2. Conic Sections - Parabola The intersection of a plane with one nappe of the cone is a parabola.
3. 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”.
4. Conic Sections - Parabola The line is called the directrix and the point is called the focus . Focus Directrix
5. 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
6. 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
8. 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
9. 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
10. 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
11. 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
12. 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
13. 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
14. 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
15. 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 ( ?, ?)
16. 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)
17. 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)
18. 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.
19. 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
20. Conic Sections - Parabola Therefore, the distance p from the vertex to the focus and the vertex to the directrix is given by the formula
21. 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.
22. Example 1 Graph a parabola. Find the vertex, focus and directrix.
23. Parabola – Example 1 Make a table of values. Graph the function. Find the vertex, focus, and directrix.
24. 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.
25. 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.
30. Conic Sections - Parabola The latus rectum is the line segment passing through the focus, perpendicular to the axis of symmetry with endpoints on the parabola. y = ax 2 Focus Vertex (0, 0) Latus Rectum
31. Conic Sections - Parabola In the previous set, we learned that the distance from the vertex to the focus is 1/(4a). Therefore, the focus is at y = ax 2 Focus Vertex (0, 0) Latus Rectum
32. Conic Sections - Parabola Using the axis of symmetry and the y-value of the focus, the endpoints of the latus rectum must be y = ax 2 Vertex (0, 0) Latus Rectum
33. Conic Sections - Parabola Since the equation of the parabola is y = ax 2 , substitute for y and solve for x.
34. Conic Sections - Parabola Replacing x, the endpoints of the latus rectum are y = ax 2 Vertex (0, 0) Latus Rectum and
35. Conic Sections - Parabola The length of the latus rectum is y = ax 2 Vertex (0, 0) Latus Rectum
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37. Example 2 Graph a parabola using the vertex, focus, axis of symmetry and latus rectum.
38. Parabola – Example 2 Find the vertex, axis of symmetry, focus, directrix, endpoints of the latus rectum and sketch the graph.
39. Parabola – Example 2 The vertex is at (1, 2) with the parabola opening down. The focus is 4 units down and the directrix is 4 units up. The length of the latus rectum is
40. Parabola – Example 2 Find the vertex, axis of symmetry, focus, directrix, endpoints of the latus rectum and sketch the graph. V(1, 2) Directrix y=6 Focus (1, -2) Latus Rectum Axis x=1
41. Parabola – Example 2 The graph of the parabola V(1, 2) Directrix y=6 Focus (1, -2) Latus Rectum Axis x=1
42. x = ay 2 Parabola Graphing and finding the vertex, focus, directrix, axis of symmetry and latus rectum.
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47. x = a(y – k) 2 + h Graphing and finding the vertex, focus, directrix, axis of symmetry and latus rectum.
48. Parabola – x = a(y – k) 2 + h When horizontal and vertical transformations are applied, a vertical shift of k units and a horizontal shift of h units will result in the equation: x = a(y – k) 2 + h We have just seen that a parabola x = ay 2 opens to the right when a is positive. When a is negative, the graph will reflect about the y-axis and open to the left. Note: In both cases of the parabola, the x always goes with h and the y always goes with k.
49. Example 3 Graphing and finding the vertex, focus, directrix, axis of symmetry and latus rectum.
50. Parabola – Example 3 Graph the parabola by finding the vertex, focus, directrix and length of the latus rectum. What is the vertex? Remember that inside the “function” we always do the opposite. So the graph moves -1 in the y direction and -2 in the x direction. The vertex is (-2, -1) What is the direction of opening? The parabola opens to the left since it is x= and “a” is negative.
51. Parabola – Example 3 Graph the parabola by finding the vertex, focus, directrix and length of the latus rectum. What is the distance to the focus and directrix? The distance is The parabola opens to the left with a vertex of (-2, -1) and a distance to the focus and directrix of ½. Begin the sketch of the parabola.
52. Parabola – Example 3 The parabola opens to the left with a vertex of (-2, -1) and a distance to the focus and directrix of ½. Begin the sketch of the parabola. Vertex? (-2, -1) Focus? (-2.5, -1) Directrix? x = -1.5
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54. Parabola – Example 3 Construct the latus rectum with a length of 2. Vertex? (-2, -1) Focus? (-2.5, -1) Directrix? x = -1.5 Latus Rectum? 2 Construct the parabola.
55. Parabola – Example 3 The parabola is: Vertex? (-2, -1) Focus? (-2.5, -1) Directrix? x = -1.5 Latus Rectum? 2
57. 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
58. 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
64. 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
65. 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
66. 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:
67. Example 4 – Satellite Receiver 8 ft 1 ft V(0, 0) (4, 1) The receiver should be placed 4 ft. above the vertex.
