Parabolas

The parabola: - A curve on which all points are equidistant from a

focusfocus and a line called the directrixdirectrix.

dPF = dPdirectrix = c c or dPF + dPdirectrix = 2c

Standard Equation of a Parabola: (Vertex at the origin)

Equation Focus Directrixx2 = 4cy (0, c) y = –c

Equation Focus Directrix

x2 = -4cy (0, -c) y = c

(If the x term is squared, the parabola opens up or down)

Equation Focus Directrixy2 = 4cx (c, 0) x = –c

Equation Focus Directrix

y2 = -4cx (-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) Directrix: y = –cFocus: (0, 1/16) 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) Directrix: x = –cFocus: (–1, 0) Directrix: x = 1

Let’s see what this parabola looks like...

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) (h, k - c) y = k + 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) (h - c, k) 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

For (x-h)2 = 4c(y-k)y = Ax2 + Bx + C

For (y-k)2 = 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).

(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)

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 + _____ = 8x + 15 + _____1 1

(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.

4c = 8 c = 2

Standardform

Example 9: Find the intersection point(s), if any, of the parabola

with equation y2 = 2x + 12 and the ellipse with equation 16436

22

yx

23043664 22 yx

2304)122(3664 2 xx

23044327264 2 xx

018727264 2 xx