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Parabolas PowerPoint Presentation

Parabolas

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Parabolas

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  1. Parabolas Warm Up Lesson Presentation Lesson Quiz Holt McDougal Algebra 2 Holt Algebra 2

  2. 1. Given , solve for p when c = Warm Up Find each distance. 2. from (0, 2) to (12, 7) 13 3. from the line y = –6 to (12, 7) 13

  3. Objectives Write the standard equation of a parabola and its axis of symmetry. Graph a parabola and identify its focus, directrix, and axis of symmetry.

  4. Vocabulary focus of a parabola directrix

  5. In Chapter 5, you learned that the graph of a quadratic function is a parabola. Because a parabola is a conic section, it can also be defined in terms of distance.

  6. A parabola is the set of all points P(x, y) in a plane that are an equal distance from both a fixed point, the focus, and a fixed line, the directrix. A parabola has a axis of symmetry perpendicular to its directrix and that passes through its vertex. The vertex of a parabola is the midpoint of the perpendicular segment connecting the focus and the directrix.

  7. Remember! The distance from a point to a line is defined as the length of the line segment from the point perpendicular to the line.

  8. Example 1: Using the Distance Formula to Write the Equation of a Parabola Use the Distance Formula to find the equation of a parabola with focus F(2, 4) and directrix y = –4. PF = PD Definition of a parabola. Distance Formula. Substitute (2, 4) for (x1, y1) and (x, –4) for (x2, y2).

  9. Example 1 Continued Simplify. Square both sides. (x – 2)2 + (y – 4)2 = (y + 4)2 Expand. (x – 2)2 + y2 – 8y + 16 = y2 + 8y + 16 Subtract y2 and 16 from both sides. (x – 2)2 – 8y = 8y Add 8y to both sides. (x – 2)2 = 16y Solve for y.

  10. Check It Out! Example 1 Use the Distance Formula to find the equation of a parabola with focus F(0, 4) and directrix y = –4. PF = PD Definition of a parabola. Distance Formula Substitute (0, 4) for (x1, y1) and (x, –4) for (x2, y2).

  11. Check It Out! Example 1 Continued Simplify. Square both sides. x2 + (y – 4)2 = (y + 4)2 Expand. x2 + y2 – 8y + 16 = y2 + 8y +16 Subtract y2 and 16 from both sides. x2 – 8y = 8y x2 = 16y Add 8y to both sides. Solve for y.

  12. Previously, you have graphed parabolas with vertical axes of symmetry that open upward or downward. Parabolas may also have horizontal axes of symmetry and may open to the left or right. The equations of parabolas use the parameter p. The |p| gives the distance from the vertex to both the focus and the directrix.

  13. Step 1 Because the axis of symmetry is vertical and the parabola opens downward, the equation is in the form 1 4p y = x2 with p < 0. Example 2A: Writing Equations of Parabolas Write the equation in standard form for the parabola.

  14. y = – x2 Step 3 The equation of the parabola is . 1 20 Example 2A Continued Step 2 The distance from the focus (0, –5) to the vertex (0, 0), is 5, so p = –5 and 4p = –20. Check Use your graphing calculator. The graph of the equation appears to match.

  15. Step 1 Because the directrix is a vertical line, the equation is in the form . The vertex is to the right of the directrix, so the graph will open to the right. Example 2B: Writing Equations of Parabolas Write the equation in standard form for the parabola. vertex (0, 0), directrix x = –6

  16. Step 3 The equation of the parabola is . x = y2 1 24 Example 2B Continued Step 2 Because the directrix is x = –6, p = 6 and 4p = 24. Check Use your graphing calculator.

  17. Step 1 Because the directrix is a vertical line, the equation is in the form of . The vertex is to the left of the directrix, so the graph will open to the left. Check It Out! Example 2a Write the equation in standard form for the parabola. vertex (0, 0), directrix x = 1.25

  18. Step 3 The equation of the parabola is Check It Out! Example 2a Continued Step 2 Because the directrix is x = 1.25, p = –1.25 and 4p = –5. Check Use your graphing calculator.

  19. Step 1 Because the axis of symmetry is vertical and the parabola opens downward, the equation is in the form Check It Out! Example 2b Write the equation in standard form for each parabola. vertex (0, 0), focus (0, –7)

  20. Step 3 The equation of the parabola is Check It Out! Example 2b Continued Step 2 The distance from the focus (0, –7) to the vertex (0, 0) is 7, so p = –7 and 4p = –28. Check Use your graphing calculator.

  21. The vertex of a parabola may not always be the origin. Adding or subtracting a value from x or y translates the graph of a parabola. Also notice that the values of p stretch or compress the graph.

  22. Find the vertex, value of p, axis of symmetry, focus, and directrix of the parabola Then graph. 1 y + 3= (x –2)2. 8 1 1 4p 8 Step 2 , so 4p = 8 and p = 2. = Example 3: Graphing Parabolas Step 1 The vertex is (2,–3).

  23. Example 3 Continued Step 3 The graph has a vertical axis of symmetry, with equation x = 2, and opens upward. Step 4 The focus is (2,–3 + 2), or (2, –1). Step 5 The directrix is a horizontal line y = –3 – 2, or y = –5.

  24. 1 1 4p 12 Step 2 , so 4p = 12 and p = 3. = Check It Out! Example 3a Find the vertex, value of p, axis of symmetry, focus, and directrix of the parabola. Then graph. Step 1 The vertex is (1,3).

  25. Check It Out! Example 3a Continued Step 3 The graph has a horizontal axis of symmetry with equation y = 3, and opens right. Step 4 The focus is (1 + 3, 3), or (4,3). Step 5 The directrix is a vertical line x = 1 – 3, or x = –2.

  26. 1 1 1 4p 2 2 Step 2 , so 4p = –2 and p = – . = – Check It Out! Example 3b Find the vertex, value of p axis of symmetry, focus, and directrix of the parabola. Then graph. Step 1 The vertex is (8,4).

  27. Step 4 The focus is or (8,3.5). Step 5 The directrix is a horizontal line or y = 4.5. Check It Out! Example 3b Continued Step 3 The graph has a vertical axis of symmetry, with equation x = 8, and opens downward.

  28. Light or sound waves collected by a parabola will be reflected by the curve through the focus of the parabola, as shown in the figure. Waves emitted from the focus will be reflected out parallel to the axis of symmetry of a parabola. This property is used in communications technology.

  29. x = y2. x = y2, 1 1 132 4p The equation for the cross section is in the form so 4p = 132 and p = 33. The focus should be 33 inches from the vertex of the cross section. Therefore, the feedhorn should be 33 inches long. Example 4: Using the Equation of a Parabola The cross section of a larger parabolic microphone can be modeled by the equation What is the length of the feedhorn?

  30. x = y2. x = y2, 1 1 44 4p The equation for the cross section is in the form so 4p = 44 and p = 11. The focus should be 11 inches from the vertex of the cross section. Therefore, the feedhorn should be 11 inches long. Check It Out! Example 4 Find the length of the feedhorn for a microphone with a cross section equation

  31. 2. Find the vertex, value of p, axis of symmetry, focus, and directrix of the parabola y –2 = (x –4)2, 1 then graph. 12 Lesson Quiz 1. Write an equation for the parabola with focus F(0, 0) and directrix y = 1. vertex: (4, 2); focus: (4,5); directrix: y = –1; p = 3; axis of symmetry: x = 4