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Parabolas

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leena
16 diapositives · 507 Views · Aug 14, 2014

Parabolas. The parabola : A curve on which all points are equidistant from a focus and a line called the directrix . dPF = dPdirectrix = c or dPF + dPdirectrix = 2c. (0, c). y = –c. Standard Equation of a Parabola: (Vertex at the origin). Equation Focus Directrix

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Parabolas
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  1. Parabolas

  2. The parabola: • A curve on which all points are equidistant from a focusand a line called the directrix. • dPF = dPdirectrix = c ordPF + dPdirectrix = 2c

  3. (0, c) y = –c Standard Equation of a Parabola:(Vertex at the origin) Equation Focus Directrix x2 = -4cy (0, -c) y = c Equation Focus Directrix x2 = 4cy (If the x term is squared, the parabola opens up or down)

  4. (c, 0) x = –c Equation Focus Directrix y2 = -4cx (-c, 0) x = c Equation Focus Directrix y2 = 4cx (If the y term is squared, the parabola opens left or right)

  5. Example 1: Determine the focus and directrix of the parabola y = 4x2: x2 = 4cy y = 4x2 4 4 x2 = 1/4y 4c = 1/4 c = 1/16 Focus: (0, c) Directrix: y = –c Focus: (0, 1/16) Directrix: y = –1/16

  6. Example 2: Graph and determine the focus and directrix of the parabola –3y2 – 12x = 0: –3y2 = 12x –3y2 = 12x –3 –3 y2 = –4x y2 = 4cx 4c = –4 c = –1 Focus: (c, 0) Directrix: x = –c Focus: (–1, 0) Directrix: x = 1 Let’s see what this parabola looks like...

  7. (h, k + c) y = k - c Standard Equation of a Parabola:(Vertex at (h,k)) Equation Focus Directrix (x-h)2 = -4c(y-k) (h, k - c) y = k + c Equation Focus Directrix (x-h)2 = 4c(y-k)

  8. (h +c , k) x = h - c Standard Equation of a Parabola:(Vertex at(h,k)) Equation Focus Directrix (y-k)2 = -4c(x-h) (h - c, k) x = h + c Equation Focus Directrix (y-k)2 = 4c(x-h)

  9. Example 3:The coordinates of vertex of a parabola are V(3, 2) and equation of directrix is y = –2. Find the coordinates of the focus.

  10. Equation of Parabola in General Form For (x-h)2 = 4c(y-k) y = Ax2 + Bx + C For (y-k)2 = 4c(x-h) x = Ay2 + By + C

  11. Example 4:Convert y = 2x2 -4x + 1 to standard form y = 2x2 -4x + 1 y - 1 = 2(x2 -2x) y – 1 + 2 = 2(x2 -2x + 1) y + 1 = 2(x -1) 2 1(y + 1) = (x -1) 22 (x -1) 2 = 1(y + 1) 2

  12. Example 5:Determine the equation of the parabola with a focus at (3, 5) and the directrix at x = 9 The distance from the focus to the directrix is 6 units, so, 2c = -6, c = -3. V(6, 5). The axis of symmetry is parallel to the x-axis: (y - k)2 = 4c(x - h) h = 6 and k = 5 (y - 5)2 = 4(-3)(x - 6) (y - 5)2 = -12(x - 6) (6, 5)

  13. Example 6:Find the equation of the parabola that has a minimum at (-2, 6) and passes through the point (2, 8). The vertex is (-2, 6),h = -2 and k = 6. (x - h)2 = 4c(y - k) x = 2 and y = 8 (2 - (-2))2 = 4c(8 - 6) 16 = 8c 2 = c (x - h)2 = 4c(y - k) (x - (-2))2 = 4(2)(y - 6) (x + 2)2 = 8(y - 6) Standard form 3.6.10

  14. Example 7:Sketch (y-2)2 ≤ 12(x-3)

  15. Example 8: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. 4c = 8 c = 2 y2 - 8x - 2y - 15 = 0 y2 - 2y + _____ = 8x + 15 + _____ 1 1 (y - 1)2 = 8x + 16 (y - 1)2 = 8(x + 2) Standard form 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.

  16. Example 9:Find the intersection point(s), if any, of the parabola with equation y2 = 2x + 12 and the ellipse with equation