Chapter 9. Conic Sections and Analytic Geometry. 9.3 The Parabola. Copyright 2014, 2010, 2007 Pearson Education, Inc.

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1 Chapter 9 Conic Sections and Analytic Geometry 9.3 The Parabola Copyright 014, 010, 007 Pearson Education, Inc. 1

2 Objectives: 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. Copyright 014, 010, 007 Pearson Education, Inc.

3 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. Copyright 014, 010, 007 Pearson Education, Inc. 3

4 Standard Forms of the Equations of a Parabola The standard form of the equation of a parabola with vertex at the origin is y = 4px or x = 4py Copyright 014, 010, 007 Pearson Education, Inc. 4

5 Example: Finding the Focus and Directrix of a Parabola Find the focus and directrix of the parabola given by y = 8x. Then graph the parabola. The given equation, y = 8x, is in the standard form y = 4px, so 4p = 8. 4p = 8 p = Because p is positive, the parabola, with its x-axis symmetry, opens to the right. The focus is units to the right of the vertex, (0, 0). The focus is (, 0). Copyright 014, 010, 007 Pearson Education, Inc. 5

6 Example: Finding the Focus and Directrix of a Parabola (continued) Find the focus and directrix of the parabola given by y = 8x. Then graph the parabola. The focus is (, 0) and directrix, x = p = 3. 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 (, 0), substitute for x in the parabola s equation, y = 8x. y 8x y 8 y 16 y 16 4 The points on the parabola above and below the focus are (, 4) and (, 4). Copyright 014, 010, 007 Pearson Education, Inc. 6

7 Example: Finding the Focus and Directrix of a Parabola (continued) Find the focus and directrix of the parabola given by y = 8x. Then graph the parabola. The focus is (, 0). The directrix is x =. Copyright 014, 010, 007 Pearson Education, Inc. 7

8 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 y = 4px and x = 4py is 4 p. Copyright 014, 010, 007 Pearson Education, Inc. 8

9 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, y = 4px. The focus is 8 units to the right of the vertex, (0, 0). Thus, p is positive and p = 8. y 4 px y 4 8x 3x The equation of the parabola is y = 3x. Copyright 014, 010, 007 Pearson Education, Inc. 9

10 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 y = 3x. Copyright 014, 010, 007 Pearson Education, Inc. 10

11 Translations of Parabolas Standard Forms of Equations of Parabolas with Vertex (h, k) Copyright 014, 010, 007 Pearson Education, Inc. 11

12 Example: Graphing a Parabola with Vertex at (h, k) Find the vertex, focus, and directrix of the parabola given by ( x ) 4( y 1). Then graph the parabola. The equation is in the form ( x h) 4 p( y k). The vertex is (, 1). 4p = 4, p = 1 The focus is located 1 unit above the vertex of (, 1). Focus: ( h, k p) (, 1 1) (,0) The directrix is located units below the vertex. Directrix: y k p y 1 1 Copyright 014, 010, 007 Pearson Education, Inc. 1

13 Example: Graphing a Parabola with Vertex at (h, k) Find the vertex, focus, and directrix of the parabola given by ( x ) 4( y 1). Then graph the parabola. The vertex is (, 1). The focus is (, 0). The directrix is y =. Copyright 014, 010, 007 Pearson Education, Inc. 13

14 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 Copyright 014, 010, 007 Pearson Education, Inc. 14

15 Example: Application (continued) We position the parabola with its vertex at the origin and opening upward. (3,4) 4 inches 6 inches 4 inches 6 inches Copyright 014, 010, 007 Pearson Education, Inc. 15

16 Example: Application (continued) The focus is on the y-axis, located at (0, p). The form of the equation is x 4 py. The point (3, 4) lies on the parabola. To find p, we let x = 3 and y = p p 9 p 16 (3,4) Copyright 014, 010, 007 Pearson Education, Inc. 16

17 Example: Application (continued) 9 We substitute in x 4 py to obtain the standard 16 form of the equation of the parabola. x 4 y x y 16 4 The equation of the parabola used to shape the mirror is x y. 4 The light source should be placed at inches above the vertex. 9 0, 16 or 9 16 Copyright 014, 010, 007 Pearson Education, Inc. 17