copyright © 2014, 2010, 2007 pearson education, inc. 1 section 9.3 the parabola
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Copyright © 2014, 2010, 2007 Pearson Education, Inc. 1
Section 9.3
The Parabola
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• Graph parabolas with vertices at the origin.• Write equations of parabolas in standard form.• Graph parabolas with vertices not at the origin.• Solve applied problems involving parabolas.
Objectives:
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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.
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Standard Forms of the Equations of a Parabola
The standard form of the equation of a parabola with vertex at the origin is
y2 = 4px or x2 = 4py
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Example: Finding the Focus and Directrix of a Parabola
Find the focus and directrix of the parabola given by
y2 = 8x. Then graph the parabola.
The given equation, y2 = 8x, is in the standard form
y2 = 4px, so 4p = 8.
4p = 8
p = 2
Because p is positive, the parabola, with its x-axis symmetry, opens to the right. The focus is 2 units to the right of the vertex, (0, 0). The focus is (2, 0).
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Example: Finding the Focus and Directrix of a Parabola (continued)
Find the focus and directrix of the parabola given by
y2 = 8x. Then graph the parabola.
The focus is (2, 0) and directrix, x = – 2.
To graph the parabola, we will use two points on the graph that are directly above and below the focus. Because the focus is at (2, 0), substitute 2 for x in the parabola’s equation, y2 = 8x.
2 8y x2 8 2y
2 16y 16 4y
The points on the parabolaabove and below the focusare (2, 4) and (2, –4).
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Example: Finding the Focus and Directrix of a Parabola (continued)
Find the focus and directrix of the parabola given by
y2 = 8x. Then graph the parabola.
The focus is (2, 0).
The directrix is x = –2.-5 -4 -3 -2 -1 1 2 3 4 5
-5
-4
-3
-2
-1
1
2
3
4
5
x
y
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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. The length of the latus rectum for the graphs of y2 = 4px and x2 = 4py is 4 .p
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Example: Finding the Equation of a Parabola from Its Focus and Directrix
Find the standard form of the equation of a parabola with focus (8, 0) and directrix x = –8.
The focus, (8, 0), is on the x-axis. We use the standard form of the equation in which there is x-axis symmetry, y2 = 4px.
The focus is 8 units to the right of the vertex, (0, 0). Thus, p is positive and p = 8.
2 4y px2 4 8 32y x x
The equation of the parabolais y2 = 32x.
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Example: Finding the Equation of a Parabola from Its Focus and Directrix (continued)
Find the standard form of the equation of a parabola with focus (8, 0) and directrix x = –8.
The equation of the parabola
is y2 = 32x.
-10 -8 -6 -4 -2 2 4 6 8 10
-10
-8
-6
-4
-2
2
4
6
8
10
x
y
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Translations of Parabolas – Standard Forms of Equations of Parabolas with Vertex (h, k)
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Example: Graphing a Parabola with Vertex at (h, k)
Find the vertex, focus, and directrix of the parabola given by Then graph the parabola.
The equation is in the form
The vertex is (2, –1).
4p = 4, p = 1
The focus is located 1 unit above the vertex of (2, –1).
Focus:
The directrix is located 2 units below the vertex.
Directrix:
2( 2) 4( 1).x y 2( ) 4 ( ). x h p y k
( , ) (2, 1 1) (2,0)h k p
y k p 1 1 2y
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Example: Graphing a Parabola with Vertex at (h, k)
Find the vertex, focus, and directrix of the parabola given by Then graph the parabola.
The vertex is (2, –1).
The focus is (2, 0).
The directrix is y = –2.
2( 2) 4( 1).x y
-8 -6 -4 -2 2 4 6 8
-5
-4
-3
-2
-1
1
2
3
4
5
x
y
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Example: Application
An engineer is designing a flashlight using a parabolic reflecting mirror and a light source. 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?
6 inches
4 inches
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Example: Application (continued)
We position the parabola with its vertex at the origin and opening upward.
6 inches
4 inches
-4 -3 -2 -1 1 2 3 4
1
2
3
4
x
y
4 inches
6 inches
(3,4)
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Example: Application (continued)
The focus is on the y-axis, located at (0, p).
The form of the equation is
The point (3, 4) lies on the parabola.
To find p, we let x = 3 and y = 4.
-4 -3 -2 -1 1 2 3 4
1
2
3
4
x
y
2 4 .x py
(3,4)
23 4 4p 9 16 p
916
p
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Example: Application (continued)
We substitute in to obtain the standard
form of the equation of the parabola.
The equation of the parabola used to shape the mirror is
The light source should be placed at or
inches above the vertex.
916
2 4x py
2 94
16x y 2 9
4x y
90,
16
2 9.
4x y
916