1. Slide 10- 1 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

2. Conic Sections 10.1Conic Sections: Parabolas and Circles 10.2Conic Sections: Ellipses 10.3Conic Sections: Hyperbolas 10.4Nonlinear Systems of Equations 10

3. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Conic Sections: Parabolas and Circles Parabolas The Distance and Midpoint Formulas Circles 10.1

4. Slide 10- 4 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley This section and the next two examine curves formed by cross sections of cones. These are all graphs of Ax 2 + By 2 + Cxy + Dx + Ey + F = 0. The constants A, B, C, D, E, and F determine which of the following shapes will serve as the graph.

5. Slide 10- 5 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

6. Slide 10- 6 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Parabolas When a cone is cut as shown in the first figure on the previous slide, the conic section formed is a parabola. Parabolas have many applications in electricity, mechanics, and optics.

7. Slide 10- 7 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Equation of a Parabola A parabola with a vertical axis of symmetry opens upward or downward and has an equation that can be written in the form y = ax 2 + bx + c. A parabola with a horizontal axis of symmetry opens right or left and has an equation that can be written in the form x = ay 2 + by + c.

8. Slide 10- 8 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Parabolas with equations of the form f (x) = ax 2 + bx + c were graphed in Chapter 8. Graph: y = x 2 – 6x + 10. y = (x 2 – 6x) + 10 y = (x 2 – 6x + 9 – 9) + 10 y = (x 2 – 6x + 9) + (–9 + 10) y = (x – 3) 2 + 1. The vertex is at (3, 1). Example Solution Complete the square to find the vertex:

9. Slide 10- 9 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued xy 3240532405 1 2 10 5 Calculate and plot some points on each side of the vertex. As expected for a positive coefficient of x 2, the graph opens upward.

10. Slide 10- 10 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley A second way to find the vertex of a parabola is to recall that the x-coordinate of the vertex of the parabola given by y = ax 2 + bx + c is –b/(2a). For the last example, y = x 2 – 6x + 10, the x-coordinate of the vertex is – (–6)/[2(1)] = 6/2 = 3. To find the y-coordinate of the vertex, we substitute 3 for x: y = 3 2 – 6(3) + 10 = 1.

11. Slide 10- 11 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley To Graph an Equation of the Form y = ax 2 + bx + c 1.Find the vertex (h, k) either by completing the square to find an equivalent equation y = a(x – h) 2 + k; or by using –b/(2a) to find the x-coordinate and substituting to find the y-coordinate.

12. Slide 10- 12 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 2.Choose other values for x on each side of the vertex, and compute the corresponding y-values. 3.The graph opens upward for a > 0 and downward for a < 0.

13. Slide 10- 13 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Equations of the form x = ay 2 + by + c represent horizontal parabolas. These parabolas open to the right for a > 0, open to the left for a < 0, and have axes of symmetry parallel to the x-axis.

14. Slide 10- 14 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solution Graph: x = –y 2 + 6y – 7. x = – y 2 + 6y – 7 x = –(y – 3) 2 + 2. x = –(y 2 – 6y + 9) – 7 – (–9) x = –(y 2 – 6y ) – 7 The vertex is at (2, 3) and the parabola opens to the left. Complete the square to find the vertex:

15. Slide 10- 15 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued xy -7 -2 2 01530153 To plot points, choose values for y and compute the values for x.

16. Slide 10- 16 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley To Graph an Equation of the Form x = ay 2 + by + c 1.Find the vertex (h, k) either by completing the square to find an equivalent equation x = a(y – k) 2 + h; or by using –b/(2a) to find the y-coordinate and substituting to find the x-coordinate.

17. Slide 10- 17 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 2.Choose other values for y that are above and below the vertex, and compute the corresponding x-values. 3.The graph opens to the right if a > 0 and to the left if a < 0.

18. Slide 10- 18 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The distance between the points (x 1, y 1 ) and (x 2, y 1 ) on a horizontal line is |x 2 – x 1 |. Similarly, the distance between the points (x 2, y 1 ) and (x 2, y 2 ) on a vertical line is |y 2 – y 1 |. The Distance Formula

19. Slide 10- 19 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Now consider any two points (x 1, y 1 ) and (x 2, y 2 ). these points, along with (x 2, y 1 ), describe a right triangle. The lengths of the legs are |x 2 – x 1 | and |y 2 – y 1 |.

20. Slide 10- 20 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley We find d, the length of the hypotenuse, by using the Pythagorean theorem: d 2 = |x 2 – x 1 | 2 + |y 2 – y 1 | 2. Since the square of a number is the same as the square of its opposite, we can replace the absolute-value signs with parentheses: d 2 = (x 2 – x 1 ) 2 + (y 2 – y 1 ) 2.

21. Slide 10- 21 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Distance Formula The distance d between any two points (x 1, y 1 ) and (x 2, y 2 ) is given by

22. Slide 10- 22 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solution Find the distance between (3, 1) and (5, –6). Find an exact answer and an approximation to three decimal places. Substitute into the distance formula: Substituting This is exact. Approximation

23. Slide 10- 23 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Midpoint Formula The distance formula is needed to verify a formula for the coordinates of the midpoint of a segment connecting two points.

24. Slide 10- 24 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Midpoint Formula If the endpoints of a segment are (x 1, y 1 ) and (x 2, y 2 ), then the coordinates of the midpoint are (To locate the midpoint, average the x-coordinates and average the y-coordinates.) (x 1, y 1 ) (x 2, y 2 ) x y

25. Slide 10- 25 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solution Find the midpoint of the segment with endpoints (3, 1) and (5, –6). Using the midpoint formula, we obtain

26. Slide 10- 26 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Circles One conic section, the circle, is a set of points in a plane that are a fixed distance r, called the radius (plural, radii), from a fixed point (h, k), called the center. If (x, y) is a point on the circle, then by the definition of a circle and the distance formula, it follows that

27. Slide 10- 27 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Squaring both sides gives the equation of a circle in standard form.

28. Slide 10- 28 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Equation of a Circle (Standard Form) The equation of a circle, centered at (h, k), with radius r, is given by

29. Slide 10- 29 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Note that for h = 0 and k = 0, the circle is centered at the origin. Otherwise, the circle is translated |h| units horizontally and |k| units vertically.

30. Slide 10- 30 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solution Find an equation of the circle having center (1, 7) and radius 4. Using the standard form, we obtain (x – 1) 2 + (y – 7) 2 = 4 2, or (x – 1) 2 + (y – 7) 2 = 16.

31. Slide 10- 31 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solution Find the center and the radius and then graph the circle x 2 + y 2 + 2x – 6y + 6 = 0. (x + 1) 2 + (y – 3) 2 = 4 x 2 + 2x + y 2 – 6y = –6 x 2 + 2x + 1 + y 2 – 6y + 9 = –6 + 1 + 9 (x – (–1)) 2 + (y – 3) 2 = 2 2 Complete the square twice to write the equation in standard form.

32. Slide 10- 32 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The center is at (–1, 3) and the radius is 2. Solution continued