#### Transcript 11.1 Parabolas

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 Standard Equation of a Parabola: (Vertex at the origin) Equation 2 x = 4cy Focus (0, c) Directrix y = –c Equation 2 x = -4cy Focus Directrix (0, -c) y=c (If the x term is squared, the parabola opens up or down) Equation 2 y = 4cx Focus (c, 0) Directrix x = –c Equation 2 y = -4cx Focus Directrix (-c, 0) x=c (If the y term is squared, the parabola opens left or right) 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) Focus: (0, 1/16) Directrix: y = –c Directrix: y = –1/16 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) Focus: (–1, 0) Directrix: x = –c Directrix: x = 1 Let’s see what this parabola looks like... Standard Equation of a Parabola: (Vertex at (h,k)) Equation Focus Directrix 2 (x-h) = 4c(y-k) (h, k + c) y = k - c Equation 2 (x-h) = -4c(y-k) Focus (h, k - c) Directrix y=k+c Standard Equation of a Parabola: (Vertex at(h,k)) Equation 2 (y-k) = 4c(x-h) Focus (h +c , k) Directrix x=h-c Equation Focus 2 (y-k) = -4c(x-h) (h - c, k) Directrix x=h+c 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. Equation of Parabola in General Form 2 For (x-h) = 4c(y-k) y = Ax2 + Bx + C 2 For (y-k) = 4c(x-h) x = Ay2 + By + C 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) 2 2 (x -1) 2 = 1(y + 1) 2 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) 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) (2 - (-2))2 = 4c(8 - 6) 16 = 8c 2=c x = 2 and y = 8 (x - h)2 = 4c(y - k) (x - (-2))2 = 4(2)(y - 6) (x + 2)2 = 8(y - 6) Standard form 3.6.10 Example 7: Sketch (y-2)2 ≤ 12(x-3) 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. y2 - 8x - 2y - 15 = 0 y2 - 2y + _____ 1 = 8x + 15 + _____ 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. 4c = 8 c=2 Example 9: Find the intersection point(s), if any, of the parabola 2 2 x y 1 with equation y2 = 2x + 12 and the ellipse with equation 36 64 64x 36y 2304 64x 2 36(2x 12) 2304 64x 2 72x 432 2304 64x 2 72x 1872 0 2 2