B. The Focus-Directrix Form of the Equation of a Parabola
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Section 8.4 The Analytic Parabola; More on Nonlinear Systems
critical. To understand these and other applications, we use the analytic definition of a
parabola first introduced in Section 8.1.
Definition of a Parabola
Given a fixed point f and fixed line D in the plane,
a parabola is the set of all points (x, y) such that the
distance from f to (x, y) is equal to the distance
from line D to (x, y). The fixed point f is the focus
of the parabola, and the fixed line is the directrix.
d1
(x, y)
f
d2
Vertex
D
d1 ϭ d2
WORTHY OF NOTE
For the analytic parabola, we use p
to designate the focus since c is so
commonly used as the constant
term in y ϭ ax2 ϩ bx ϩ c.
The general equation of a parabola can be
obtained by combining this definition with the distance formula. With no loss of generality, we can
assume the parabola shown in the definition box is
oriented in the plane with the vertex at (0, 0) and
the focus at (0, p). As the diagram in Figure 8.39
indicates, this gives the directrix an equation of
y ϭ Ϫp with all points on D having coordinates of
1x, Ϫp2. Using d1 ϭ d2 the distance formula yields
Figure 8.39
y
1x Ϫ 02 ϩ 1y Ϫ p2 ϭ 1x Ϫ x2 ϩ 1y ϩ p2
2
2
2
2
2
d2
y ϭ Ϫp (0, 0)
D
2
x ϩ y Ϫ 2py ϩ p ϭ 0 ϩ y ϩ 2py ϩ p
2
P(x, y)
F
21x Ϫ 02 2 ϩ 1y Ϫ p2 2 ϭ 21x Ϫ x2 2 ϩ 1y ϩ p2 2
2
d1
(0, p)
2
x Ϫ 2py ϭ 2py
2
(0, Ϫp)
x
(x, Ϫp)
from the definition
square both sides
simplify; expand binomials
subtract p 2 and y 2
x2 ϭ 4py
isolate x 2
The resulting equation is called the focus-directrix form of a vertical parabola
with center at (0, 0). If we had begun by orienting the parabola so it opened to the right,
we would have obtained the equation of a horizontal parabola with center (0, 0):
y2 ϭ 4px.
The Equation of a Parabola in Focus-Directrix Form
Vertical Parabola
Horizontal Parabola
x ϭ 4py
y2 ϭ 4px
2
focus (0, p), directrix: y ϭ Ϫp
If p 7 0, opens upward.
If p 6 0, opens downward.
focus at ( p, 0), directrix: x ϭ Ϫp
If p 7 0, opens to the right.
If p 6 0, opens to the left.
For a parabola, note there is only one second-degree term.
EXAMPLE 3
ᮣ
Locating the Focus and Directrix of a Parabola
Find the vertex, focus, and directrix for the parabola defined by x2 ϭ Ϫ12y. Then
sketch the graph, including the focus and directrix.
Solution
ᮣ
Since the x-term is squared and no shifts have been applied, the graph will be a
vertical parabola with a vertex of (0, 0). Use a direct comparison between the given
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equation and the focus-directrix form to determine the value of p:
x2 ϭ Ϫ12y
T
x2 ϭ 4py
y
10
xϭ0
4p ϭ Ϫ12
p ϭ Ϫ3
yϭ3
(0, 0)
Ϫ10
f
focus-directrix form
This shows:
D
(Ϫ6, Ϫ3)
given equation
10 x
(6, Ϫ3)
(0, Ϫ3)
Ϫ10
Since p ϭ Ϫ3 1p 6 02, the parabola opens downward, with the focus at 10, Ϫ32
and directrix y ϭ 3. To complete the graph we need a few additional points. Since
36 is divisible by 12, we can use inputs of x ϭ 6 and x ϭ Ϫ6 3 1Ϫ62 2 ϭ 62 ϭ 36 4,
giving the points 16, Ϫ32 and 1Ϫ6, Ϫ32. Note the axis of symmetry is x ϭ 0. The
graph is shown.
Now try Exercises 37 through 48
Figure 8.40
y
2p
f
d1
Vertex
y ϭ Ϫp
P
(x, y)
d2 p
x
p
D
EXAMPLE 4
ᮣ
ᮣ
As an alternative to calculating additional points to sketch the graph, we can use what
is called the focal chord of the parabola. Similar to the ellipse and hyperbola, the focal
chord is the line segment that contains the focus, is parallel to the directrix, and has its endpoints on the graph. Using the definition of a parabola and the diagram in Figure 8.40, we
note the vertical distance from (x, y) to the directrix y ϭ Ϫp is 2p. Since d1 ϭ d2 a line
segment parallel to the directrix from the focus to the graph will also have a length of Ϳ2pͿ,
and the focal chord of any parabola has a total length of Ϳ4pͿ. Note that in Example 3, the
points we happened to choose were actually the endpoints of the focal chord.
Finally, if the vertex of a vertical parabola is shifted to (h, k), the equation will have
the form 1x Ϯ h2 2 ϭ 4p1y Ϯ k2. As with the other conic sections, both the horizontal
and vertical shifts are “opposite the sign.”
Locating the Focus and Directrix of a Parabola
Find the vertex, focus, and directrix for the parabola whose equation is given,
then sketch the graph, including the focus, focal chord, and directrix:
x2 Ϫ 6x ϩ 12y Ϫ 15 ϭ 0.
Solution
ᮣ
Since only the x-term is squared, the graph will be a vertical parabola. To find the
end-behavior, vertex, focus, and directrix, we complete the square in x and use a
direct comparison between the shifted form and the focus-directrix form:
x2 Ϫ 6x ϩ 12y Ϫ 15 ϭ 0
x2 Ϫ 6x ϩ ___ ϭ Ϫ12y ϩ 15
x2 Ϫ 6x ϩ 9 ϭ Ϫ12y ϩ 24
1x Ϫ 32 2 ϭ Ϫ121y Ϫ 22
y
10
xϭ3
yϭ5
(3, 2)
(Ϫ3, Ϫ1)
(9, Ϫ1)
Ϫ10
10
(3, Ϫ1)
Ϫ10
x
given equation
complete the square in x
add 9
factor
Notice the parabola has been shifted 3 units right and 2 up, so all features of the
parabola will likewise be shifted. Since we have 4p ϭ Ϫ12 (the coefficient of the
linear term), we know p ϭ Ϫ3 1p 6 02 and the parabola opens downward. If the
parabola were in standard position, the vertex would be at (0, 0), the focus at (0, Ϫ3)
and the directrix a horizontal line at y ϭ 3. But since the parabola is shifted 3 right
and 2 up, we add 3 to all x-values and 2 to all y-values to locate the features of the
shifted parabola. The vertex is at 10 ϩ 3, 0 ϩ 22 ϭ 13, 22. The focus is
10 ϩ 3, Ϫ3 ϩ 22 ϭ 13, Ϫ12 and the directrix is y ϭ 3 ϩ 2 ϭ 5. Finally, the
horizontal distance from the focus to the graph is Ϳ2pͿ ϭ 6 units (since Ϳ4pͿ ϭ 12),
giving us the additional points 1Ϫ3, Ϫ12 and 19, Ϫ12 as endpoints of the focal
chord. The graph is shown.
Now try Exercises 49 through 60
ᮣ
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In many cases, we need to construct the equation of the parabola when only
partial information in known, as illustrated in Example 5.
EXAMPLE 5
ᮣ
Constructing the Equation of a Parabola
Find the equation of the parabola with vertex (4, 4) and focus (1, 4). Then graph
the parabola using the equation and focal chord.
Solution
ᮣ
B. You’ve just seen how
we can identify and use the
focus-directrix form of the
equation of a parabola
As the vertex and focus are on a horizontal line, we
have a horizontal parabola with general equation
1y Ϯ k2 2 ϭ 4p1x Ϯ h2 . The distance p from vertex to
focus is 3 units, and with the focus to the left of the
vertex, the parabola opens left so p ϭ Ϫ3. Using the
focal chord, the vertical distance from (1, 4) to the
graph is Ϳ2pͿ ϭ Ϳ21Ϫ32 Ϳ ϭ 6, giving points (1, 10)
and (1, Ϫ2). The vertex is shifted 4 units right
and 4 units up from (0, 0), showing h ϭ 4 and
k ϭ 4, and the equation of the parabola must be
1y Ϫ 422 ϭ Ϫ121x Ϫ 42 , with directrix x ϭ 7. The graph
is shown.
y
14
Ϫ10
10
x
Ϫ6
Now try Exercises 61 through 76
ᮣ
C. Nonlinear Systems and the Conic Sections
Similar to our work with nonlinear systems in Section 6.3, the graphing, substitution,
or elimination method can still be used when the system involves a conic section.
When both equations in the system have at least one second degree term, it is generally
easier to use the elimination method.
EXAMPLE 6
ᮣ
Solving a System of Nonlinear Equations
Solve the system using elimination: e
Solution
y
5
(Ϫ1, 3)
ᮣ
The first equation represents a vertical and central hyperbola, while the second
represents a horizontal and central ellipse. After writing the system with the x- and
Ϫ5x2 ϩ 2y2 ϭ 13
y-terms in the same order, we obtain e
. Using Ϫ2R1 ϩ R2 will
3x2 ϩ 4y2 ϭ 39
eliminate the y-term.
10x2 Ϫ 4y2 ϭ Ϫ26
3x2 ϩ 4y2 ϭ 39
13x2 ϩ 0 ϭ 13
x2 ϭ 1
x ϭ Ϫ1 or x ϭ 1
e
Ϫ5x2 ϩ 2y2 ϭ 13
(1, 3)
Ϫ5
5
x
3x ϩ 4y2 ϭ 39
(1, Ϫ3)
2
(Ϫ1, Ϫ3)
Ϫ5
2y2 Ϫ 5x2 ϭ 13
3x2 ϩ 4y2 ϭ 39
Ϫ2R1
ϩR2
sum
divide by 13
square root property
Substituting x ϭ 1 and x ϭ Ϫ1 into the second equation we obtain
3112 2 ϩ 4y2 ϭ 39
3 ϩ 4y2 ϭ 39
4y2 ϭ 36
y2 ϭ 9
y ϭ Ϫ3 or y ϭ 3
31Ϫ12 2 ϩ 4y2 ϭ 39
3 ϩ 4y2 ϭ 39
4y2 ϭ 36
y2 ϭ 9
y ϭ Ϫ3 or y ϭ 3
Since Ϫ1 and 1 each generated two outputs, there are a total of four ordered
pair solutions: (1, Ϫ3), (1, 3), (Ϫ1, Ϫ3), and (Ϫ1, 3). The graph is shown and
supports our results.
Now try Exercises 77 through 82
ᮣ
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Nonlinear systems like the one in Example 6 can also be solved by graphing the
system on a graphing calculator and looking for points of intersection. Solving each
equation for y yields the following results.
2y2 Ϫ 5x2 ϭ 13
2y2 ϭ 13 ϩ 5x2
y2 ϭ
13 ϩ 5x2
2
yϭϮ
Y1 ϭ
13 ϩ 5x2
B
2
3x2 ϩ 4y2 ϭ 39
original equation
4y2 ϭ 39 Ϫ 3x2
isolate y-term
y2 ϭ
divide by coefficient
yϭϮ
take square roots
13 ϩ 5X2
13 ϩ 5X2
, Y2 ϭ Ϫ
B
2
B
2
Y3 ϭ
39 Ϫ 3x2
4
39 Ϫ 3x2
B
4
39 Ϫ 3X2
39 Ϫ 3X2
, Y4 ϭ Ϫ
B
4
B
4
The equations and graphs are shown in Figures 8.41 and 8.42, and verify that the
point (1, 3) is a solution to the system. Using symmetry, the other solutions are (Ϫ1, 3),
(Ϫ1, Ϫ3), and (1, Ϫ3). See Exercises 83 through 86.
Figure 8.42
Figure 8.41
6.2
Ϫ9.4
C. You’ve just seen how
we can solve nonlinear
systems involving the conic
sections
9.4
Ϫ6.2
D. Application of the Analytic Parabola
Here is just one of the many ways the analytic definition of a parabola can be applied.
There are several others in the Exercise Set. Many applications use the parabolic property
that light or sound coming in parallel to the axis of a parabola will be reflected to the focus.
EXAMPLE 7
ᮣ
Locating the Focus of a Parabolic Receiver
The diagram shows the cross section of
a radio antenna dish. Engineers have
located a point on the cross section that is
0.75 m above and 6 m to the right of the
vertex. At what coordinates should the
engineers build the focus of the antenna?
Solution
ᮣ
Focus
(6, 0.75)
(0, 0)
By inspection we see this is a vertical parabola with center at (0, 0). This means its
equation must be of the form x2 ϭ 4py. Because we know (6, 0.75) is a point on
this graph, we can substitute (6, 0.75) in this equation and solve for p:
x2 ϭ 4py
162 2 ϭ 4p10.752
36 ϭ 3p
p ϭ 12
equation for vertical parabola, vertex at (0, 0)
substitute 6 for x and 0.75 for y
simplify
result
With p ϭ 12, we see that the focus must be located at (0, 12), or 12 m directly
above the vertex.
Now try Exercises 89 through 96
D. You’ve just seen how
we can solve applications of
the analytic parabola
ᮣ
Note that in many cases, the focus of a parabolic dish may be above the rim of
the dish.