Presentation is loading. Please wait.

Presentation is loading. Please wait.

Parabola Conic section. Warm-up Graph the following parabola using: I Finding the solution of the equations (Factoring) II Finding the VERTEX (Using formula)

Similar presentations


Presentation on theme: "Parabola Conic section. Warm-up Graph the following parabola using: I Finding the solution of the equations (Factoring) II Finding the VERTEX (Using formula)"— Presentation transcript:

1 Parabola Conic section

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

5 The equation of a parabola with vertex (0, 0) and focus on the y-axis is x 2 = 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 Standard equation of a PARABOLA The equation of a parabola with vertex (0, 0) and focus on the x-axis is y 2 = 4px. 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 A parabola has vertex (0, 0) and the focus on an axis. Write the equation of each parabola. Since the focus is (-6, 0), the equation of the parabola is y 2 = 4px. p is equal to the distance from the vertex to the focus, therefore p = -6. The equation of the parabola is y 2 = -24x. b) The directrix is defined by x = 5. The equation of the parabola is y 2 = -20x. Finding the Equation of a Parabola with Vertex (0, 0) The equation of the directrix is x = -p, therefore -p = 5 or p = -5. Since the focus is on the x-axis, the equation of the parabola is y 2 = 4px. c) The focus is (0, 3). a) The focus is (-6, 0). Since the focus is (0, 3), the equation of the parabola is x 2 = 4py. p is equal to the distance from the vertex to the focus, therefore p = 3. The equation of the parabola is x 2 = 12y.

8 Practice A parabola has vertex (0, 0) and the focus on an axis. Write the equation of each parabola. Finding the Equation of a Parabola with Vertex (0, 0) b) The directrix is defined by x = 3. c) The focus is (0, -5). a) The focus is (8, 0). The equation of the parabola is y 2 = 32x. The equation of the parabola is y 2 = -12x. The equation of the parabola is x 2 = -20y.

9 Finding the FOCUS DIRECTRIX y = 4(4py) y = 16py 1 = 16p 1/16 = p FOCUS: (0, 1 / 16 ) Directrix Y = - 1 / 16 y = 4x 2 x = -3y 2 x = -3(4px) x = -12px 1 = -12p -1/12 = p FOCUS: (-1/12, 0) Directrix x = 1 / 12

10 Practice y = 8x 2 x = -4y 2 FOCUS: (0, 1 / 32 ) Directrix Y = - 1 / 32 FOCUS: (- 1 / 16, 0) Directrix x = 1 / 16

11 Parabola Conic section

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

13 (y-k) 2 =4p(x-h),p0 Horizontal Axis, directrix: x = h-p 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.

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 Standard form y y + 25 = -12x + 72 y x - 10y - 47 = 0 General form (y - 5) 2 = 4(-3)(x - 6) (y - 5) 2 = -12(x - 6)

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

16 Standard Equation of a Parabola with vertex at (h,k) (x-h) 2 =4p(y-k),p0 Vertical Axis, directrix: y = k-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. The general form of the parabola is Ax 2 + Cy 2 + 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) (2 - (-2)) 2 = 4p(8 - 6) 16 = 8p 2 = p x = 2 and y = 8 (x - h) 2 = 4p(y - k) (x - (-2)) 2 = 4(2)(y - 6) (x + 2) 2 = 8(y - 6) Standard form x 2 + 4x + 4 = 8y - 48 x 2 + 4x - 8y + 52 = 0 General form

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

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 X 2 - 4x -12y + 16 = 0 General Form

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 Find the coordinates of the vertex and focus, the equation of the directrix, the axis of symmetry, and the direction of opening of y 2 - 8x - 2y - 15 = 0. y 2 - 8x - 2y - 15 = 0 y 2 - 2y + _____ = 8x _____ 11 (y - 1) 2 = 8x + 16 (y - 1) 2 = 8(x + 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. 4p = 8 p = 2 Standard form Analyzing a Parabola

22 Finding the FOCUS DIRECTRIX y = 4(4py) y = 16py 1 = 16p 1/16 = p FOCUS: (0, 1 / 16 ) Directrix Y = - 1 / 16 y = 4x 2 x = -3y 2 x = -3(4px) x = -12px 1 = -12p -1/12 = p FOCUS: (-1/12, 0) Directrix x = 1 / 12

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) (2 - (-2)) 2 = 4p(8 - 6) 16 = 8p 2 = p x = 2 and y = 8 (x - h) 2 = 4p(y - k) (x - (-2)) 2 = 4(2)(y - 6) (x + 2) 2 = 8(y - 6) Standard form x 2 + 4x + 4 = 8y - 48 x 2 + 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 Standard form y y + 25 = -12x + 72 y x - 10y - 47 = 0 General form (y - 5) 2 = 4(-3)(x - 6) (y - 5) 2 = -12(x - 6)


Download ppt "Parabola Conic section. Warm-up Graph the following parabola using: I Finding the solution of the equations (Factoring) II Finding the VERTEX (Using formula)"

Similar presentations


Ads by Google