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Introduction

Introduction

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Introduction

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  1. Introduction Earlier we studied the circle, which is the set of all points in a plane that are equidistant from a given point in that plane. We have investigated how to translate between geometric descriptions and algebraic descriptions of circles. Now we will investigate how to translate between geometric descriptions and algebraic descriptions of parabolas. 6.1.2: Deriving the Equation of a Parabola

  2. Key Concepts A quadratic function is a function that can be written in the form f(x)= ax2 + bx + c, where a ≠ 0. The graph of any quadratic function is a parabola that opens up or down. A parabola is the set of all points that are equidistant from a given fixed point and a given fixed line that are both in the same plane as the parabola. That given fixed line is called the directrix of the parabola. The fixed point is called the focus. 6.1.2: Deriving the Equation of a Parabola

  3. Key Concepts, continued The parabola, directrix, and focus are all in the same plane. The vertex of the parabola is the point on the parabola that is closest to the directrix. The illustration on the following slide shows the parts of a parabola. 6.1.2: Deriving the Equation of a Parabola

  4. Key Concepts, continued 6.1.2: Deriving the Equation of a Parabola

  5. Key Concepts, continued Parabolas can open in any direction. In this lesson, we will work with parabolas that open up, down, right, and left. As with circles, there is a standard form for the equation of a parabola; however, that equation differs depending on which direction the parabola opens (right/left or up/down). 6.1.2: Deriving the Equation of a Parabola

  6. Key Concepts, continued Parabolas That Open Up or Down The standard form of an equation of a parabola that opens up or down and has vertex (h, k) is (x – h)2 = 4p(y – k), where p ≠ 0. 6.1.2: Deriving the Equation of a Parabola

  7. Key Concepts, continued 6.1.2: Deriving the Equation of a Parabola

  8. Key Concepts, continued Parabolas That Open Right or Left The standard form of an equation of a parabola that opens right or left and has vertex (h, k) is (y – k)2 = 4p(x – h), where p ≠ 0. 6.1.2: Deriving the Equation of a Parabola

  9. Key Concepts, continued 6.1.2: Deriving the Equation of a Parabola

  10. Key Concepts, continued All Parabolas For any parabola, the focus and directrix are each |p|units from the vertex. Also, the focus and directrix are 2|p|units from each other. If the vertex is (0, 0), then the standard equation of the parabola has a simple form. (x – 0)2 = 4p(y – 0) is equivalent to the simpler form x2= 4py. (y – 0)2 = 4p(x – 0) is equivalent to the simpler form y2= 4px. 6.1.2: Deriving the Equation of a Parabola

  11. Key Concepts, continued Either of the following two methods can be used to write an equation of a parabola: Apply the geometric definition to derive the equation. Or, substitute the vertex coordinates and the value of p directly into the standard form. 6.1.2: Deriving the Equation of a Parabola

  12. Common Errors/Misconceptions confusing vertical and horizontal when graphing one-variable linear equations subtracting incorrectly when negative numbers are involved using the wrong sign for the last term when writing the result of squaring a binomial omitting the middle term when writing the result of squaring a binomial using the incorrect standard form of the equation based on the opening direction of the parabola 6.1.2: Deriving the Equation of a Parabola

  13. Guided Practice Example 2 Derive the standard equation of the parabola with focus (–1, 2) and directrix x = 7 from the definition of a parabola. Then write the equation by substituting the vertex coordinates and the value of p directly into the standard form. 6.1.2: Deriving the Equation of a Parabola

  14. Guided Practice: Example 2, continued To derive the equation, begin by plotting the focus. Label it F (–1, 2). Graph the directrix and label it x = 7. Sketch the parabola. Label the vertex V. 6.1.2: Deriving the Equation of a Parabola

  15. Guided Practice: Example 2, continued Let A (x, y) be any point on the parabola. Point A is equidistant from the focus and the directrix. The distance from A to the directrix is the horizontal distance AB, where B is on the directrix directly to the right of A. The x-value of B is 7 because the directrix is at x = 7. Because B is directly to the right of A, it has the same y-coordinate as A. So, B has coordinates (7, y). 6.1.2: Deriving the Equation of a Parabola

  16. Guided Practice: Example 2, continued 6.1.2: Deriving the Equation of a Parabola

  17. Guided Practice: Example 2, continued Apply the definition of a parabola to derive the standard equation using the distance formula. Since the definition of a parabola tells us that AF= AB, use the graphed points for AF and AB to apply the distance formula to this equation. 6.1.2: Deriving the Equation of a Parabola

  18. Guided Practice: Example 2, continued 6.1.2: Deriving the Equation of a Parabola

  19. Guided Practice: Example 2, continued 6.1.2: Deriving the Equation of a Parabola

  20. Guided Practice: Example 2, continued The standard equation is (y – 2)2 = –16(x – 3). 6.1.2: Deriving the Equation of a Parabola

  21. Guided Practice: Example 2, continued Write the equation using standard form. To write the equation using the standard form, first determine the coordinates of the vertex and the value of p. 6.1.2: Deriving the Equation of a Parabola

  22. Guided Practice: Example 2, continued For any parabola, the distance between the focus and the directrix is 2|p|. In this case, 2|p|= 7−(−1) = 8, thus |p|= 4 . The parabola opens left, so p is negative, and therefore p = –4. For any parabola, the distance between the focus and the vertex is |p|. Since |p|= 4 , the vertex is 4 units right of the focus. Therefore, the vertex coordinates are (3, 2). 6.1.2: Deriving the Equation of a Parabola

  23. Guided Practice: Example 2, continued Use the results found in step 4 to write the equation. 6.1.2: Deriving the Equation of a Parabola

  24. Guided Practice: Example 2, continued The standard equation is (y – 2)2 = –16(x – 3). The results shown in steps 3 and 5 match; so, either method of finding the equation (deriving it using the definition or writing the equation using the standard form) will yield the same equation. ✔ 6.1.2: Deriving the Equation of a Parabola

  25. Guided Practice: Example 2, continued 6.1.2: Deriving the Equation of a Parabola

  26. Guided Practice Example 4 Write the standard equation of the parabola with focus (–5, –6) and directrix y = 3.4. Then use a graphing calculator to graph your equation. 6.1.2: Deriving the Equation of a Parabola

  27. Guided Practice: Example 4, continued Plot the focus and graph the directrix. Sketch the parabola. Label the vertex V. 6.1.2: Deriving the Equation of a Parabola

  28. Guided Practice: Example 4, continued To write the equation, first determine the coordinates of the vertex and the value of p. 6.1.2: Deriving the Equation of a Parabola

  29. Guided Practice: Example 4, continued The distance between the focus and the directrix is 2|p|. So 2|p| = 3.4 − (−6) = 9.4, and |p| = 4.7. The parabola opens down, so p is negative, and therefore p = –4.7. The distance between the focus and the vertex is |p|, so the vertex is 4.7 units above the focus. Add to find the y-coordinate of the vertex: –6 + 4.7 = –1.3. The vertex coordinates are (–5, –1.3). 6.1.2: Deriving the Equation of a Parabola

  30. Guided Practice: Example 4, continued Use the results found in step 2 to write the equation. The standard equation is (x + 5)2 = –18.8(y + 1.3). 6.1.2: Deriving the Equation of a Parabola

  31. Guided Practice: Example 4, continued Solve the standard equation for y to obtain a function that can be graphed. 6.1.2: Deriving the Equation of a Parabola

  32. Guided Practice: Example 4, continued Graph the function using a graphing calculator. On a TI-83/84: Step 1: Press [Y=]. Step 2: At Y1, type in [(][(–)][1][ ÷ ][18.8][)][(][X,T,θ,n][+][5][)][x2][–][1.3]. Step 3: Press [WINDOW] to change the viewing window. Step 4: At Xmin, enter [(–)][12]. Step 5: At Xmax, enter [12]. 6.1.2: Deriving the Equation of a Parabola

  33. Guided Practice: Example 4, continued Step 6: AtXscl, enter [1]. Step 7: AtYmin, enter [(–)] [8]. Step 8: AtYmax, enter [8]. Step 9: AtYscl, enter [1]. Step 10: Press [GRAPH]. 6.1.2: Deriving the Equation of a Parabola

  34. Guided Practice: Example 4, continued On a TI-Nspire: Step 1: Press the [home] key. Step 2: Arrow tothegraphingiconand press [enter]. Step 3: At the blinking cursor at the bottom of the screen, enter [(][(–)][1][ ÷ ][18.8][)][(][x][+][5][)][x2][–][1.3]. Step 4: Change the viewing window by pressing [menu], arrowing down to number 4: Window/Zoom, and clicking the center button of the navigation pad. 6.1.2: Deriving the Equation of a Parabola

  35. Guided Practice: Example 4, continued Step 5: Choose 1: Window settings by pressing the center button. Step 6: Enter in an appropriate XMin value, –12, by pressing [(–)] and [12], then press [tab]. Step 7: Enter in an appropriate XMax value, [12], then press [tab]. Step 8: Leave the XScale set to “Auto.” Press [tab] twice to navigate to YMin and enter an appropriate YMin value, –8, by pressing [(–)] and [8]. 6.1.2: Deriving the Equation of a Parabola

  36. Guided Practice: Example 4, continued Step 9: Press [tab] to navigate to YMax. Enter [8]. Press [tab] twice to leave YScale set to “auto” and to navigate to “OK.” Step 10: Press [enter]. Step 11: Press [menu] and select 2: View and 5: Show Grid. 6.1.2: Deriving the Equation of a Parabola

  37. Guided Practice: Example 4, continued ✔ 6.1.2: Deriving the Equation of a Parabola

  38. Guided Practice: Example 4, continued 6.1.2: Deriving the Equation of a Parabola

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