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Conic Sections

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

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Conic Sections

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  1. Conic Sections Parabola

  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 Focus Directrix The line is called the directrix and the point is called the focus.

  5. Conic Sections - Parabola Axis of Symmetry Focus Vertex Directrix 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.

  6. Conic Sections - Parabola Focus d1 Directrix d2 The definition of the parabola is the set of points the same distance from the focus and directrix. Therefore, d1 = d2 for any point (x, y) on the parabola.

  7. Finding the Focus and Directrix Parabola

  8. Conic Sections - Parabola Focus y = ax2 p Directrix p We know that a parabola has a basic equation y = ax2. The vertex is at (0, 0). The distance from the vertex to the focus and directrix is the same. Let’s call it p.

  9. Conic Sections - Parabola Focus( ?, ?) y = ax2 p Directrix ??? ( 0, 0) p Find the point for the focus and the equation of the directrix if the vertex is at (0, 0).

  10. Conic Sections - Parabola Focus( 0, p) y = ax2 p Directrix ??? ( 0, 0) p The focus is p units up from (0, 0), so the focus is at the point (0, p).

  11. Conic Sections - Parabola Focus( 0, p) y = ax2 p Directrix ??? ( 0, 0) p The directrix is a horizontal line p units below the origin. Find the equation of the directrix.

  12. Conic Sections - Parabola Focus( 0, p) y = ax2 p Directrixy = -p ( 0, 0) p 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.

  13. Conic Sections - Parabola ( x, y) Focus( 0, p) d1 y = ax2 d2 Directrix y = -p ( 0, 0) The definition of the parabola indicates the distance d1 from any point (x, y) on the curve to the focus and the distance d2 from the point to the directrix must be equal.

  14. Conic Sections - Parabola ( x, ax2) Focus( 0, p) d1 y = ax2 d2 Directrix y = -p ( 0, 0) However, the parabola is y = ax2. We can substitute for y in the point (x, y). The point on the curve is (x, ax2).

  15. Conic Sections - Parabola ( x, ax2) Focus( 0, p) d1 y = ax2 d2 Directrix y = -p ( 0, 0) ( ?, ?) What is the coordinates of the point on the directrix immediately below the point (x, ax2)?

  16. Conic Sections - Parabola ( x, ax2) Focus( 0, p) d1 y = ax2 d2 Directrix y = -p ( 0, 0) ( x, -p) The x value is the same as the point (x, ax2) and the y value is on the line y = -p, so the point must be (x, -p).

  17. Conic Sections - Parabola ( x, ax2) Focus( 0, p) d1 y = ax2 d2 Directrix y = -p ( 0, 0) ( x, -p) d1 is the distance from (0, p) to (x, ax2). d2 is the distance from (x, ax2) to (x, -p) and d1 = d2. Use the distance formula to solve for p.

  18. Conic Sections - Parabola d1 is the distance from (0, p) to (x, ax2). d2 is the distance from (x, ax2) to (x, -p) and d1 = d2. Use the distance formula to solve for p. d1 = d2 You finish the rest.

  19. Conic Sections - Parabola d1 is the distance from (0, p) to (x, ax2). d2 is the distance from (x, ax2) to (x, -p) and d1 = d2. Use the distance formula to solve for p. d1 = d2

  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=ax2 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 x y -2 -1 0 1 2 3 4 -3 Make a table of values. -1 Plot the points on the graph!Use the line of symmetry to plot the other side of the graph.

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

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

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

  29. Latus Rectum Parabola

  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. LatusRectum Focus y = ax2 Vertex(0, 0)

  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 LatusRectum Focus y = ax2 Vertex(0, 0)

  32. Conic Sections - Parabola Using the axis of symmetry and the y-value of the focus, the endpoints of the latus rectum must be LatusRectum Vertex(0, 0) y = ax2

  33. Conic Sections - Parabola Since the equation of the parabola is y = ax2, substitute for y and solve for x.

  34. Conic Sections - Parabola Replacing x, the endpoints of the latus rectum are and LatusRectum Vertex(0, 0) y = ax2

  35. Conic Sections - Parabola The length of the latus rectum is LatusRectum Vertex(0, 0) y = ax2

  36. Conic Sections - Parabola Given the value of “a” in the quadratic equation y = a (x – h)2 + k, the length of the latus rectum is • An alternate method to graphing a parabola with the latus rectum is to: • Plot the vertex and axis of symmetry • Plot the focus and directrix. • Use the length of the latus rectum to plot two points on the parabola. • Draw the parabola.

  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. Directrixy=6 Axisx=1 V(1, 2) Focus(1, -2) Latus Rectum

  41. Parabola – Example 2 The graph of the parabola Axisx=1 Directrixy=6 V(1, 2) Focus(1, -2) Latus Rectum

  42. x = ay2 Parabola Graphing and finding the vertex, focus, directrix, axis of symmetry and latus rectum.

  43. Parabola – Graphing x = ay2 Graph x = 2y2 by constructing a table of values. x y 18 -3 -2 -1 0 1 2 3 8 2 0 Graph x = 2y2 by plotting the points in the table. 2 8 18

  44. Parabola – Graphing x = ay2 Graph the table of values.

  45. Parabola – Graphing x = ay2 One could follow a similar proof to show the distancefrom the vertex to the focus and directrix to be . Similarly, the length of the latus rectum can be shown to be .

  46. Parabola – Graphing x = ay2 Graphing the axis of symmetry, vertex, focus, directrix and latus rectum. Directrix V(0,0) Focus Axis y=0

  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 We have just seen that a parabola x = ay2 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. 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 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.

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