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2. Conic Sections Parabola

3. Conic Sections - Parabola The intersection of a plane with one nappe of the cone is a parabola.

4. OBJECTIVES:-

6. Conic Sections - Parabola The parabola has the characteristic shape shown above. A parabola is defined to be the “set of points the same distance from a point and a line”.

7. Conic Sections - Parabola The line is called the directrix and the point is called the focus . Focus Directrix

8. Conic Sections - Parabola The line perpendicular to the directrix passing through the focus is the axis of symmetry . The vertex is the point of intersection of the axis of symmetry with the parabola. Focus Directrix Axis of Symmetry Vertex

9. Conic Sections - Parabola The definition of the parabola is the set of points the same distance from the focus and directrix. Therefore, d 1 = d 2 for any point (x, y) on the parabola. Focus Directrix d 1 d 2

10. Finding the Focus and Directrix Parabola

11. Conic Sections - Parabola We know that a parabola has a basic equation y = ax 2 . The vertex is at (0, 0). The distance from the vertex to the focus and directrix is the same. Let’s call it p . Focus Directrix p p y = ax 2

12. Conic Sections - Parabola Find the point for the focus and the equation of the directrix if the vertex is at (0, 0). Focus ( ?, ?) Directrix ??? p p ( 0, 0) y = ax 2

13. Conic Sections - Parabola The focus is p units up from (0, 0), so the focus is at the point (0, p). Focus ( 0, p) Directrix ??? p p ( 0, 0) y = ax 2

14. Conic Sections - Parabola The directrix is a horizontal line p units below the origin. Find the equation of the directrix. Focus ( 0, p) Directrix ??? p p ( 0, 0) y = ax 2

15. Conic Sections - Parabola The directrix is a horizontal line p units below the origin or a horizontal line through the point (0, -p). The equation is y = -p. Focus ( 0, p) Directrix y = -p p p ( 0, 0) y = ax 2

16. Conic Sections - Parabola The definition of the parabola indicates the distance d 1 from any point (x, y) on the curve to the focus and the distance d 2 from the point to the directrix must be equal. Focus ( 0, p) Directrix y = -p ( 0, 0) ( x, y) y = ax 2 d 1 d 2

17. Conic Sections - Parabola However, the parabola is y = ax 2 . We can substitute for y in the point (x, y). The point on the curve is (x, ax 2 ). Focus ( 0, p) Directrix y = -p ( 0, 0) ( x, ax 2 ) y = ax 2 d 1 d 2

18. Conic Sections - Parabola What is the coordinates of the point on the directrix immediately below the point (x, ax 2 )? Focus ( 0, p) Directrix y = -p ( 0, 0) ( x, ax 2 ) y = ax 2 d 1 d 2 ( ?, ?)

19. Conic Sections - Parabola The x value is the same as the point (x, ax 2 ) and the y value is on the line y = -p, so the point must be (x, -p). Focus ( 0, p) Directrix y = -p ( 0, 0) ( x, ax 2 ) y = ax 2 d 1 d 2 ( x, -p)

20. Conic Sections - Parabola d 1 is the distance from (0, p) to (x, ax 2 ). d 2 is the distance from (x, ax 2 ) to (x, -p) and d 1 = d 2 . Use the distance formula to solve for p. Focus ( 0, p) Directrix y = -p ( 0, 0) ( x, ax 2 ) y = ax 2 d 1 d 2 ( x, -p)

21. Conic Sections - Parabola d1 is the distance from (0, p) to (x, ax 2 ). d2 is the distance from (x, ax 2 ) to (x, -p) and d1 = d2. Use the distance formula to solve for p. d 1 = d 2 You finish the rest.

22. Conic Sections - Parabola d1 is the distance from (0, p) to (x, ax 2 ). d2 is the distance from (x, ax 2 ) to (x, -p) and d1 = d2. Use the distance formula to solve for p. d 1 = d 2

23. Conic Sections - Parabola Therefore, the distance p from the vertex to the focus and the vertex to the directrix is given by the formula

24. Conic Sections - Parabola Using transformations, we can shift the parabola y=ax 2 horizontally and vertically. If the parabola is shifted h units right and k units up, the equation would be The vertex is shifted from (0, 0) to (h, k). Recall that when “a” is positive, the graph opens up. When “a” is negative, the graph reflects about the x-axis and opens down.

25. Example 1 Graph a parabola. Find the vertex, focus and directrix.

26. Parabola – Example 1 Make a table of values. Graph the function. Find the vertex, focus, and directrix.

27. Parabola – Example 1 The vertex is (-2, -3). Since the parabola opens up and the axis of symmetry passes through the vertex, the axis of symmetry is x = -2.

28. Parabola – Example 1 Make a table of values. x y -2 -1 0 1 2 3 4 -3 -1 Plot the points on the graph! Use the line of symmetry to plot the other side of the graph.

29. Parabola – Example 1 Find the focus and directrix.

30. Parabola – Example 1 The focus and directrix are “p” units from the vertex where The focus and directrix are 2 units from the vertex.

31. Parabola – Example 1 Focus: (-2, -1) Directrix: y = -5 2 Units

32. Building a Table of Rules Parabola

33. Table of Rules - y = a(x - h) 2 + k a > 0 a < 0 Opens Vertex Focus Axis Directrix Latus Rectum Up Down ( h , k ) ( h , k ) x = h x = h (h, k) (h, k) x = h x = h

34. Table of Rules - x = a(y - k) 2 + h a > 0 a < 0 Opens Vertex Focus Axis Directrix Latus Rectum Right Left ( h , k ) ( h , k ) y = k y = k (h, k) (h, k) y = k y = k

35. Paraboloid Revolution Parabola

39. Example 4 – Satellite Receiver 8 ft 1 ft With a vertex of (0, 0), the point on the diameter would be (4, 1). Fit a parabolic equation passing through these two points. V(0, 0) (4, 1) y = a(x – h) 2 + k Since the vertex is (0, 0), h and k are 0. y = ax 2

40. Example 4 – Satellite Receiver 8 ft 1 ft V(0, 0) (4, 1) y = a x 2 The parabola must pass through the point (4, 1). 1 = a (4) 2 Solve for a . 1 = 16 a

41. Example 4 – Satellite Receiver 8 ft 1 ft V(0, 0) (4, 1) The model for the parabola is: The receiver should be placed at the focus. Locate the focus of the parabola. Distance to the focus is:

42. Example 4 – Satellite Receiver 8 ft 1 ft V(0, 0) (4, 1) The receiver should be placed 4 ft. above the vertex.

43. Sample Problems Parabola

49. Sample Problems Vertex (1, -3) Opens Right Axis y = -3 Focus (4, -3) Directrix x = -2