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Section 10.2 Introduction to Conics: Parabolas 735
Conics
Conic sections were discovered during the classical Greek period, 600 to 300 B.C.
The early Greeks were concerned largely with the geometric properties of conics.
It was not until the 17th century that the broad applicability of conics became
apparent and played a prominent role in the early development of calculus.
A conic section (or simply conic ) is the intersection of a plane and a double-
napped cone. Notice in Figure 10.8 that in the formation of the four basic
conics, the intersecting plane does not pass through the vertex of the cone. When
the plane does pass through the vertex, the resulting figure is a degenerate conic,
as shown in Figure 10.9.
Circle Ellipse Parabola Hyperbola FIGURE 10.8 Basic Conics
Point Line Two Intersecting FIGURE 10.9 Degenerate Conics Lines
There are several ways to approach the study of conics. You could begin by
defining conics in terms of the intersections of planes and cones, as the Greeks
did, or you could define them algebraically, in terms of the general second-
degree equation
However, you will study a third approach, in which each of the conics is defined
as a locus (collection) of points satisfying a geometric property. For example, in
Section 1.2, you learned that a circle is defined as the collection of all points
that are equidistant from a fixed point This leads to the standard form
of the equation of a circle
x h ^2 y k ^2 r^2. Equation of circle
x , y h , k .
Ax^2 Bxy Cy^2 Dx Ey F 0.
What you should learn
- Recognize a conic as the intersection of a plane and a double-napped cone.
- Write equations of parabolas in standard form and graph parabolas.
- Use the reflective property of parabolas to solve real-life problems.
Why you should learn it Parabolas can be used to model and solve many types of real-life problems. For instance, in Exercise 62 on page 742, a parabola is used to model the cables of the Golden Gate Bridge.
Introduction to Conics: Parabolas
Cosmo Condina/Getty Images
10.
This study of conics is from a locus-of- points approach, which leads to the development of the standard equation for each conic. Your students should know the standard equations of all conics well. Make sure they understand the relationship of h and k to the horizontal and vertical shifts.
Parabolas
In Section 2.1, you learned that the graph of the quadratic function
is a parabola that opens upward or downward. The following definition of a
parabola is more general in the sense that it is independent of the orientation of
the parabola.
The midpoint between the focus and the directrix is called the vertex, and the
line passing through the focus and the vertex is called the axis of the parabola.
Note in Figure 10.10 that a parabola is symmetric with respect to its axis. Using
the definition of a parabola, you can derive the following standard form of the
equation of a parabola whose directrix is parallel to the -axis or to the -axis.
For a proof of the standard form of the equation of a parabola, see Proofs in
Mathematics on page 807.
x y
f x ax^2 bx c
736 Chapter 10 Topics in Analytic Geometry
Definition of Parabola
A parabola is the set of all points in a plane that are equidistant from
a fixed line (directrix) and a fixed point (focus) not on the line.
x , y
Standard Equation of a Parabola
The standard form of the equation of a parabola with vertex at is as
follows.
Vertical axis, directrix:
Horizontal axis, directrix:
The focus lies on the axis units ( directed distance ) from the vertex. If the
vertex is at the origin the equation takes one of the following forms.
Vertical axis
Horizontal axis
See Figure 10.11.
y^2 4 px
x^2 4 py
p
y k ^2 4 p x h , p 0 x h p
x h ^2 4 p y k , p 0 y k p
h , k
Focus
Vertex
Directrix x
d 1
d 1 d 2
d 2
y
FIGURE 10.10 Parabola
Focus: ( , h k + p )
Directrix: y = k − p
Vertex: ( , h k )
p > 0
x = h
Axis:
(a) Vertical axis: FIGURE 10.
p > 0
x h^2 4 py k
p < 0
Focus: ( h , k + p )
Vertex: ( h , k )
Directrix: y = k − p
Axis: x = h
(b) Vertical axis: p < 0
x h^2 4 py k
y = k h p
Focus: ( + , k )
x = h − p
Vertex: ( , h k )
p > 0
Axis:
Directrix:
(c) Horizontal axis: p > 0
y k^2 4 px h
p < 0
Axis: y = k
Focus: ( h + p , k )
Vertex: ( h , k )
Directrix: x = h − p
(d) Horizontal axis: p < 0
y k^2 4 px h
Finding the Standard Equation of a Parabola
Find the standard form of the equation of the parabola with vertex and
focus
Solution
Because the axis of the parabola is vertical, passing through and
consider the equation
where and So, the standard form is
You can obtain the more common quadratic form as follows.
Write original equation.
Multiply.
Add 12 to each side.
Divide each side by 12.
The graph of this parabola is shown in Figure 10.14.
Now try Exercise 45.
Application
A line segment that passes through the focus of a parabola and has endpoints on
the parabola is called a focal chord. The specific focal chord perpendicular to the
axis of the parabola is called the latus rectum.
Parabolas occur in a wide variety of applications. For instance, a parabolic
reflector can be formed by revolving a parabola around its axis. The resulting
surface has the property that all incoming rays parallel to the axis are reflected
through the focus of the parabola. This is the principle behind the construction of
the parabolic mirrors used in reflecting telescopes. Conversely, the light rays
emanating from the focus of a parabolic reflector used in a flashlight are all
parallel to one another, as shown in Figure 10.15.
A line is tangent to a parabola at a point on the parabola if the line intersects,
but does not cross, the parabola at the point. Tangent lines to parabolas have spe-
cial properties related to the use of parabolas in constructing reflective surfaces.
x^2 4 x 16 y
x^2 4 x 16 12 y
x^2 4 x 4 12 y 12
x 2 ^2 12 y 1
x 2 ^2 12 y 1 .
h 2, k 1, p 4 1 3.
x h ^2 4 p y k
738 Chapter 10 Topics in Analytic Geometry
Example 3
Reflective Property of a Parabola
The tangent line to a parabola at a point makes equal angles with the
following two lines (see Figure 10.16).
1. The line passing through and the focus
2. The axis of the parabola
P
P
Parabolic reflector: Light is reflected in parallel rays.
Focus Axis
Light source at focus
FIGURE 10.
Tangent line
Focus
P
α
α
Axis
FIGURE 10.
− 4
− 4
− 2 2 4 6 8 − 2
4
6
8
Focus (2, 4)
Vertex (2, 1)
( x − 2) 2 = 12( y − 1)
x
y
FIGURE 10.
Finding the Tangent Line at a Point on a Parabola
Find the equation of the tangent line to the parabola given by at the point
Solution
For this parabola, and the focus is as shown in Figure 10.17. You
can find the -intercept of the tangent line by equating the lengths of the
two sides of the isosceles triangle shown in Figure 10.17:
and
Note that rather than The order of subtraction for the distance
is important because the distance must be positive. Setting produces
So, the slope of the tangent line is
and the equation of the tangent line in slope-intercept form is
Now try Exercise 55.
y 2 x 1.
m
b 1.
b
d 1 d 2
d 1 14 b b 14.
d 2 1 0 ^2 1
4 ^
2
d 1
b
y 0, b
p 14 0, 14 ,
y x^2
Section 10.2 Introduction to Conics: Parabolas 739
Example 4
WRITING ABOUT^ MATHEMATICS
Television Antenna Dishes Cross sections of television antenna dishes are parabolic
in shape. Use the figure shown to write a paragraph explaining why these dishes are parabolic.
Amplifier
Dish reflector
Cable to radio or TV
Activities
- Find the vertex, focus, and directrix of the parabola Answer: Vertex Focus Directrix
- Find the standard form of the equa- tion of the parabola with vertex and directrix Answer:
- Find an equation of the tangent line to the parabola at the point
Answer: y 4 x 2
y 2 x^2
y^2 4 x 4
x 5.
y 2
x^2 6 x 4 y 5 0.
d
d
1
2
(0, b )
y = x^2
− 1 1
1
α
α
y
x
FIGURE 10.
Use a graphing utility to confirm the result of Example 4. By graphing
and
in the same viewing window, you should be able to see that the line touches the parabola at the point 1, 1.
y 1 x^2 y 2 2 x 1
Te c h n o l o g y
Section 10.2 Introduction to Conics: Parabolas 741
In Exercises 29–40, find the standard form of the equation of the parabola with the given characteristic(s) and vertex at the origin.
29. 30. 31. Focus: 32. Focus: 33. Focus: 34. Focus: 35. Directrix: 36. Directrix: 37. Directrix: 38. Directrix: 39. Horizontal axis and passes through the point 40. Vertical axis and passes through the point
In Exercises 41–50, find the standard form of the equation of the parabola with the given characteristics.
41. 42.
45. Vertex: focus: 46. Vertex: focus: 47. Vertex: directrix: 48. Vertex: directrix: 49. Focus: directrix: 50. Focus: directrix:
In Exercises 51 and 52, change the equation of the parabola so that its graph matches the description.
51. upper half of parabola 52. lower half of parabola
In Exercises 53 and 54, the equations of a parabola and a tangent line to the parabola are given. Use a graphing utility to graph both equations in the same viewing win- dow. Determine the coordinates of the point of tangency.
Parabola Tangent Line 53. 54.
In Exercises 55–58, find an equation of the tangent line to the parabola at the given point, and find the -intercept of the line.
**55.
58.**
59. Revenue The revenue (in dollars) generated by the sale of units of a patio furniture set is given by
Use a graphing utility to graph the function and approxi- mate the number of sales that will maximize revenue.
60. Revenue The revenue (in dollars) generated by the sale of units of a digital camera is given by
Use a graphing utility to graph the function and approxi- mate the number of sales that will maximize revenue.
61. Satellite Antenna The receiver in a parabolic television dish antenna is 4.5 feet from the vertex and is located at the focus (see figure). Write an equation for a cross section of the reflector. (Assume that the dish is directed upward and the vertex is at the origin.)
4.5 ft
Receiver
x
y
x 135 ^2
R 25,515.
x
R
x 106 ^2
R 14,045.
x
R
y 2 x^2 , 2, 8
y 2 x^2 , 1, 2
x^2 2 y , 3, (^92)
x^2 2 y , 4, 8
x
x^2 12 y 0 x y 3 0
y^2 8 x 0 x y 2 0
y 1 ^2 2 x 4 ;
y 3 ^2 6 x 1 ;
0, 0; y 8
2, 2; x 2
2, 1; x 1
0, 4; y 2
− 4 8
8
12
− 4 (3,^ −3)
y
x
4 8
8
− 8
y
x
x 2 4
2
4
y
x
2 4 6 − 2
− 4
y (2, 0) 2
x 3
x 2
y 3
y 1
^52 , 0
0,^32
x − 8 − 4 4
− 8
8 (−2, 6)
y
x − 4 − 2 2 4
2
4
y
63. Road Design Roads are often designed with parabolic surfaces to allow rain to drain off. A particular road that is 32 feet wide is 0.4 foot higher in the center than it is on the sides (see figure).
Cross section of road surface
(a) Find an equation of the parabola that models the road surface. (Assume that the origin is at the center of the road.) (b) How far from the center of the road is the road surface 0.1 foot lower than in the middle?
64. Highway Design Highway engineers design a parabolic curve for an entrance ramp from a straight street to an interstate highway (see figure). Find an equation of the parabola.
FIGURE FOR 64
65. Satellite Orbit A satellite in a 100-mile-high circular orbit around Earth has a velocity of approximately 17, miles per hour. If this velocity is multiplied by the satellite will have the minimum velocity necessary to escape Earth’s gravity and it will follow a parabolic path with the center of Earth as the focus (see figure).
(a) Find the escape velocity of the satellite. (b) Find an equation of the parabolic path of the satellite (assume that the radius of Earth is 4000 miles).
66. Path of a Softball The path of a softball is modeled by
where the coordinates and are measured in feet, with corresponding to the position from which the ball was thrown. (a) Use a graphing utility to graph the trajectory of the softball. (b) Use the trace feature of the graphing utility to approx- imate the highest point and the range of the trajectory.
Projectile Motion In Exercises 67 and 68, consider the path of a projectile projected horizontally with a velocity of feet per second at a height of feet, where the model for the path is
In this model (in which air resistance is disregarded), is the height (in feet) of the projectile and is the horizontal distance (in feet) the projectile travels.
x
y
x^2
v^2 16
y s.
v s
x 0
x y
12.5 y 7.125 x 6.25^2
Parabolic path
t
4100 miles x
y
Not drawn to scale
Circular orbi
400 800 1200 1600
400
800
− 400
− (^800) Street
Interstate
y
x
32 ft (^) 0.4 ft Not drawn to scale
742 Chapter 10 Topics in Analytic Geometry
62. Suspension Bridge Each cable of the Golden Gate Bridge is suspended (in the shape of a parabola) between two towers that are 1280 meters apart. The top of each tower is 152 meters above the roadway. The cables touch the roadway midway between the towers. (a) Draw a sketch of the bridge. Locate the origin of a rectangular coordinate system at the center of the roadway. Label the coordinates of the known points. (b) Write an equation that models the cables. (c) Complete the table by finding the height of the suspension cables over the roadway at a distance of x meters from the center of the bridge.
y
Model It
Distance, x Height, y
