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

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Presentation on theme: "Copyright © 2011 Pearson Education, Inc. Slide 6.1-1."— Presentation transcript:

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

2 Copyright © 2011 Pearson Education, Inc. Slide 6.1-2 Chapter 6: Analytic Geometry 6.1Circles and Parabolas 6.2Ellipses and Hyperbolas 6.3Summary of the Conic Sections 6.4Parametric Equations

3 Copyright © 2011 Pearson Education, Inc. Slide 6.1-3 6.1 Circles and Parabolas Conic Sections –Parabolas, circles, ellipses, hyperbolas

4 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.

5 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

6 Copyright © 2011 Pearson Education, Inc. Slide 6.1-6 6.1 Finding the Equation of a Circle ExampleFind 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. SolutionSubstitute h = –3, k = 4, and r = 6 into the equation of a circle.

7 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

8 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

9 Copyright © 2011 Pearson Education, Inc. Slide 6.1-9 6.1 Graphing Circles with the Graphing Calculator TECHNOLOGY NOTES: –Graphs in a nondecimalwindow may not be connected. –Graphs in a rectangular (non-square) window look like an ellipse.

10 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.

11 Copyright © 2011 Pearson Education, Inc. Slide 6.1-11 6.1 Finding the Center and Radius of a Circle ExampleFind the center and radius of the circle with equation SolutionOur goal is to obtain an equivalent equation of the form We complete the square in both x and y. The circle has center (3, –2) with radius 3.

12 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.

13 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.

14 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 x 2 = 4cy. The parabola has vertex (0, 0), vertical axis x = 0, and opens upward if c > 0 or downward if c < 0.

15 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 y 2 = 4cx for y: Let and graph each half of the parabola. The parabola with focus (c, 0) and directrix x = –c has equation y 2 = 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.

16 Copyright © 2011 Pearson Education, Inc. Slide 6.1-16 6.1 Determining Information about Parabolas from Equations ExampleFind the focus, directrix, vertex, and axis of each parabola. (a) Solution (a) 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.

17 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.

18 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.

19 Copyright © 2011 Pearson Education, Inc. Slide 6.1-19 6.1 Writing Equations of Parabolas ExampleWrite an equation for the parabola with vertex (1, 3) and focus (–1, 3). SolutionFocus lies left of the vertex implies the parabola has -a horizontal axis, and -opens to the left. Distance between vertex and focus is 1–(–1) = 2, so c = –2.

20 Copyright © 2011 Pearson Education, Inc. Slide 6.1-20 6.1 An Application of Parabolas ExampleThe 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.

21 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 = ax 2 (vertex at the origin) and pass through the point

22 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.


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