copyright © 2011 pearson education, inc. slide 6.1-1

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Copyright © 2011 Pearson Education, Inc. Slide 6.1-1

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Page 1: Copyright © 2011 Pearson Education, Inc. Slide 6.1-1

Copyright © 2011 Pearson Education, Inc. Slide 6.1-1

Page 2: Copyright © 2011 Pearson Education, Inc. Slide 6.1-1

Copyright © 2011 Pearson Education, Inc. Slide 6.1-2

Chapter 6: Analytic Geometry

6.1 Circles and Parabolas

6.2 Ellipses and Hyperbolas

6.3 Summary of the Conic Sections

6.4 Parametric Equations

Page 3: Copyright © 2011 Pearson Education, Inc. Slide 6.1-1

Copyright © 2011 Pearson Education, Inc. Slide 6.1-3

6.1 Circles and Parabolas

• Conic Sections– Parabolas, circles, ellipses, hyperbolas

Page 4: Copyright © 2011 Pearson Education, Inc. Slide 6.1-1

Copyright © 2011 Pearson Education, Inc. Slide 6.1-4

• A circle with center (h, k) and radius r has length to some point (x, y)

on the circle. • Squaring both sides yields the center-radius

form of the equation of a circle.

6.1 Circles

A circle is a set of points in a plane that are equidistant from a fixed point. The distance is called the radius of the circle, and the fixed point is called the center.

22 )()( kyhxr

22 )()( kyhxr 2

Page 5: Copyright © 2011 Pearson Education, Inc. Slide 6.1-1

Copyright © 2011 Pearson Education, Inc. Slide 6.1-5

Notice that a circle is the graph of a relation that is

not a function, since it does not pass the vertical line

test.

6.1 Center-Radius Form of the Equation of a Circle

The center-radius form of the equation of a circle with center (h, k) and radius r is

. 2 2( ) ( ) 2x h y k r

Page 6: Copyright © 2011 Pearson Education, Inc. Slide 6.1-1

Copyright © 2011 Pearson Education, Inc. Slide 6.1-6

6.1 Finding the Equation of a Circle

Example Find the center-radius form of the equation of a circle with radius 6 and center (–3, 4). Graph the circle and give the domain and range of the relation.

Solution Substitute h = –3, k = 4, and r = 6 into the equation of a circle.

22

222

)4()3(36

)4())3((6

yx

yx

Page 7: Copyright © 2011 Pearson Education, Inc. Slide 6.1-1

Copyright © 2011 Pearson Education, Inc. Slide 6.1-7

6.1 Equation of a Circle with Center at the Origin

A circle with center (0, 0) and radius r has equation

. 2 2 2x y r

Page 8: Copyright © 2011 Pearson Education, Inc. Slide 6.1-1

Copyright © 2011 Pearson Education, Inc. Slide 6.1-8

6.1 Graphing Circles with the Graphing Calculator

Example Use the graphing calculator to graph the circle in a square viewing window.

Solution

922 yx

.9 and 9Let

9

9

9

22

21

2

22

22

xyxy

xy

xy

yx

Page 9: Copyright © 2011 Pearson Education, Inc. Slide 6.1-1

Copyright © 2011 Pearson Education, Inc. Slide 6.1-9

6.1 Graphing Circles with the Graphing Calculator

• TECHNOLOGY NOTES:– Graphs in a nondecimal window may not be

connected.

– Graphs in a rectangular (non-square) window look like an ellipse.

Page 10: Copyright © 2011 Pearson Education, Inc. Slide 6.1-1

Copyright © 2011 Pearson Education, Inc. Slide 6.1-10

6.1 General Form of the Equation of a Circle

For real numbers c, d, and e, the equation

can have a graph that is a circle, a point, or is empty.

22x y cx dy e 0

Page 11: Copyright © 2011 Pearson Education, Inc. Slide 6.1-1

Copyright © 2011 Pearson Education, Inc. Slide 6.1-11

6.1 Finding the Center and Radius of a Circle

Example Find the center and radius of the circle with equation

Solution Our goal is to obtain an equivalent equation of the formWe complete the square in both x and y.

2 26 10 25 0.x x y y

.)()( 222 kyhxr

2 2

2 2

2 2

2 2 2

6 10 25

( 6 9) ( 10 25) 25 9 25

( 3) ( 5) 9

( 3) ( 2) 3

x x y y

x x y y

x y

x y

The circle has center (3, –2) with radius 3.

Page 12: Copyright © 2011 Pearson Education, Inc. Slide 6.1-1

Copyright © 2011 Pearson Education, Inc. Slide 6.1-12

6.1 Equations and Graphs of Parabolas

• For example, let the directrix be the line y = –c and the focus be the point F with coordinates (0, c).

A parabola is a set of points in a plane equidistant from a fixed point and a fixed line. The fixed point is called the focus, and the fixed line the directrix, of the parabola.

Page 13: Copyright © 2011 Pearson Education, Inc. Slide 6.1-1

Copyright © 2011 Pearson Education, Inc. Slide 6.1-13

6.1 Equations and Graphs of Parabolas

• To get the equation of the set of points that are the same distance from the line y = –c and the point (0, c), choose a point P(x, y) on the parabola. The distance from the focus, F, to P, and the point on the directrix, D, to P, must have the same length.

cyxcycycycyx

cycycycyx

cyxxcyx

DPdFPd

422

22

))(()()()0(

),(),(

2

22222

22222

2222

Page 14: Copyright © 2011 Pearson Education, Inc. Slide 6.1-1

Copyright © 2011 Pearson Education, Inc. Slide 6.1-14

6.1 Parabola with a Vertical Axis

and Vertex (0, 0)

• The focal chord through the focus and perpendicular to the axis of symmetry of a parabola has length |4c|.– Let y = c and solve for x.

The endpoints of the chord are ( x, c), so the length is |4c|.

The parabola with focus (0, c) and directrix y = –c has equation x2 = 4cy. The parabola has vertex (0, 0), vertical axis x = 0, and opens upward if c > 0 or downward if c < 0.

c cxcx

cyx

2or 24

422

2

Page 15: Copyright © 2011 Pearson Education, Inc. Slide 6.1-1

Copyright © 2011 Pearson Education, Inc. Slide 6.1-15

6.1 Parabola with a Horizontal Axis

and Vertex (0, 0)

• Note: a parabola with a horizontal axis is not a function.

• The graph can be obtained using a graphing calculator by solving y2 = 4cx for y:

Let and graph each half of the parabola.

The parabola with focus (c, 0) and directrix x = –c has equation y2 = 4cx. The parabola has vertex (0, 0),

horizontal axis y = 0, and opens to the right if c > 0 or to the left if c < 0.

.2 cxy cxycxy 2 and 2 21

Page 16: Copyright © 2011 Pearson Education, Inc. Slide 6.1-1

Copyright © 2011 Pearson Education, Inc. Slide 6.1-16

6.1 Determining Information about Parabolas from Equations

Example Find the focus, directrix, vertex, and axis

of each parabola.(a)

Solution(a)

xyyx 28 (b)8 22

284

cc

Since the x-term is squared, the parabola is vertical, with focus at (0, c) = (0, 2) and directrix y = –2. The vertex is (0, 0), and the axis is the y-axis.

Page 17: Copyright © 2011 Pearson Education, Inc. Slide 6.1-1

Copyright © 2011 Pearson Education, Inc. Slide 6.1-17

6.1 Determining Information about Parabolas from Equations

(b)

The parabola is horizontal, with focus (–7, 0), directrix x = 7, vertex (0, 0), and x-axis as axis of the parabola. Since c is negative, the graph opens to the left.

7284

cc

Page 18: Copyright © 2011 Pearson Education, Inc. Slide 6.1-1

Copyright © 2011 Pearson Education, Inc. Slide 6.1-18

6.1 Translations of Parabolas

A parabola with vertex (h, k) has an equation of the form

or

where the focus is a distance |c| from the vertex.

24x h c y k

24y k c x h

Page 19: Copyright © 2011 Pearson Education, Inc. Slide 6.1-1

Copyright © 2011 Pearson Education, Inc. Slide 6.1-19

6.1 Writing Equations of Parabolas

Example Write an equation for the parabola with vertex (1, 3) and focus (–1, 3).

Solution Focus lies left of the vertex implies theparabola has

- a horizontal axis, and- opens to the left.

Distance between vertex and focus is 1–(–1) = 2, so c = –2.

)1(8)3()1)(2(4)3(

2

2

xyxy

Page 20: Copyright © 2011 Pearson Education, Inc. Slide 6.1-1

Copyright © 2011 Pearson Education, Inc. Slide 6.1-20

6.1 An Application of Parabolas

Example The Parkes radio tele-scope has a parabolic dish shape with diameter 210 feet and depth 32 feet. Because of this parabolic shape, distant rays hitting the dish are reflected directly toward the focus.

Page 21: Copyright © 2011 Pearson Education, Inc. Slide 6.1-1

Copyright © 2011 Pearson Education, Inc. Slide 6.1-21

6.1 An Application of Parabolas

(a) Determine an equation describing the cross section.(b) The receiver must be placed at the focus of the parabola.

How far from the vertex of the parabolic dish should the receiver be placed?

Solution(a) The parabola will have the form y = ax2 (vertex at the

origin) and pass through the point ).32,105(32,2210

2

2

2

32 (105)32 32

The cross section can be described by105 11,025

32.

11,025

a

a

y x

Page 22: Copyright © 2011 Pearson Education, Inc. Slide 6.1-1

Copyright © 2011 Pearson Education, Inc. Slide 6.1-22

6.1 An Application of Parabolas

(b) Since

The receiver should be placed at (0, 86.1), or 86.1 feet above the vertex.

,025,11

32 2xy

.1.86128

025,1132025,11

4

14

c

c

ac