College Algebra Fifth Edition James Stewart  Lothar Redlin  Saleem Watson

1. College Algebra Fifth Edition James StewartLothar RedlinSaleem Watson

2. Conic Sections 9

3. Chapter Overview • Conic sections are the curves we get when we make a straight cut in a cone. • For example, if a cone is cut horizontally, the cross section is a circle. • So, a circle is a conic section.

4. Chapter Overview • Other ways of cutting a cone produce parabolas, ellipses, and hyperbolas.

5. Chapter Overview • Our goal in this chapter is to find equations whose graphs are the conic sections. • We already know from Section 2.2 that the graph of the equation x2 + y2 = r2 is a circle. • We will find equations for each of the other conic sections by analyzing their geometric properties.

6. Parabolas 8.1

7. Parabolas • We saw in Section 4.1 that the graph of the equation y =ax2 + bx + c is a U-shaped curve called a parabolathat opens either upward or downward—depending on whether the sign of a is positive or negative. • Here, we study parabolas from a geometric rather than an algebraic point of view.

8. Parabolas • We begin with the geometric definition of a parabola and show how this leads to the algebraic formula that we are already familiar with.

9. Parabola—Geometric Definition • A parabola is the set of points in the plane equidistant from a fixed point F (focus) and a fixed line l(directrix). • The vertex V lies halfway between the focus and the directrix. • The axis of symmetry is the line that runs through the focus perpendicular to the directrix.

10. Parabolas • In this section, we restrict our attention to parabolas that: • Are situated with the vertex at the origin. • Have a vertical or horizontal axis of symmetry. • Parabolas in more general positions will be considered in Section 8.4.

11. Parabolas • If the focus of such a parabola is the point F(0, p), the axis of symmetry must be vertical and the directrix has the equation y = –p. • The figure illustrates the case p > 0.

12. Parabolas • If P(x, y) is any point on the parabola, • The distance from P to the focus F (using the Distance Formula) is: • The distance from P to the directrix is: | y – (–p) | = | y +p |

13. Parabolas • By the definition of a parabola, these two distances must be equal:

14. Parabolas • If p > 0, then the parabola opens upward. • If p < 0, it opens downward. • When x is replaced by –x, the equation remains unchanged. • So, the graph is symmetric about the y-axis.

15. Equations and Graphs of Parabola

16. Equations and Graphs of Parabola • We now summarize what we have just proved about the equation and features of a parabola with a vertical axis.

17. Parabola with Vertical Axis • The graph of the equation x2 = 4py is a parabola with these properties.

18. Parabola with Vertical Axis • The parabola opens: • Upward if p > 0. • Downward if p < 0.

19. E.g. 1—Finding the Equation of a Parabola • Find the equation of the parabola with vertex V(0, 0) and focus F(0, 2), and sketch its graph. • Since the focus is F(0, 2), we conclude that p = 2 (and so the directrix is y = –2). • Thus, the equation is: x2 = 4(2)y x2 = 8y

20. E.g. 1—Finding the Equation of a Parabola • Since p = 2 > 0, the parabola opens upward.

21. E.g. 2—Finding Focus and Directrix from Equation • Find the focus and directrix of the parabola y = –x2, and sketch the graph. • To find the focus and directrix, we put the given equation in the standard form x2= –y. • Comparing this to the general equation x2 = 4py, we see that 4p = –1; so, p = –¼. • The focus is F(0, –¼) and the directrix is y = ¼.

22. E.g. 2—Finding Focus and Directrix from Equation • Here’s the graph of the parabola, together with the focus and the directrix.

23. E.g. 2—Finding Focus and Directrix from Equation • We can also draw the graph using a graphing calculator.

24. Equations and Graphs of Parabola • Reflecting the graph in this figure about the diagonal line y =x has the effect of interchanging the roles of x and y. • This results in a parabola with horizontal axis. • By the same method as before, we can prove the following properties.

25. Parabola with Horizontal Axis • The graph of the equation y2 = 4px is a parabola with these properties.

26. Parabola with Horizontal Axis • The parabola opens: • To the right if p > 0. • To the left if p < 0.

27. E.g. 3—Parabola with Horizontal Axis • A parabola has the equation 6x +y2 = 0 (a) Find the focus and directrix of the parabola, and sketch the graph. (b) Use a graphing calculator to draw the graph.

28. Example (a) E.g. 3—Parabola with Horiz. Axis • We put the given equation in the standard form y2 = –6x. • Comparing this to the general equation y2 = 4px, we see that 4p = –6; so, p = (–3/2). • The focus is F(–3/2, 0) and the directrix is x = 3/2.

29. Example (a) E.g. 3—Parabola with Horiz. Axis • Since p < 0, the parabola opens to the left.

30. Example (b) E.g. 3—Parabola with Horiz. Axis • To draw the graph using a graphing calculator, we need to solve for y. • 6x + y2 = 0 • y2 = –6x

31. Example (b) E.g. 3—Parabola with Horiz. Axis • To obtain the graph of the parabola, we graph both functions.

32. Note • The equation y2 = 4px does not define y as a function of x. • So, to use a graphing calculator to graph a parabola with horizontal axis, we must first solve for y. • This leads to two functions • We need to graph both to get the complete graph of the parabola.

33. Width of Parabola • We can use the coordinates of the focus to estimate the “width” of a parabola when sketching its graph. • The line segment that runs through the focus perpendicular to the axis—with endpoints on the parabola—is called the latus rectum. • Its length is the focal diameter of the parabola.

34. Width of Parabola • From the figure, we can see that the distance from an endpoint Q of the latus rectum to the directrix is |2p|. • So, the distance from Q to the focus must be |2p| as well (by the definition of a parabola). • Hence, the focal diameter is |4p|.

35. Width of Parabola • In the next example, we use the focal diameter to determine the “width” of a parabola when graphing it.

36. E.g. 4—Focal Diameter of a Parabola • Find the focus, directrix, and focal diameter of the parabola y = ½x2, and sketch its graph. • We put the equation in the form x2 = 4py. y = ½x2 x2 = 2y • We see that 4p = 2. • So, the focal diameter is 2.

37. E.g. 4—Focal Diameter of a Parabola • Solving for p gives p = ½. • So, the focus is (0, ½) and the directrix is y = –½ . • Since the focal diameter is 2, the latus rectum extends 1 unit to the left and 1 unit to the right of the focus.

38. E.g. 4—Focal Diameter of a Parabola • Here’s the graph.

39. Family of Parabolas • In the next example, we graph a family of parabolas—to show how changing the distance between the focus and the vertex affects the “width” of a parabola.

40. E.g. 5—Family of Parabolas • (a) Find equations for the parabolas with vertex at the origin and foci • (b) Draw the graphs of the parabolas in (a). • What do you conclude?

41. Example (a) E.g. 5—Family of Parabolas • Since the foci are on the positive y-axis, the parabolas open upward and have equations of the form x2 = 4py. • This leads to the following equations.

42. Example (b) E.g. 5—Family of Parabolas • We see that the closer the focus to the vertex, the narrower the parabola.

43. Applications

44. Applications • Parabolas have an important property that makes them useful as reflectors for lamps and telescopes.

45. Applications • Light from a source placed at the focus of a surface with parabolic cross section will be reflected in such a way that it travels parallel to the axis of the parabola. • Thus, a parabolic mirror reflects the light into a beam of parallel rays.

46. Reflection Property • Conversely, light approaching the reflector in rays parallel to its axis of symmetry is concentrated to the focus. • This reflection property—which can be proved using calculus—is used in the construction of reflecting telescopes.

47. E.g. 6—Focal Point of a Searchlight Reflector • A searchlight has a parabolic reflector that forms a “bowl,” 12 in. wide from rim to rim and 8 in. deep. • If the filament of the light bulb is located at the focus, how far from the vertex of the reflector is it?

48. E.g. 6—Focal Point of a Searchlight Reflector • We introduce a coordinate system and place a parabolic cross section of the reflector so that: • Its vertex is at the origin. • Its axis is vertical.

49. E.g. 6—Focal Point of a Searchlight Reflector • Then, the equation of this parabola has the form x2 = 4py. • We see that the point (6, 8) lies on the parabola. • We use this to find p.

50. E.g. 6—Focal Point of a Searchlight Reflector • 62 = 4p(8) 36 = 32p p = (9/8) • The focus is F(0, (9/8)). • So, the distance between the vertex and the focus is in. • Since the filament is positioned at the focus, it is located in. from the vertex of the reflector.