# Parabola: Standard Form of the Equation of a Parabola

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This pdf includes the following topics:-
The Parabola
Definition of Parabola
Standard Form of the Equation of a Parabola
Using the Standard Form of the Equation of a Parabola
The Latus Rectum and Graphing Parabolas
Examples
1. P-BLTZMC09_873-950-hr 21-11-2008 13:28 Page 900
900 Chapter 9 Conic Sections and Analytic Geometry
79. Write 4x2 - 6xy + 2y2 - 3x + 10y - 6 = 0 as a quadratic In Exercises 85–88, determine whether each statement is true or false.
equation in y and then use the quadratic formula to express y If the statement is false, make the necessary change(s) to produce a
in terms of x. Graph the resulting two equations using a true statement.
graphing utility in a 3 - 50, 70, 104 by 3- 30, 50, 104 viewing 85. If one branch of a hyperbola is removed from a graph, then
rectangle. What effect does the xy-term have on the graph of the branch that remains must define y as a function of x.
the resulting hyperbola? What problems would you 86. All points on the asymptotes of a hyperbola also satisfy the
encounter if you attempted to write the given equation in hyperbola’s equation.
standard form by completing the square? x2 y2 2
87. The graph of - = 1 does not intersect the line y = - x.
x2 y2 xƒxƒ yƒyƒ 9 4 3
80. Graph - = 1 and - = 1 in the same viewing 88. Two different hyperbolas can never share the same asymptotes.
16 9 16 9
rectangle. Explain why the graphs are not the same. x2 y2
89. What happens to the shape of the graph of 2 - 2 = 1 as
c a b
: q , where c2 = a2 + b2?
a
Critical Thinking Exercises 90. Find the standard form of the equation of the hyperbola with
Make Sense? In Exercises 81–84, determine whether each vertices 15, -62 and (5, 6), passing through (0, 9).
statement makes sense or does not make sense, and explain your 91. Find the equation of a hyperbola whose asymptotes are
reasoning. perpendicular.
81. I changed the addition in an ellipse’s equation to subtraction
and this changed its elongation from horizontal to vertical.
Preview Exercises
82. I noticed that the definition of a hyperbola closely resembles
the next section.
that of an ellipse in that it depends on the distances between
a set of points in a plane to two fixed points, the foci. In Exercises 92–93, graph each parabola with the given equation.
92. y = x2 + 4x - 5 93. y = - 31x - 122 + 2
83. I graphed a hyperbola centered at the origin that had
y-intercepts, but no x-intercepts. 94. Isolate the terms involving y on the left side of the equation:
84. I graphed a hyperbola centered at the origin that was y2 + 2y + 12x - 23 = 0.
symmetric with respect to the x-axis and also symmetric Then write the equation in an equivalent form by completing
with respect to the y-axis. the square on the left side.
Section 9.3 The Parabola
Objectives At first glance, this image looks like columns of smoke rising from a
��� Graph parabolas with vertices fire into a starry sky. Those are, indeed, stars in the background, but
you are not looking at ordinary smoke columns. These stand almost
at the origin. 6 trillion miles high and are 7000 light-years from Earth—more
��� Write equations of parabolas than 400 million times as far away as the sun.
in standard form.
��� Graph parabolas with vertices
not at the origin.
T his NASA photograph is one of a series of
stunning images captured from the ends of
the universe by the Hubble Space
��� Solve applied problems Telescope. The image shows infant
involving parabolas. star systems the size of our solar
system emerging from the gas and
dust that shrouded their creation.
Using a parabolic mirror that is 94.5
inches in diameter, the Hubble has
provided answers to many of the
profound mysteries of the cosmos:
How big and how old is the
universe? How did the galaxies
come to exist? Do other Earth-like planets orbit other sun-like stars? In this section, we
study parabolas and their applications, including parabolic shapes that gather distant
rays of light and focus them into spectacular images.
Definition of a Parabola
In Chapter 2, we studied parabolas, viewing them as graphs of quadratic functions in
the form
y = a1x - h22 + k or y = ax2 + bx + c.
2. P-BLTZMC09_873-950-hr 21-11-2008 13:28 Page 901
Section 9.3 The Parabola 901
Study Tip
Here is a summary of what you should already know about graphing parabolas.
Graphing y ⴝ a(x ⴚ h)2 ⴙ k and y ⴝ ax 2 ⴙ bx ⴙ c
1. If a 7 0, the graph opens upward. If a 6 0, the graph opens downward.
2. The vertex of y = a1x - h22 + k is 1h, k2.
y y
x=h y = a(x − h)2 + k (h, k)
a<0
x x
y = a(x − h)2 + k
a>0
(h, k) x=h
b
3. The x-coordinate of the vertex of y = ax2 + bx + c is x = - .
2a
Parabola Parabolas can be given a geometric definition that enables us to include
graphs that open to the left or to the right, as well as those that open obliquely. The
Directrix
definitions of ellipses and hyperbolas involved two fixed points, the foci. By
Axis of contrast, the definition of a parabola is based on one point and a line.
Focus symmetry
Vertex
Definition of a Parabola
A parabola is the set of all points in a plane that are equidistant from a fixed
line, the directrix, and a fixed point, the focus, that is not on the line
(see Figure 9.29).
Figure 9.29 In Figure 9.29, find the line passing through the focus and perpendicular to the
directrix. This is the axis of symmetry of the parabola. The point of intersection of
the parabola with its axis of symmetry is called the vertex. Notice that the vertex is
midway between the focus and the directrix.
Standard Form of the Equation of a Parabola
The rectangular coordinate system enables us y
to translate a parabola’s geometric definition d1
M(−p, y)
into an algebraic equation. Figure 9.30 is our P(x, y)
starting point for obtaining an equation. We
d2
place the focus on the x-axis at the point
1p, 02. The directrix has an equation given by
x
x = - p. The vertex, located midway between
the focus and the directrix, is at the origin. Focus (p, 0)
Directrix: x = −p
What does the definition of a parabola
tell us about the point 1x, y2 in Figure 9.30?
For any point 1x, y2 on the parabola, the
distance d1 to the directrix is equal to the
distance d2 to the focus. Thus, the point 1x, y2 Figure 9.30
is on the parabola if and only if
d1 = d2 .
41x + p2 + 1y - y2 = 41x - p2 + 1y - 02
2 2 2 2
Use the distance formula.
1x + p22 = 1x - p22 + y2 Square both sides of the
equation.
3. P-BLTZMC09_873-950-hr 21-11-2008 13:28 Page 902
902 Chapter 9 Conic Sections and Analytic Geometry
x2 + 2px + p2 = x2 - 2px + p2 + y2 Square x + p and x - p.
2px = - 2px + y2 Subtract x 2 + p 2 from both
sides of the equation.
y2 = 4px Solve for y 2.
This last equation is called the standard form of the equation of a parabola with its
vertex at the origin. There are two such equations, one for a focus on the x-axis and
one for a focus on the y-axis.
Standard Forms of the Equations of a Parabola
The standard form of the equation of a parabola with vertex at the origin is
y 2 = 4px or x2 = 4py.
Figure 9.31(a) illustrates that for the equation on the left, the focus is on the
x-axis, which is the axis of symmetry. Figure 9.31(b) illustrates that for the
equation on the right, the focus is on the y-axis, which is the axis of symmetry.
y y
x2 = 4py
y2 = 4px Focus (0, p)
Vertex
Vertex
x x
Study Tip
Focus (p, 0)
It is helpful to think of p as the Directrix: x = −p
directed distance from the vertex to Directrix: y = −p
the focus. If p 7 0, the focus lies p
units to the right of the vertex or p
units above the vertex. If p 6 0, the
focus lies ƒ p ƒ units to the left of the Figure 9.31(a) Parabola with the x-axis as the Figure 9.31(b) Parabola with the y-axis
axis of symmetry. If p 7 0, the graph opens to the as the axis of symmetry. If p 7 0, the graph
vertex or ƒ p ƒ units below the vertex.
right. If p 6 0, the graph opens to the left. opens upward. If p 6 0, the graph opens
downward.
��� Graph parabolas with vertices at Using the Standard Form of the Equation of a Parabola
the origin.
We can use the standard form of the equation of a parabola to find its focus and
directrix. Observing the graph’s symmetry from its equation is helpful in locating
the focus.
y2=4px x2=4py
The equation does not change if The equation does not change if
y is replaced with −y. There is x is replaced with −x. There is
x-axis symmetry and the focus is y-axis symmetry and the focus is
on the x-axis at (p, 0). on the y-axis at (0, p).
Although the definition of a parabola is given in terms of its focus and its
directrix, the focus and directrix are not part of the graph. The vertex, located at the
origin, is a point on the graph of y2 = 4px and x2 = 4py. Example 1 illustrates how
you can find two additional points on the parabola.
EXAMPLE 1 Finding the Focus and Directrix of a Parabola
Find the focus and directrix of the parabola given by y 2 = 12x. Then graph the
parabola.
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Section 9.3 The Parabola 903
y
Solution The given equation, y2 = 12x, is in the standard form y2 = 4px,
7 so 4p = 12.
6
(3, 6) No change if y is
Directrix: x = −3 5 replaced with −y.
4 The parabola has
y2=12x
3 x-axis symmetry.
2 Focus (3, 0) This is 4p.
1
x
−5 −4 −2 −1−1 1 2 3 4 5 We can find both the focus and the directrix by finding p.
−2 Vertex (0, 0) 4p = 12
−3
−4 p = 3 Divide both sides by 4.
−5
(3, −6) Because p is positive, the parabola, with its x-axis symmetry, opens to the right. The
−6
−7 focus is 3 units to the right of the vertex, (0, 0).
Figure 9.32 The graph of y2 = 12x Focus: 1p, 02 = 13, 02
Directrix: x = - p; x = - 3
Technology The focus, (3, 0), and directrix, x = - 3, are shown in Figure 9.32.
We graph y2 = 12x with a graphing To graph the parabola, we will use two points on the graph that lie directly
utility by first solving for y. The above and below the focus. Because the focus is at (3, 0), substitute 3 for x in the
screen shows the graphs of parabola’s equation, y 2 = 12x.
y = 212x and y = - 212x. The
graph fails the vertical line test. y2 = 12 # 3 Replace x with 3 in y 2 = 12x.
Because y2 = 12x is not a function, y2 = 36 Simplify.
you were not familiar with this form
y = ; 236 = ; 6 Apply the square root property.
of the parabola’s equation in
Chapter 2. The points on the parabola above and below the focus are (3, 6) and 13, - 62. The
graph is sketched in Figure 9.32.
y1 = 兹12x
Check Point 1
Find the focus and directrix of the parabola given by y2 = 8x.
Then graph the parabola.
In general, the points on a parabola y2 = 4px that lie above and below the focus,
y2 = −兹12x
1p, 02, are each at a distance ƒ 2p ƒ from the focus. This is because if x = p, then
y2 = 4px = 4p2, so y = ; 2p. The line segment joining these two points is called the
latus rectum; its length is ƒ 4p ƒ .
[−6, 6, 1] by [−8, 8, 1]
The Latus Rectum and Graphing Parabolas
The latus rectum of a parabola is a line segment that passes through its focus, is
parallel to its directrix, and has its endpoints on the parabola. Figure 9.33
shows that the length of the latus rectum for the graphs of y2 = 4px and
x2 = 4py is ƒ 4p ƒ .
y y
x2 = 4py
Focus (0, p)
y2 = 4px
Latus rectum
length: ⎥ 4p⎥
x x
Focus (p, 0)
Directrix: x = −p Latus rectum
Directrix: y = −p
length: ⎥ 4p⎥
Figure 9.33 Endpoints of the latus rectum are helpful in determining a parabola’s “width,” or how
it opens.
5. P-BLTZMC09_873-950-hr 21-11-2008 13:28 Page 904
904 Chapter 9 Conic Sections and Analytic Geometry
y EXAMPLE 2 Finding the Focus and Directrix of a Parabola
5
4 Find the focus and directrix of the parabola given by x2 = - 8y. Then graph the
Directrix: y = 2
3 parabola.
Vertex (0, 0) 1 Latus Rectum Solution The given equation, x2 = - 8y, is in the standard form x2 = 4py, so
x 4p = - 8.
−5 −4 −3 −2 −1−1 1 2 3 4 5
(−4, −2) (4, −2) No change if x is
−2 replaced with −x.
−3 The parabola has
x2=–8y
−4 Focus (0, −2) y-axis symmetry.
−5 This is 4p.
Figure 9.34 The graph of We can find both the focus and the directrix by finding p.
x2 = - 8y
4p = - 8
p = -2 Divide both sides by 4.
Technology
Because p is negative, the parabola, with its y-axis symmetry, opens downward. The
2
Graph x = - 8y by first solving for focus is 2 units below the vertex, (0, 0).
x2
y: y = - . The graph passes the Focus: 10, p2 = 10, - 22
8
Directrix: y = - p; y = 2
vertical line test. Because x 2 = - 8y
is a function, you were familiar with The focus and directrix are shown in Figure 9.34.
the parabola’s alternate algebraic To graph the parabola, we will use the vertex, (0, 0), and the two endpoints of
1 the latus rectum. The length of the latus rectum is
form, y = - x2, in Chapter 2.
8
The form is y = ax2 + bx + c, with ƒ 4p ƒ = ƒ 41-22 ƒ = ƒ - 8 ƒ = 8.
1
a = - , b = 0, and c = 0.
8 Because the graph has y-axis symmetry, the latus rectum extends 4 units to the left and
4 units to the right of the focus, 10, -22. The endpoints of the latus rectum are
1- 4, -22 and 14, -22. Passing a smooth curve through the vertex and these two
points, we sketch the parabola, shown in Figure 9.34.
Check Point 2
Find the focus and directrix of the parabola given by
x2 = - 12y. Then graph the parabola.
[−6, 6, 1] by [−6, 6, 1] In Examples 1 and 2, we used the equation of a parabola to find its focus and
directrix. In the next example, we reverse this procedure.
��� Write equations of parabolas in EXAMPLE 3 Finding the Equation of a Parabola from Its Focus
standard form. and Directrix
y
Find the standard form of the equation of a parabola
with focus (5, 0) and directrix x = - 5, shown in 7
6
Figure 9.35.
Directrix: 5
Solution The focus is (5, 0). Thus, the focus is x = −5 4
3
on the x-axis. We use the standard form of the
2 Focus (5, 0)
equation in which there is x-axis symmetry, namely 1
y2 = 4px. x
−4 −3 −2 −1−1 1 2 3 4 5
We need to determine the value of p.
Figure 9.35 shows that the focus is 5 units to the −2
−3
right of the vertex, (0, 0). Thus, p is positive and −4
p = 5. We substitute 5 for p in y 2 = 4px to obtain −5
the standard form of the equation of the parabola. −6
The equation is −7
y2 = 4 # 5x or y2 = 20x. Figure 9.35
6. P-BLTZMC09_873-950-hr 21-11-2008 13:28 Page 905
Section 9.3 The Parabola 905
Check Point 3
Find the standard form of the equation of a parabola with focus
(8, 0) and directrix x = - 8.
��� Graph parabolas with vertices Translations of Parabolas
The graph of a parabola can have its vertex at 1h, k2, rather than at the origin.
not at the origin.
Horizontal and vertical translations are accomplished by replacing x with x - h and
y with y - k in the standard form of the parabola’s equation.
Table 9.3 gives the standard forms of equations of parabolas with vertex at
1h, k2. Figure 9.36 shows their graphs.
Table 9.3 Standard Forms of Equations of Parabolas with Vertex at (h, k)
Equation Vertex Axis of Symmetry Focus Directrix Description
1y - k22 = 4p1x - h2 1h, k2 1h + p, k2
If p 7 0, opens to the right.
Horizontal x = h - p If p 6 0, opens to the left.
1x - h22 = 4p1y - k2 1h, k2 1h, k + p2
If p 7 0, opens upward.
Vertical y = k - p If p 6 0, opens downward.
y y
(y − k)2 = 4p(x − h)
(x − h)2 = 4p(y − k)
Focus
Directrix: x = h − p (h, k + p)
Study Tip
If y is the squared term, there is Vertex (h, k)
horizontal symmetry and the Vertex (h, k)
parabola’s equation is not a function. x x
If x is the squared term, there is Focus (h + p, k)
vertical symmetry and the parabola’s
equation is a function. Continue to Directrix: y = k − p
think of p as the directed distance
from the vertex, 1h, k2, to the focus.
Figure 9.36 Graphs of parabolas with vertex at 1h, k2 and p 7 0
The two parabolas shown in Figure 9.36 illustrate standard forms of equations
for p 7 0. If p 6 0, a parabola with a horizontal axis of symmetry will open to the left
and the focus will lie to the left of the directrix. If p 6 0, a parabola with a vertical axis
of symmetry will open downward and the focus will lie below the directrix.
EXAMPLE 4 Graphing a Parabola with Vertex at 1h, k2
Find the vertex, focus, and directrix of the parabola given by
1x - 322 = 81y + 12.
Then graph the parabola.
Solution In order to find the focus and directrix, we need to know the vertex. In
the standard forms of equations with vertex at 1h, k2, h is the number subtracted
from x and k is the number subtracted from y.
(x-3)2=8(y-(–1))
This is (x − h)2, This is y − k,
with h = 3. with k = −1.
We see that h = 3 and k = - 1. Thus, the vertex of the parabola is 1h, k2 = 13, -12.
7. P-BLTZMC09_873-950-hr 21-11-2008 13:28 Page 906
906 Chapter 9 Conic Sections and Analytic Geometry
y Now that we have the vertex, 13, - 12, we can find both the focus and directrix
5 by finding p.
The equation is in the standard form 1x - h22 = 4p1y - k2.
4 Latus Rectum
3 (x-3)2=8(y+1)
2 Focus (3, 1) Because x is the squared term, there is vertical symmetry and
(−1, 1) (7, 1) This is 4p. the parabola’s equation is a function.
x
−2 −1
−1 1 2 4 5 6 7 8
Because 4p = 8, p = 2. Based on the standard form of the equation, the axis of
−2 Vertex (3, −1) symmetry is vertical. With a positive value for p and a vertical axis of symmetry, the
−4 Directrix: y = −3
parabola opens upward. Because p = 2, the focus is located 2 units above the
−5 vertex, 13, - 12. Likewise, the directrix is located 2 units below the vertex.
Figure 9.37 The graph of Focus: (h, k+p)=(3, –1+2)=(3, 1)
1x - 32 = 81y + 12
2
The vertex, (h, k), The focus is 2 units
is (3, −1). above the vertex, (3, −1).
Technology Directrix: y=k-p
Graph 1x - 32 = 81y + 12 by first
2 y=–1-2=–3
solving for y:
8 1x
1
The directrix is 2 units
- 322 = y + 1 below the vertex, (3, −1).
y = 181x - 322 - 1. Thus, the focus is (3, 1) and the directrix is y = - 3. They are shown in Figure 9.37.
To graph the parabola, we will use the vertex, 13, -12, and the two endpoints of the
The graph passes the vertical line test. latus rectum. The length of the latus rectum is
Because 1x - 322 = 81y + 12 is a
function, you were familiar with the ƒ 4p ƒ = ƒ 4 # 2 ƒ = ƒ 8 ƒ = 8.
parabola’s alternate algebraic form,
y = 181x - 322 - 1, in Chapter 2. The Because the graph has vertical symmetry, the latus rectum extends 4 units to the left
2 and 4 units to the right of the focus, (3, 1). The endpoints of the latus rectum are
13 - 4, 12, or 1- 1, 12, and 13 + 4, 12, or (7, 1). Passing a smooth curve through the
form is y = a1x - h2 + k with
1
a = 8 , h = 3, and k = - 1.
vertex and these two points, we sketch the parabola, shown in Figure 9.37.
Check Point 4
Find the vertex, focus, and directrix of the parabola given by
1x - 222 = 41y + 12. Then graph the parabola.
In some cases, we need to convert the equation of a parabola to standard form
by completing the square on x or y, whichever variable is squared. Let’s see how this
is done.
[−3, 9, 1] by [−6, 6, 1]
EXAMPLE 5 Graphing a Parabola with Vertex at 1h, k2
Find the vertex, focus, and directrix of the parabola given by
y2 + 2y + 12x - 23 = 0.
Then graph the parabola.
Solution We convert the given equation to standard form by completing the
square on the variable y. We isolate the terms involving y on the left side.
y2 + 2y + 12x - 23 = 0 This is the given equation.
y2 + 2y = - 12x + 23 Isolate the terms involving y.
y2 + 2y + 1 = - 12x + 23 + 1 Complete the square by adding the
square of half the coefficient of y.
1y + 122 = - 12x + 24 Factor.
To express the equation 1y + 122 = - 12x + 24 in the standard form
1y - k22 = 4p1x - h2, we factor out - 12 on the right. The standard form of the
parabola’s equation is
1y + 122 = - 121x - 22.
8. P-BLTZMC09_873-950-hr 21-11-2008 13:28 Page 907
Section 9.3 The Parabola 907
Directrix: x = 5 We use 1y + 122 = - 12(x - 2) to identify the vertex, 1h, k2, and the value for p
y
needed to locate the focus and the directrix.
5
(−1, 5)
4 [y-(–1)]2=–12(x-2) The equation is in the standard form
1y - k22 = 4p1x - h2. Because y is the squared
Latus
3
Rectum
2 This is (y − k)2, This is This is x − h, term, there is horizontal symmetry and the parabola’s
1 with k = −1. 4p. with h = 2.
x equation is not a function.
−3 −2 1 3 4 6 7
−2 We see that h = 2 and k = - 1. Thus, the vertex of the parabola is 1h, k2 = 12, -12.
Focus −3 Because 4p = - 12, p = - 3. Based on the standard form of the equation, the axis of
(−1, −1) −4 Vertex
(2, −1) symmetry is horizontal. With a negative value for p and a horizontal axis of symmetry,
−5
the parabola opens to the left. Because p = - 3, the focus is located 3 units to the left
of the vertex, 12, - 12. Likewise, the directrix is located 3 units to the right of the
−6
(−1, −7)
−7
vertex.
Figure 9.38 The graph of Focus:
y2 + 2y + 12x - 23 = 0, or (h+p, k)=(2+(–3), –1)=(–1, –1)
1y + 122 = - 121x - 22
The vertex, (h, k), The focus is 3 units
is (2, −1). to the left of the
Technology vertex, (2, −1).
Graph y 2 + 2y + 12x - 23 = 0 by
solving the equation for y. Directrix: x=h-p
x=2-(–3)=5
y2+2y+(12x-23)=0
a=1 b=2 c = 12x − 23 The directrix is 3 units
to the right of the
Use the quadratic formula to solve for vertex, (2, −1).
Thus, the focus is 1-1, - 12 and the directrix is x = 5. They are shown in Figure 9.38.
y and enter the resulting equations.
-2 + 44 - 4112x - 232 To graph the parabola, we will use the vertex, 12, -12, and the two endpoints
y1 = of the latus rectum. The length of the latus rectum is
2
-2 - 44 - 4112x - 232 ƒ 4p ƒ = ƒ 41- 32 ƒ = ƒ -12 ƒ = 12.
y2 =
2 Because the graph has horizontal symmetry, the latus rectum extends 6 units above
and 6 units below the focus, 1- 1, -12. The endpoints of the latus rectum are
y1 1- 1, -1 + 62, or 1- 1, 52, and 1-1, - 1 - 62, or 1-1, -72. Passing a smooth curve
through the vertex and these two points, we sketch the parabola shown in
Figure 9.38.
y2 Check Point 5
Find the vertex, focus, and directrix of the parabola given by
y 2 + 2y + 4x - 7 = 0. Then graph the parabola.
[−4, 8, 1] by [−8, 6, 1]
��� Solve applied problems involving Applications
parabolas.
Parabolas have many applications. Cables hung between structures to form
suspension bridges form parabolas. Arches constructed of steel and concrete, whose
main purpose is strength, are usually parabolic in shape.
Parabola
Parabola
Suspension bridge Arch bridge
We have seen that comets in our solar system travel in orbits that are ellipses
and hyperbolas. Some comets follow parabolic paths. Only comets with elliptical
orbits, such as Halley’s Comet, return to our part of the galaxy.
9. P-BLTZMC09_873-950-hr 21-11-2008 13:28 Page 908
908 Chapter 9 Conic Sections and Analytic Geometry
If a parabola is rotated about its axis of symmetry, a parabolic surface is
formed. Figure 9.39(a) shows how a parabolic surface can be used to reflect light.
Light originates at the focus. Note how the light is reflected by the parabolic surface,
so that the outgoing light is parallel to the axis of symmetry. The reflective proper-
ties of parabolic surfaces are used in the design of searchlights [see Figure 9.39(b)],
The Hubble Space
Telescope
Outgoing light
Axis of symmetry
Light at focus
Focus
Figure 9.39(b) Light from the
The Hubble Space Telescope Figure 9.39(a) Parabolic surface focus is reflected parallel to the axis
reflecting light of symmetry.
to create an observatory above
the atmosphere that would pro- Figure 9.40(a) shows how a parabolic surface can be used to reflect incoming
vide an unobscured view of the light. Note that light rays strike the surface and are reflected to the focus. This
universe. This vision was realized principle is used in the design of reflecting telescopes, radar, and television satellite
with the 1990 launching of the
dishes. Reflecting telescopes magnify the light from distant stars by reflecting the
Hubble Space Telescope. The tele-
light from these bodies to the focus of a parabolic mirror [see Figure 9.40(b)].
due to problems with its parabolic
ground two millionths of a meter Eyepiece
smaller than design specifications.
In 1993, astronauts from the Incoming light
Space Shuttle Endeavor equipped Axis of symmetry
the telescope with optics to cor-
rect the blurred vision. “A small
change for a mirror, a giant leap
for astronomy,” Christopher J.
Focus Parabolic surface
Burrows of the Space Telescope
Science Institute said when clear
images from the ends of the uni-
verse were presented to the public Figure 9.40(b) Incoming light rays are
Figure 9.40(a) Parabolic surface
after the repair mission. reflected to the focus.
reflecting incoming light
EXAMPLE 6 Using the Reflection Property of Parabolas
y
An engineer is designing a flashlight using a parabolic
(2, 2) reflecting mirror and a light source, shown in Figure 9.41.
2 The casting has a diameter of 4 inches and a depth of 2 inches
2 inches. What is the equation of the parabola used to
1 2 inches
shape the mirror? At what point should the light source
x be placed relative to the mirror’s vertex?
−3 −2 −1 1 2 3 4 inches
Solution We position the parabola with its vertex at Figure 9.41 Designing a
4 inches the origin and opening upward (Figure 9.42). Thus, the
focus is on the y-axis, located at 10, p2. We use the
flashlight
Figure 9.42
10. P-BLTZMC09_873-950-hr 21-11-2008 13:28 Page 909
Section 9.3 The Parabola 909
standard form of the equation in which there is y-axis symmetry, namely x2 = 4py.
We need to find p. Because (2, 2) lies on the parabola, we let x = 2 and y = 2 in
x2 = 4py.
2 2 = 4p # 2 Substitute 2 for x and 2 for y in x 2 = 4py.
4 = 8p Simplify.
1
p = 2 Divide both sides of the equation by 8 and reduce
the resulting fraction.
We substitute 12 for p in x2 = 4py to obtain the standard form of the equation of the
parabola. The equation of the parabola used to shape the mirror is
x 2 = 4 # 12 y or x2 = 2y.
The light source should be placed at the focus, 10, p2. Because p = 12 , the light should
be placed at A 0, 12 B , or 12 inch above the vertex.
6
Check Point In Example 6, suppose that the casting has a diameter of
6 inches and a depth of 4 inches. What is the equation of the parabola used
to shape the mirror? At what point should the light source be placed relative to
the mirror’s vertex?
Two
intersecting
Degenerate Conic Sections Point Line lines
We opened the chapter by noting that conic
sections are curves that result from the
intersection of a cone and a plane. However,
these intersections might not result in a
conic section. Three degenerate cases occur
when the cutting plane passes through the
vertex. These degenerate conic sections are a
point, a line, and a pair of intersecting lines,
illustrated in Figure 9.43. Figure 9.43 Degenerate conics
Exercise Set 9.3
Practice Exercises c. d.
y y
In Exercises 1–4, find the focus and directrix of each parabola with
the given equation. Then match each equation to one of the graphs 4 4
that are shown and labeled (a)–(d). 3 3
2 2
1. y 2 = 4x 2. x2 = 4y 1 1
x x
3. x2 = - 4y 4. y 2 = - 4x −4 −3 −2 −1−1 1 2 3 4 −4 −3 −2 −1−1 1 2 3 4
−2 −2
−3 −3
a. b. −4 −4
y y
4 4 In Exercises 5–16, find the focus and directrix of the parabola with
3 3 the given equation. Then graph the parabola.
2 2 5. y2 = 16x 6. y 2 = 4x
1 1
x x 7. y2 = - 8x 8. y 2 = - 12x
−4 −3 −2 −1−1 1 2 3 4 −4 −3 −2 −1−1 1 2 3 4 9. x 2 = 12y 10. x2 = 8y
−2 −2 11. x2 = - 16y 12. x2 = - 20y
−3 −3
−4 −4
13. y2 - 6x = 0 14. x2 - 6y = 0
15. 8x2 + 4y = 0 16. 8y2 + 4x = 0
11. P-BLTZMC09_873-950-hr 21-11-2008 13:28 Page 910
910 Chapter 9 Conic Sections and Analytic Geometry
In Exercises 17–30, find the standard form of the equation of each In Exercises 43–48, convert each equation to standard form by
parabola satisfying the given conditions. completing the square on x or y. Then find the vertex, focus, and
17. Focus: (7, 0); Directrix: x = -7 directrix of the parabola. Finally, graph the parabola.
18. Focus: (9, 0); Directrix: x = -9 43. x2 - 2x - 4y + 9 = 0 44. x2 + 6x + 8y + 1 = 0
19. Focus: 1- 5, 02; Directrix: x = 5 45. y2 - 2y + 12x - 35 = 0 46. y2 - 2y - 8x + 1 = 0
20. Focus: 1- 10, 02; Directrix: x = 10 47. x2 + 6x - 4y + 1 = 0 48. x2 + 8x - 4y + 8 = 0
21. Focus: (0, 15); Directrix: y = - 15 Practice Plus
22. Focus: (0, 20); Directrix: y = - 20
In Exercises 49–54, use the vertex and the direction in which the
23. Focus: 10, - 252; Directrix: y = 25 parabola opens to determine the relation’s domain and range. Is
24. Focus: 10, - 152; Directrix: y = 15 the relation a function?
25. Vertex: 12, - 32; Focus: 12, - 52 49. y2 + 6y - x + 5 = 0 50. y2 - 2y - x - 5 = 0
26. Vertex: 15, - 22; Focus: 17, - 22 51. y = - x2 + 4x - 3 52. y = - x2 - 4x + 4
2
27. Focus: (3, 2); Directrix: x = -1 53. x = - 41y - 12 + 3 54. x = - 31y - 122 - 2
28. Focus: (2, 4); Directrix: x = -4 In Exercises 55–60, find the solution set for each system by
1 - 3, 42; Directrix:
graphing both of the system’s equations in the same rectangular
29. Focus: y = 2
coordinate system and finding points of intersection. Check all
30. Focus: 17, - 12; Directrix: y = -9 solutions in both equations.
In Exercises 31–34, find the vertex, focus, and directrix of each 1y - 222 = x + 4 1y - 322 = x - 2
parabola with the given equation. Then match each equation to 55. 56. b
c 1 x + y = 5
one of the graphs that are shown and labeled (a)–(d). y = - x
31. 1y - 122 = 41x - 12
2
32. 1x + 122 = 41y + 12 x = y2 - 3 x = y2 - 5
57. b 58. b
33. 1x + 122 = - 41y + 12 x = y2 - 3y x2 + y2 = 25
34. 1y - 122 = - 41x - 12 x = 1y + 222 - 1 x = 2y2 + 4y + 5
59. b 60. b
a. b. 1x - 22 + 1y + 22 = 1
2 2
1x + 122 + 1y - 222 = 1
y y
4 5 Application Exercises
3 4 61. The reflector of a flashlight is in the shape of a parabolic
2 3 surface. The casting has a diameter of 4 inches and a depth of 1
1 2
inch. How far from the vertex should the light bulb be placed?
x 1
−5 −4 −3 −2 −1−1 1 2 3 x 62. The reflector of a flashlight is in the shape of a parabolic
−2 −4 −3 −2 −1−1 1 2 3 4
surface. The casting has a diameter of 8 inches and a depth of 1
−3 −2 inch. How far from the vertex should the light bulb be placed?
−4 −3
63. A satellite dish, like the one shown below, is in the shape of a
c. d. parabolic surface. Signals coming from a satellite strike the
surface of the dish and are reflected to the focus, where the
y y receiver is located. The satellite dish shown has a diameter of
5 4 12 feet and a depth of 2 feet. How far from the base of the
4 3 dish should the receiver be placed?
3 2 y
2 1
1 x
x −5 −4 −3 −2 −1−1 1 2 3 12 feet
−4 −3 −2 −1−1 1 2 3 4 −2 Receiver
−2 −3
−3 −4 (6, 2)
2 feet
x
In Exercises 35–42, find the vertex, focus, and directrix of each
parabola with the given equation. Then graph the parabola.
35. 1x - 222 = 81y - 12 36. 1x + 222 = 41y + 12
37. 1x + 122 = - 81y + 12 38. 1x + 222 = - 81y + 22
39. 1y + 322 = 121x + 12 40. 1y + 422 = 121x + 22
64. In Exercise 63, if the diameter of the dish is halved and the
depth stays the same, how far from the base of the smaller
41. 1y + 122 = - 8x 42. 1y - 122 = - 8x dish should the receiver be placed?
12. P-BLTZMC09_873-950-hr 21-11-2008 13:28 Page 911
Section 9.3 The Parabola 911
65. The towers of the Golden Gate Bridge connecting 76. Use a graphing utility to graph any three of the parabolas
San Francisco to Marin County are 1280 meters apart and rise that you graphed by hand in Exercises 35–42. First solve the
160 meters above the road. The cable between the towers has given equation for y, possibly using the square root property.
the shape of a parabola and the cable just touches the sides of Enter each of the two resulting equations to produce the
the road midway between the towers. What is the height of the complete graph.
cable 200 meters from a tower? Round to the nearest meter. Use a graphing utility to graph the parabolas in Exercises 77–78.
y Write the given equation as a quadratic equation in y and use the
Parabolic (640, 160) quadratic formula to solve for y. Enter each of the equations to
Cable
produce the complete graph.
x 77. y2 + 2y - 6x + 13 = 0 78. y2 + 10y - x + 25 = 0
160 meters In Exercises 79–80, write each equation as a quadratic equation in y
rs
1280 mete and then use the quadratic formula to express y in terms of x. Graph
the resulting two equations using a graphing utility. What effect does
the xy-term have on the graph of the resulting parabola?
79. 16x2 - 24xy + 9y2 - 60x - 80y + 100 = 0
66. The towers of a suspension bridge are 800 feet apart and rise 80. x2 + 223xy + 3y2 + 823x - 8y + 32 = 0
160 feet above the road. The cable between the towers has
the shape of a parabola and the cable just touches the sides Critical Thinking Exercises
of the road midway between the towers. What is the height of
the cable 100 feet from a tower? Make Sense? In Exercises 81–84, determine whether each state-
ment makes sense or does not make sense, and explain your reasoning.
Parabolic (400, 160) 81. I graphed a parabola that opened to the right and contained
Cable a maximum point.
82. Knowing that a parabola opening to the right has a vertex at
1 - 1, 12 gives me enough information to determine its graph.
800 feet 83. I noticed that depending on the values for A and B, assuming
that they are both not zero, the graph of Ax2 + By2 = C can
represent any of the conic sections other than a parabola.
67. The parabolic arch shown in the figure is 50 feet above the 84. I’m using a telescope in which light from distant stars is
water at the center and 200 feet wide at the base. Will a boat reflected to the focus of a parabolic mirror.
that is 30 feet tall clear the arch 30 feet from the center?
In Exercises 85–88, determine whether each statement is true or false.
If the statement is false, make the necessary change(s) to produce a
true statement.
85. The parabola whose equation is x = 2y - y2 + 5 opens to
100 ft 100 ft the right.
50 ft 86. If the parabola whose equation is x = ay2 + by + c has its
vertex at (3, 2) and a 7 0, then it has no y-intercepts.
87. Some parabolas that open to the right have equations that
define y as a function of x.
68. A satellite dish in the shape of a parabolic surface has a
88. The graph of x = a1y - k2 + h is a parabola with vertex at
1h, k2.
diameter of 20 feet. If the receiver is to be placed 6 feet from
the base, how deep should the dish be?
89. Find the focus and directrix of a parabola whose equation is
Writing in Mathematics of the form Ax2 + Ey = 0, A Z 0, E Z 0.
90. Write the standard form of the equation of a parabola whose
points are equidistant from y = 4 and 1- 1, 02.
69. What is a parabola?
70. Explain how to use y2 = 8x to find the parabola’s focus and
directrix. Group Exercise
71. If you are given the standard form of the equation of a 91. Consult the research department of your library or the
parabola with vertex at the origin, explain how to determine Internet to find an example of architecture that incorporates
if the parabola opens to the right, left, upward, or downward. one or more conic sections in its design. Share this example
72. Describe one similarity and one difference between the with other group members. Explain precisely how conic
graphs of y2 = 4x and 1y - 122 = 41x - 12. sections are used. Do conic sections enhance the appeal of
73. How can you distinguish parabolas from other conic sections the architecture? In what ways?
by looking at their equations?
74. Look at the satellite dish shown in Exercise 63. Why must the
Preview Exercises
receiver for a shallow dish be farther from the base of the Exercises 92–94 will help you prepare for the material covered in
dish than for a deeper dish of the same diameter? the next section.
92. Simplify and write the equation in standard form in terms of
Technology Exercises x¿ and y¿:
75. Use a graphing utility to graph any five of the parabolas that 22 22
you graphed by hand in Exercises 5–16. B 1x¿ - y¿2 R B 1x¿ + y¿2 R = 1.
2 2
13. P-BLTZMC09_873-950-hr 21-11-2008 13:28 Page 912
912 Chapter 9 Conic Sections and Analytic Geometry
7
93. a. Make a sketch showing that cot 2u = - for d. In part (c), why did we not write ; before the radical in
24 each formula?
90° 6 2u 6 180°.
b. Use your sketch from part (a) to determine the value of 94. The equation 3x2 - 223xy + y2 + 2x + 223y = 0 is in
cos 2u. the form Ax2 + Bxy + Cy2 + Dx + Ey + F = 0. Use the
c. Use the value of cos 2u from part (b) and the identities equation to determine the value of B2 - 4AC.
1 - cos 2u 1 + cos 2u
sin u = and cos u =
A 2 A 2
to determine the values of sin u and cos u.
Chapter 9 Mid-Chapter Check Point
What You Know: We learned that the four conic sections In Exercises 14–21, graph each equation.
are the circle, the ellipse, the hyperbola, and the parabola. 14. x2 + y2 = 4 15. x + y = 4
Prior to this chapter, we graphed circles with center 1h, k2 2 2
16. x - y = 4 17. x 2 + 4y2 = 4
18. 1x + 122 + 1y - 122 = 4
1x - h22 + 1y - k22 = r2.
19. x 2 + 41y - 122 = 4
20. 1x - 122 - 1y - 122 = 4 21. 1y + 122 = 41x - 12
In this chapter, you learned to graph ellipses centered at
the origin and ellipses centered at 1h, k2: In Exercises 22–27, find the standard form of the equation of the
1x - h22 1y - k22 conic section satisfying the given conditions.
22. Ellipse; Foci: 1-4, 02, (4, 0); Vertices: 1 - 5, 02, (5, 0)
+ = 1 or
a2 b2
1x - h22 1y - k22 23. Ellipse; Endpoints of major axis: 1- 8, 22, (10, 2);
+ = 1, a2 7 b2. Foci: 1 -4, 22, (6, 2)
24. Hyperbola; Foci: 10, -32, (0, 3); Vertices: 10, - 22, (0, 2)
b2 a2
We saw that the larger denominator 1a22 determines 25. Hyperbola; Foci: 1-4, 52, (2, 5); Vertices: 1- 3, 52, (1, 5)
whether the major axis is horizontal or vertical. We used 26. Parabola; Focus: (4, 5); Directrix: y = - 1
27. Parabola; Focus: 1-2, 62; Directrix: x = 8
vertices and asymptotes to graph hyperbolas centered at
the origin and hyperbolas centered at 1h, k2:
28. A semielliptical archway over a one-way road has a height of
1x - h22 1y - k22 1y - k22 1x - h22 10 feet and a width of 30 feet. A truck has a width of 10 feet
- = 1 or - = 1. and a height of 9.5 feet. Will this truck clear the opening of
a2 b2 a2 b2 the archway?
We used c2 = a2 - b2 to locate the foci of an ellipse.We used 29. A lithotriper is used to disentegrate kidney stones.The patient is
c2 = a2 + b2 to locate the foci of a hyperbola. Finally, we placed within an elliptical device with the kidney centered at
used the vertex and the latus rectum to graph parabolas with one focus, while ultrasound waves from the other focus hit the
vertices at the origin and parabolas with vertices at 1h, k2: walls and are reflected to the kidney stone, shattering the stone.
1y - k22 = 4p1x - h2 or 1x - h22 = 4p1y - k2.
Suppose that the length of the major axis of the ellipse is 40 cen-
timeters and the length of the minor axis is 20 centimeters. How
In Exercises 1–5, graph each ellipse. Give the location of the foci. far from the kidney stone should the electrode that sends the
x2 y2 ultrasound waves be placed in order to shatter the stone?
1. + = 1 2. 9x 2 + 4y2 = 36 30. An explosion is recorded by two forest rangers, one at a
25 4
1x - 222 1y + 122 1x + 222 1y - 122
primary station and the other at an outpost 6 kilometers
3. + = 1 4. + = 1 away. The ranger at the primary station hears the explosion
16 25 25 16 6 seconds before the ranger at the outpost.
5. x2 + 9y2 - 4x + 54y + 49 = 0 a. Assuming sound travels at 0.35 kilometer per second,
In Exercises 6–11, graph each hyperbola. Give the location of the write an equation in standard form that gives all the
foci and the equations of the asymptotes. possible locations of the explosion. Use a coordinate
system with the two ranger stations on the x-axis and the
x2 y2
6. - y2 = 1 7. - x2 = 1 midpoint between the stations at the origin.
9 9 b. Graph the equation that gives the possible locations of
8. y2 - 4x2 = 16 9. 4x 2 - 49y2 = 196 the explosion. Show the locations of the ranger stations
1x - 222 1y + 222 in your drawing.
10. - = 1 31. A domed ceiling is a parabolic surface. Ten meters down
9 16
from the top of the dome, the ceiling is 15 meters wide. For
11. 4x2 - y2 + 8x + 6y + 11 = 0
the best lighting on the floor, a light source should be placed
In Exercises 12–13, graph each parabola. Give the location of the at the focus of the parabolic surface. How far from the top of
focus and the directrix. the dome, to the nearest tenth of a meter, should the light
12. 1x - 222 = - 121y + 12 13. y2 - 2x - 2y - 5 = 0 source be placed?