 0
This is a hyperbola. In standard form, we have
y2
x2
−
= 1.
12
(1/2)2
The focal points lie at (0, p) and (0, −p), where p2 = 12 + (1/2)2 . The
asymptotes are y = 2x and y = −2x.
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