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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.