1. Copyright © 2011 Pearson Education, Inc. Conic Sections CHAPTER 13.1Parabolas and Circles 13.2Ellipses and Hyperbolas 13.3Nonlinear Systems of Equations 13.4Nonlinear Inequalities and Systems of Inequalities 13

2. Copyright © 2011 Pearson Education, Inc. Parabolas and Circles 13.1 1.Graph parabolas of the form x = a ( y – k ) 2 + h. 2.Find the distance and midpoint between two points. 3.Graph circles of the form ( x – h ) 2 + ( y – k ) 2 = r 2. 4.Find the equation of a circle with a given center and radius. 5.Graph circles of the form x 2 + y 2 + dx + ey + f = 0.

3. Slide 13- 3 Copyright © 2011 Pearson Education, Inc. Conic Section: A curve in a plane that is the result of intersecting the plane with a cone. More specifically, a circle, ellipse, parabola, or hyperbola.

4. Slide 13- 4 Copyright © 2011 Pearson Education, Inc. Example 1 Given determine whether the graph opens upward or downward, find the vertex and axis of symmetry, and draw the graph. Solution a = 2, h = 2, and k =  3. a is positive the parabola opens upward. The vertex is at (2,  3) and the axis of symmetry is x = 2.

5. Slide 13- 5 Copyright © 2011 Pearson Education, Inc. Equations of Parabolas Opening Left or Right The graph of an equation in the form x = a(y – k) 2 + h is a parabola with vertex at (h, k). The parabola opens to the right if a > 0 and to the left if a < 0. The equation of the axis of symmetry is y = k. y = k Vertex (h, k) x = a(y – k) 2 + h, where a > 0 x = a(y – k) 2 + h, where a < 0

6. Slide 13- 6 Copyright © 2011 Pearson Education, Inc. Example 2b Determine whether the graphs opens left or right, find the vertex and axis of symmetry, and draw the graph. x = 2y 2 + 4y – 1 Solution Parabola opens to the right because a = 2, which is positive. To find the vertex and axis of symmetry we need to write the equation in the form x = a(y – k) 2 + h. x = 2y 2 + 4y – 1 x + 1 = 2y 2 + 4y Add 1 to both sides. x + 1 = 2(y 2 + 2y) Factor out the common factor, 2. x + 1 + 2 = 2(y 2 + 2y + 1) Complete the square. x + 3 = 2(y + 1) 2 Simplify the left side and factor the right. x = 2(y + 1) 2 – 3 Subtract 3 from both sides of the equation.

7. Slide 13- 7 Copyright © 2011 Pearson Education, Inc. continued x = 2(y + 1) 2 – 3 The vertex is at the point with coordinates (–3, –1). The axis of symmetry is y = –1. xy –10 –2 51 5–3

8. Slide 13- 8 Copyright © 2011 Pearson Education, Inc. We can use the Pythagorean theorem along with the square root principle to derive a formula for finding the distance between two points in the coordinate plane. The Distance Formula The distance, d, between two points with coordinates (x 1, y 1 ) and (x 2, y 2 ), can be found using the formula:

9. Slide 13- 9 Copyright © 2011 Pearson Education, Inc. Example 3 Find the distance between (–3, 7) and (–11, 9). If the distance is an irrational number, also give a decimal approximation rounded to three places. Solution Use the distance formula with (–3, 7) as (x 1, y 1 ) and (–11, 9) as (x 2, y 2 ). It does not matter which pair is (x 1, y 1 ) and which is (x 2, y 2 ).

10. Slide 13- 10 Copyright © 2011 Pearson Education, Inc. continued Replace the variables with the corresponding numbers. Subtract within parentheses. Evaluate exponential forms, then add. Exact answer Approximate answer (–3, 7) as (x 1, y 1 ) and (–11, 9) as (x 2, y 2 )

11. Slide 13- 11 Copyright © 2011 Pearson Education, Inc. Midpoint Formula If the coordinates of the endpoints of a line segment are (x 1, y 1 ) and (x 2, y 2 ), the coordinates of the midpoint are To find the coordinates of the midpoint of a line segment, we average the x-coordinates and the y- coordinates of the endpoints.

12. Slide 13- 12 Copyright © 2011 Pearson Education, Inc. Example 4 Find the midpoint of the line segment whose endpoints are (–4, 3) and (6, 7). Solution

13. Slide 13- 13 Copyright © 2011 Pearson Education, Inc. Circle: A set of points that are equally distant from a central point. The central point is the center. Radius: The distance from the center of a circle to any point on the circle.

14. Slide 13- 14 Copyright © 2011 Pearson Education, Inc. Standard Form of the Equation of a Circle The equation of a circle with center at (h, k) and radius r is (x – h) 2 + (y – k) 2 = r 2.

15. Slide 13- 15 Copyright © 2011 Pearson Education, Inc. Example 5a Find the center and radius of the circle and draw the graph. (x + 2) 2 + (y  1) 2 = 9 Solution Since h = –2 and k = 1, the center is (  2, 1). Since 9 = 3 2, the radius is 3. r = 3 (-2, 1)

16. Slide 13- 16 Copyright © 2011 Pearson Education, Inc. Example 6a Write the equation of the circle in standard form. center: (–5, 3); radius 4 Solution The center is at (–5, 3), h = –5 and k = 3. (x – h) 2 + (y – k) 2 = r 2 (x  (  5)) 2 + (y – 3) 2 = 4 2 (x + 5) 2 + (y – 3) 2 = 16

17. Slide 13- 17 Copyright © 2011 Pearson Education, Inc. Example 6b Write the equation of the circle in standard form. center: (0, 0); radius 5 Solution The center is at (0, 0). x 2 + y 2 = r 2 x 2 + y 2 = 5 2 x 2 + y 2 = 25

18. Slide 13- 18 Copyright © 2011 Pearson Education, Inc. Example 7 Find the center and radius of the circle whose equation is x 2 + 8x + y 2  6y + 16 = 0 and draw the graph. Solution The center is at (  4, 3) and the radius is 3.

19. Slide 13- 19 Copyright © 2011 Pearson Education, Inc. Find the direction the parabola opens, the coordinates of the vertex, and the equation of the axis of symmetry. x = 2y 2 – 8y + 13. a) opens right; vertex (5, –2); y = –2 b) opens right; vertex (5, 2); y = 2 c) opens right; vertex (  5, 2); y = 2 d) opens left; vertex (5, 2); y = 2

20. Slide 13- 20 Copyright © 2011 Pearson Education, Inc. Find the direction the parabola opens, the coordinates of the vertex, and the equation of the axis of symmetry. x = 2y 2 – 8y + 13. a) opens right; vertex (5, –2); y = –2 b) opens right; vertex (5, 2); y = 2 c) opens right; vertex (  5, 2); y = 2 d) opens left; vertex (5, 2); y = 2

21. Slide 13- 21 Copyright © 2011 Pearson Education, Inc. Find the approximate distance between the points (5, 9) and (10, –1). a) 8.660 b) 9.434 c) 11.180 d) 17

22. Slide 13- 22 Copyright © 2011 Pearson Education, Inc. Find the approximate distance between the points (5, 9) and (10, –1). a) 8.660 b) 9.434 c) 11.180 d) 17

23. Slide 13- 23 Copyright © 2011 Pearson Education, Inc. Find the coordinates of the center of the circle and the length of its radius (x – 1) 2 + (y – 6) 2 = 4 a) center (1, 6); radius 2 b) center (  1,  6); radius 2 c) center (1, 6); radius 4 d) center (1,  6); radius 2

24. Slide 13- 24 Copyright © 2011 Pearson Education, Inc. Find the coordinates of the center of the circle and the length of its radius (x – 1) 2 + (y – 6) 2 = 4 a) center (1, 6); radius 2 b) center (  1,  6); radius 2 c) center (1, 6); radius 4 d) center (1,  6); radius 2

25. Copyright © 2011 Pearson Education, Inc. Ellipses and Hyperbolas 13.2 1.Graph ellipses. 2.Graph hyperbolas.

26. Slide 13- 26 Copyright © 2011 Pearson Education, Inc. Ellipse: The set of all points the sum of whose distances from two fixed points is constant.

27. Slide 13- 27 Copyright © 2011 Pearson Education, Inc. The Equation of an Ellipse Centered at (0, 0) The equation of an ellipse with center (0, 0), x-intercepts (a, 0) and (  a, 0), and y-intercepts (0, b) and (0,  b) is

28. Slide 13- 28 Copyright © 2011 Pearson Education, Inc. Example 1a Solution Graph the ellipse and label the x- and y-intercepts. Can rewrite the equation as a = 4 and b = 2 The x-intercepts are (–4, 0) and (4, 0). The y-intercepts are (0, –2) and (0, 2).

29. Slide 13- 29 Copyright © 2011 Pearson Education, Inc. Example 1b Graph the ellipse Solution Write the equation in standard form. a = 6 b = 7 (0, 7) (6, 0) (0, -7) (-6, 0) The x-intercepts are (  6, 0) and (6, 0). The y-intercepts are (0, 7) and (0,  7).

30. Slide 13- 30 Copyright © 2011 Pearson Education, Inc. General Equation for an Ellipse The equation of an ellipse with center (h, k) is The ellipse passes through two points that are a units to the right and left of the center, and two points that are b units above and below the center.

31. Slide 13- 31 Copyright © 2011 Pearson Education, Inc. Example 2 Solution h =  2 and k = 1 The center is at (  2, 1). Also, a = 5, b = 4 Passes through two points 4 units to the right and left of (  2, 1). These points are (  6, 1) and (2, 1). Passes through two points 5 units above and below the center. These points are (  2, 6) and (  2,  4). Graph the ellipse. Label the center and the points above and below, to the left and to the right of the center.

32. Slide 13- 32 Copyright © 2011 Pearson Education, Inc. Hyperbola: The set of all points the difference of whose distances from two fixed points remains constant.

33. Slide 13- 33 Copyright © 2011 Pearson Education, Inc. Equations of Hyperbolas in Standard Form The equation of a hyperbola with center (0, 0), x-intercepts (a, 0) and (  a, 0), and no y-intercepts is The equation of a hyperbola with center (0, 0), y-intercepts (0, b) and (0,  b), and no x-intercepts is

34. Slide 13- 34 Copyright © 2011 Pearson Education, Inc. Graphing a Hyperbola in Standard Form 1. Find the intercepts. If the x 2 term is positive, the x- intercepts are (a, 0) and (  a, 0) and there are no y- intercepts. If the y 2 term is positive, the y- intercepts are (0, b) and (0,  b), and there are no x- intercepts. 2. Draw the fundamental rectangle. The vertices are (a, b), (  a, b), (a,  b), and (  a,  b). 3. Draw the asymptotes, which are the extended diagonals of the fundamental rectangle. 4. Draw the graph so that each branch passes through an intercept and approaches the asymptotes the farther the branches are from the origin.

35. Slide 13- 35 Copyright © 2011 Pearson Education, Inc. Example 3a Graph the hyperbola Also show the fundamental rectangle with its corner points labeled, the asymptotes, and the intercepts. Solution The equation can be written as a = 2 and b = 4 The graph has x-intercepts at (  2, 0) and (2, 0). The fundamental rectangle has vertices at (  2, 4), (2, 4), (  2,  4) and (2,  4).

36. Slide 13- 36 Copyright © 2011 Pearson Education, Inc. Example 3b Solution Graph the hyperbola Also show the fundamental rectangle with its corner points labeled, the asymptotes, and the intercepts. The equation can be written as The graph has y-intercepts at (0,  4) and (0, 4). The fundamental rectangle has vertices at (1, 4), (  1, 4), (  1,  4) and (1,  4).

37. Slide 13- 37 Copyright © 2011 Pearson Education, Inc. Match the equation with the correct graph. a)b) c)d)

38. Slide 13- 38 Copyright © 2011 Pearson Education, Inc. Match the equation with the correct graph. a)b) c)d)

39. Slide 13- 39 Copyright © 2011 Pearson Education, Inc. Match the equation with the correct graph. a)b) c)d)

40. Slide 13- 40 Copyright © 2011 Pearson Education, Inc. Match the equation with the correct graph. a)b) c)d)

41. Copyright © 2011 Pearson Education, Inc. Nonlinear Systems of Equations 13.3 1.Solve nonlinear systems of equations using substitution. 2.Solve nonlinear systems of equations using elimination.

42. Slide 13- 42 Copyright © 2011 Pearson Education, Inc. Nonlinear system of equations: A system of equations that contains at least one nonlinear equation. No points of intersection: no solutions One point of intersection: One solution Two points of intersection: two solutions

43. Slide 13- 43 Copyright © 2011 Pearson Education, Inc. Example 1 Solution Solve using the substitution method. x 2 + y 2 = 20, (1) (The graph is a circle.) y – 2x = 0. (2) (The graph is a line.) Solve the second equation for y: y = 2x. (3) Then substitute 2x for y in equation (1) and solve for x: x 2 + (2x) 2 = 20 x 2 + 4x 2 = 20 5x 2 = 20

44. Slide 13- 44 Copyright © 2011 Pearson Education, Inc. continued x 2 = 4 Now substitute these numbers for x in equation (3) and solve for y: For x = 2, y = 2(2) = 4; For x = –2, y = 2(–2) = –4. The solutions are (2, 4) and (–2, –4). Using the principle of square roots

45. Slide 13- 45 Copyright © 2011 Pearson Education, Inc. Example 2 Solution Solve using the elimination method. 3x 2 + 2y 2 = 66, (1) x 2 – y 2 = 7. (2) Multiply equation (2) by 2 and add: 5x 2 = 80 x 2 = 16 3x 2 + 2y 2 = 66 2x 2 – 2y 2 = 14 x =

46. Slide 13- 46 Copyright © 2011 Pearson Education, Inc. x 2 = 16 where x = 4 or –4, thus we can substitute 4 and –4. ( ) 2 – y 2 = 7 16 – y 2 = 7 –y 2 = – 9 y 2 = 9 y =. The system has four solutions: (4, 3), (4, –3), (–4, 3), and (–4, –3). continued

47. Slide 13- 47 Copyright © 2011 Pearson Education, Inc. Solve the system of equations. a) (3, 22), (10, 9) b) (3, 16), (10, 9) c) (  3, 22), (  10, 29) d) (7, 12)

48. Slide 13- 48 Copyright © 2011 Pearson Education, Inc. Solve the system of equations. a) (3, 22), (10, 9) b) (3, 16), (10, 9) c) (  3, 22), (  10, 29) d) (7, 12)

49. Slide 13- 49 Copyright © 2011 Pearson Education, Inc. Solve the system of equations. a) (8,  4), (8, 4) b) c) d) No solution

50. Slide 13- 50 Copyright © 2011 Pearson Education, Inc. Solve the system of equations. a) (8,  4), (8, 4) b) c) d) No solution

51. Copyright © 2011 Pearson Education, Inc. Nonlinear Inequalities and Systems of Inequalities 13.4 1.Graph nonlinear inequalities. 2.Graph the solution set of a system of nonlinear inequalities.

52. Slide 13- 52 Copyright © 2011 Pearson Education, Inc. Graphing Nonlinear Inequalities 1. Graph the related equation. If the inequality symbol is  or , draw the graph as a solid curve. If the inequality symbol is, draw a dashed curve. 2. The graph divides the coordinate plane into at least two regions. Test an ordered pair from each region by substituting it into the inequality. If the ordered pair satisfies the inequality, then shade the region containing that ordered pair.

53. Slide 13- 53 Copyright © 2011 Pearson Education, Inc. Example 1a Graph the inequality. y > 3(x – 2) 2 + 1 Solution We first graph the related equation y = 3(x – 2) 2 + 1. The parabola is dashed because the inequality is >. The graph divides the coordinate plane into two regions. Region 1 Region 2

54. Slide 13- 54 Copyright © 2011 Pearson Education, Inc. continued Test an ordered pair from each region. Region 1: (2, 4) 4 > 3(2 – 2) 2 + 1 4 > 3(0) 2 + 1 4 > 1 Region 1 Region 2 Region 2: (0, 0) 0 > 3(0 – 2) 2 + 1 0 > 3(–2 ) 2 + 1 0 > 13 False, so Region 2 is not the solution set. True, so region 1 is the solution set y > 3(x – 2) 2 + 1

55. Slide 13- 55 Copyright © 2011 Pearson Education, Inc. Example 2a Graph the solution set of the system of inequalities. Solution We begin by graphing The intersection of the two regions is the solution set.

56. Slide 13- 56 Copyright © 2011 Pearson Education, Inc. Graph the inequality. y > (x – 5) 2 – 6 a)b) c)d)

57. Slide 13- 57 Copyright © 2011 Pearson Education, Inc. Graph the inequality. y > (x – 5) 2 – 6 a)b) c)d)

58. Slide 13- 58 Copyright © 2011 Pearson Education, Inc. Graph the solution set of the system of inequalities. a)b) c)d)

59. Slide 13- 59 Copyright © 2011 Pearson Education, Inc. Graph the solution set of the system of inequalities. a)b) c)d)