1. 6.1 - 1 10 TH EDITION LIAL HORNSBY SCHNEIDER COLLEGE ALGEBRA

2. 6.1 - 2 6.1 Parabolas Conic Sections Horizontal Parabolas Geometric Definition and Equations of Parabolas An Application of Parabolas

3. 6.1 - 3 Conic Sections Parabolas, circles, ellipses, and hyperbolas form a group of curves known as conic sections, because they are the results of intersecting a cone with a plane. The graphics shown on the next slide illustrate these curves. We studied circles and some parabolas (those that open up or down, that is, vertical parabolas) in Chapters 2 and 3.

4. 6.1 - 4 Conic Sections

5. 6.1 - 5 Horizontal Parabolas From Chapter 3, we know that the graph of the equation is a parabola with vertex (h,k) and the vertical line x = h as axis. If we subtract k from each side, this equation becomes Subtract k. (1)

6. 6.1 - 6 Horizontal Parabolas Subtract k. (1) The equation also has a parabola as its graph. While the graph of y – k = a(x – h) 2 has a vertical axis, the graph of x – h = a(y – k) 2 has a horizontal axis. The graph of the first equation is the graph of a function (specifically a quadratic function), while the graph of the second equation is not.

7. 6.1 - 7 Parabola with Horizontal Axis The parabola with vertex (h, k) and the horizontal line y = k as axis has an equation of the form The parabola opens to the right if a > 0 and to the left if a < 0.

8. 6.1 - 8 Note When the vertex (h, k) is (0, 0) and a = 1 in (1) (2) and the equations y = x 2 and x = y 2 result. Notice that the graphs are mirror images of each other with respect to the line y = x.

9. 6.1 - 9 Example 1 GRAPHING A PARABOLA WITH A HORIZONTAL AXIS Graph x + 3 = (y – 2) 2. Give the domain and range. Solution The graph of x + 3 = (y – 2) 2 or x – (–3) = (y – 2) 2 has vertex (–3, 2) and opens to the right because a = 1, and 1 > 0. Plotting a few additional points gives the graph shown in the next slide.

10. 6.1 - 10 Example 1 GRAPHING A PARABOLA WITH A HORIZONTAL AXIS Note that the graph is symmetric about its axis, The domain is [–3,  ), and the range is (– ,  ). xy – 3– 3 2 – 2– 2 3 – 2– 2 1 14 10

11. 6.1 - 11 GRAPHING A PARABOLA WITH A HORIZONTAL AXIS When an equation of a horizontal parabola is given in the form completing the square on y allows us to write the equation in the form and more easily find the vertex.

12. 6.1 - 12 Example 2 GRAPHING A PARABOLA WITH A HORIZONTAL AXIS Graph. Give the domain and range. Solution Factor out 2. Complete the square;

13. 6.1 - 13 Example 2 GRAPHING A PARABOLA WITH A HORIZONTAL AXIS Distributive property Factor; simplify.

14. 6.1 - 14 Example 2 GRAPHING A PARABOLA WITH A HORIZONTAL AXIS Subtract ½. The vertex of the parabola is The axis is the horizontal line y = k, or y = –3/2. Using the vertex and the axis and plotting a few additional points gives the graph. The domain is [½,  ) and the range is (– ,  ).

15. 6.1 - 15 Parabolas A parabola is the 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 is called the directrix of the parabola.

16. 6.1 - 16 Geometric Definition and Equations of Parabolas The axis of symmetry of a parabola passes through the focus and is perpendicular to the directrix. The vertex is the midpoint of the line segment joining the focus and directrix on the axis.

17. 6.1 - 17 Geometric Definition and Equations of Parabolas We can find an equation of a parabola from the preceding definition. Let p represent the directed distance from the vertex to the focus. Then the directrix is the line y = – p and the focus is the point F(0, p).

18. 6.1 - 18 Geometric Definition and Equations of Parabolas To find the equation of the set of points that are the same distance from the line y = – p and the point (0, p), choose one such point P and give it coordinates (x, y). Since d(P, F) and d(P, D) must be equal, using the distance formula gives the following:

19. 6.1 - 19 Geometric Definition and Equations of Parabolas Distance formula Square both sides and multiply. Remember the middle term.

20. 6.1 - 20 Parabola with Vertical Axis and Vertex (0, 0) The parabola with focus (0, p) and directrix y = – p has equation The parabola has vertical axis x = 0 and opens up if p > 0 or down if p < 0.

21. 6.1 - 21 The parabola with focus (p, 0) and directrix x = – p has equation Parabola with Horizontal Axis and Vertex (0, 0) The parabola has horizontal axis y = 0 and opens to the right if p > 0 or to the left if p < 0.

22. 6.1 - 22 Example 3 DETERMINING INFORMATION ABOUT PARABOLAS FROM THEIR EQUATIONS Find the focus, directrix, vertex, and axis of each parabola. Then use this information to graph the parabola. a. Solution The equation x 2 = 8y has the form x 2 = 4py, so 4p = 8, from which p = 2. Since the x-term is squared, the parabola is vertical, with focus (0, p) = (0, 2) and directrix y = – 2. The vertex is (0, 0) and the axis of the parabola is the y-axis.

23. 6.1 - 23 Example 3 DETERMINING INFORMATION ABOUT PARABOLAS FROM THEIR EQUATIONS

24. 6.1 - 24 Example 3 DETERMINING INFORMATION ABOUT PARABOLAS FROM THEIR EQUATIONS Find the focus, directrix, vertex, and axis of each parabola. Then use this information to graph the parabola. b. Solution The equation y 2 = – 28x has the form y 2 = 4px, with 4p = – 28, so p = – 7. The parabola is horizontal, with focus (– 7, 0), directrix x = 7, vertex (0, 0) and x-axis as axis of the parabola. Since p is negative, the graph opens to the left.

25. 6.1 - 25 Example 3 DETERMINING INFORMATION ABOUT PARABOLAS FROM THEIR EQUATIONS

26. 6.1 - 26 Example 4 WRITING EQUATIONS OF PARABOLAS Write an equation for each parabola. Solution a. focus (⅔, 0) and vertex at the origin Since the focus (⅔, 0) and the vertex (0, 0) are both on the x-axis, the parabola is horizontal. It opens to the right because p = ⅔ is positive. The equation, which will have the form y 2 = 4px is

27. 6.1 - 27 Example 4 WRITING EQUATIONS OF PARABOLAS

28. 6.1 - 28 Example 4 WRITING EQUATIONS OF PARABOLAS Write an equation for each parabola. Solution b. The parabola will have an equation of the form x 2 = 4py because the axis is vertical and the vertex is (0, 0). Since the point (– 2, 12) is on the graph, it must satisfy the equation. vertical axis, vertex at the origin, through the point (– 2, 12)

29. 6.1 - 29 Example 4 WRITING EQUATIONS OF PARABOLAS Let x = – 2 and y = 12. Multiply. Solve for p. Thus, Let p =. which gives the equation

30. 6.1 - 30 Equation Forms for Translated Parabolas A parabola with vertex (h, k) has an equation of the form Vertical axis or Horizontal axis where the focus is distance  p  from the vertex.

31. 6.1 - 31 Example 5 WRITING AN EQUATION OF A PARABOLA Write an equation for the parabola with vertex (1, 3) and focus (– 1, 3) and graph it. Give the domain and range. Solution Since the focus is to the left of the vertex, the axis is horizontal and the parabola opens to the left. The distance between the vertex and the focus is – 1 – 1 or – 2, so p = – 2 (since the parabola opens to the left). The equation of the parabola is shown on the next slide.

32. 6.1 - 32 Example 5 Parabola with horizontal axis Substitute for p, h, and k. Multiply. The domain is (– , 1], and the range is (– ,  ). WRITING AN EQUATION OF A PARABOLA

33. 6.1 - 33 Example 6 MODELING THE REFLECTIVE PROPERTY OF PARABOLAS The Parkes radio telescope has a parabolic dish shape with diameter 210 ft and depth 32 ft. Because of this parabolic shape, distant rays hitting the dish will be reflected directly toward the focus. A cross section of the dish is shown here.

34. 6.1 - 34 Example 6 MODELING THE REFLECTIVE PROPERTY OF PARABOLAS a. Determine an equation that models this cross section by placing the vertex at the origin with the parabola opening up.

35. 6.1 - 35 Example 6 MODELING THE REFLECTIVE PROPERTY OF PARABOLAS Solution Locate the vertex at the origin as shown below. The form of the parabola is x 2 = 4py. The parabola must pass through the point

36. 6.1 - 36 Example 6 MODELING THE REFLECTIVE PROPERTY OF PARABOLAS Solution Vertical parabola Let x = 105 and y = 32. Multiply. Solve for p.

37. 6.1 - 37 Example 6 MODELING THE REFLECTIVE PROPERTY OF PARABOLAS Solution Substitute for p. The cross section can be modeled by the equation

38. 6.1 - 38 Example 6 MODELING THE REFLECTIVE PROPERTY OF PARABOLAS 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 located?

39. 6.1 - 39 Example 6 MODELING THE REFLECTIVE PROPERTY OF PARABOLAS b. The distance between the vertex and the focus is p. In part (a), we found so the receiver should be located at (0, 86.1) or 86.1 ft above the vertex.