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Parabola

Parabola. Conic section. Warm-up. Graph the following parabola using:. I Finding the solution of the equations (Factoring). 1. y = x 2 - 6 x + 8. II Finding the VERTEX (Using formula). 2. y = –x 2 + 4 x – 4. III Graphing on y-axis (using vertex). 3. y = x 2 - 5.

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Parabola

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  1. Parabola • Conic section

  2. Warm-up Graph the following parabola using: I Finding the solution of the equations (Factoring) 1. y = x2- 6x + 8 II Finding the VERTEX (Using formula) 2. y = –x2+ 4x – 4 III Graphing on y-axis (using vertex) 3. y = x2- 5

  3. Parabola: the set of points in a plane that are the same distance from a given point called the focus and a given line called the directrix.

  4. Note the line through the focus, perpendicular to the directrix Axis of symmetry • Note the point midway between the directrix and the focus • Vertex 4

  5. The equation of a parabola with vertex (0, 0) and focus on the y-axis is x2 = 4py. The coordinates of the focus are (0, p). The equation of the directrix is y = -p. If p > 0, the parabola opens up. If p < 0, the parabola opens down.

  6. The equation of a parabola with vertex (0, 0) and focus on the x-axis is y2 = 4px. Standard equation of a PARABOLA The coordinates of the focus are (p, 0). The equation of the directrix is x = -p. If p > 0, the parabola opens right. If p < 0, the parabola opens left.

  7. Since the focus is (-6, 0), the equation of the parabola is y2 = 4px. Since the focus is on the x-axis, the equation of the parabola is y2 = 4px. Since the focus is (0, 3), the equation of the parabola is x2 = 4py. p is equal to the distance from the vertex to the focus, therefore p = -6. p is equal to the distance from the vertex to the focus, therefore p = 3. The equation of the directrix is x = -p, therefore -p = 5or p = -5. Finding the Equation of a Parabola with Vertex (0, 0) A parabola has vertex (0, 0) and the focus on an axis. Write the equation of each parabola. a) The focus is (-6, 0). The equation of the parabola isy2 = -24x. b) The directrix is defined by x = 5. The equation of the parabola isy2 = -20x. c) The focus is (0, 3). The equation of the parabola isx2 = 12y.

  8. Practice Finding the Equation of a Parabola with Vertex (0, 0) A parabola has vertex (0, 0) and the focus on an axis. Write the equation of each parabola. a) The focus is (8, 0). The equation of the parabola isy2 = 32x. b) The directrix is defined by x = 3. The equation of the parabola isy2 = -12x. c) The focus is (0, -5). The equation of the parabola isx2 = -20y.

  9. Finding the FOCUS DIRECTRIX x = -3y2 y = 4x2 y = 4(4py) y = 16py 1 = 16p 1/16 = p x = -3(4px) x = -12px 1 = -12p -1/12 = p FOCUS: (-1/12, 0) FOCUS: (0, 1/16) Directrix x = 1/12 Directrix Y = - 1/16

  10. Practice y = 8x2 x = -4y2 FOCUS: (0, 1/32) FOCUS: (-1/16, 0) Directrix Y = - 1/32 Directrix x = 1/16

  11. Parabola • Conic section

  12. Find the focus and directrix of the following: WARM -UP 2. (try this one on your own) x = 8y2 1. (try this one on your own) y = -6x2 FOCUS (0, -1/24) Directrix y = 1/24 FOCUS (1/32, 0) Directrix x = -32

  13. The equation of the • axis of symmetry is y = k. • The coordinates of the focus are (h + p, k). • The equation of the directrix is x = h - p. (y-k)2 =4p(x-h),p≠0 Horizontal Axis, directrix: x = h-p

  14. Ex: Write the equation of the parabola with a focus at (3, 5) and the directrix at x = 9, in standard form and general form. The distance from the focus to the directrix is 6 units, therefore, 2p = -6, p = -3. Thus, the vertex is (6, 5). The axis of symmetry is parallel to the x-axis: (y - k)2 = 4p(x - h) h = 6 and k = 5 (y - 5)2 = 4(-3)(x - 6) (y - 5)2 = -12(x - 6) Standard form y2 - 10y + 25 = -12x + 72 y2 + 12x - 10y - 47 = 0 General form

  15. Practice Write the equation of the parabola with a focus at (4, 6) and the directrix at x = 8, in standard form and general form. Vertex: (6,6) (y-6)2 = -8(x-6) Standard Form y2 + 8x -12y -12 General Form

  16. The equation of the • axis of symmetry is x = h. • The coordinates of the focus are (h, k + p). • The equation of the directrix is y = k - p. Standard Equation of a Parabola with vertex at (h,k) (x-h)2 =4p(y-k),p≠0 Vertical Axis, directrix: y = k-p The general form of the parabola is Ax2 + Cy2 + Dx + Ey + F = 0 where A = 0 or C = 0.

  17. Find the equation of the parabola that has a min at (-2, 6)and passes through the point (2, 8). The axis of symmetry is parallel to the y-axis. The vertex is (-2, 6), therefore, h = -2 and k = 6. Substitute into the standard form of the equation and solve for p: (x - h)2 = 4p(y - k) x = 2 and y = 8 (2 - (-2))2 = 4p(8 - 6) 16 = 8p 2 = p (x - h)2 = 4p(y - k) (x - (-2))2 = 4(2)(y - 6) (x + 2)2 = 8(y - 6) Standard form x2 + 4x + 4 = 8y - 48 x2 + 4x - 8y + 52 = 0 General form

  18. homework Find the equation of the parabola that has a maximum at (3, 6)and passes through the point (9, 5). Vertex: (3,6) (x-3)2 = -36(y-6) Standard Form x2 - 6x +36y -207 General Form

  19. Find the equation of the parabola that has a vertex at (2,1)and focus (2,4). (x-h)2 = 4p (y-k) h=2, k=1, p= 4-1 = 3 (x-2)2 = 4(3) (y-1) (x-2)2 = 12 (y-1) Standard Form General Form X2 - 4x -12y + 16 = 0

  20. The equation of the • axis of symmetry is y = k. • The coordinates of the focus are (h + p, k). • The equation of the directrix is x = h - p. • The equation of the • axis of symmetry is x = h. • The coordinates of the focus are (h, k + p). • The equation of the directrix is y = k - p.

  21. Analyzing a Parabola Find the coordinates of the vertex and focus, the equation of the directrix, the axis of symmetry, and the direction of opening of y2 - 8x - 2y - 15 = 0. y2 - 8x - 2y - 15 = 0 y2 - 2y + _____ = 8x + 15 + _____ 1 1 (y - 1)2 = 8x + 16 (y - 1)2 = 8(x + 2) Standard form 4p = 8 p = 2 The vertex is (-2, 1). The focus is (0, 1). The equation of the directrix is x + 4 = 0. The axis of symmetry is y - 1 = 0. The parabola opens to the right.

  22. Finding the FOCUS DIRECTRIX x = -3y2 y = 4x2 y = 4(4py) y = 16py 1 = 16p 1/16 = p x = -3(4px) x = -12px 1 = -12p -1/12 = p FOCUS: (-1/12, 0) FOCUS: (0, 1/16) Directrix x = 1/12 Directrix Y = - 1/16

  23. Find the equation of the parabola that has a min at (-2, 6)and passes through the point (2, 8). The axis of symmetry is parallel to the y-axis. The vertex is (-2, 6), therefore, h = -2 and k = 6. Substitute into the standard form of the equation and solve for p: (x - h)2 = 4p(y - k) x = 2 and y = 8 (2 - (-2))2 = 4p(8 - 6) 16 = 8p 2 = p (x - h)2 = 4p(y - k) (x - (-2))2 = 4(2)(y - 6) (x + 2)2 = 8(y - 6) Standard form x2 + 4x + 4 = 8y - 48 x2 + 4x - 8y + 52 = 0 General form

  24. Ex: Write the equation of the parabola with a focus at (3, 5) and the directrix at x = 9, in standard form and general form. The distance from the focus to the directrix is 6 units, therefore, 2p = -6, p = -3. Thus, the vertex is (6, 5). The axis of symmetry is parallel to the x-axis: (y - k)2 = 4p(x - h) h = 6 and k = 5 (y - 5)2 = 4(-3)(x - 6) (y - 5)2 = -12(x - 6) Standard form y2 - 10y + 25 = -12x + 72 y2 + 12x - 10y - 47 = 0 General form

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