ParabolaConic section

Warm-upGraph 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 = x2 - 6x + 8

2. y = –x2 + 4x – 4

3. y = x2 - 5

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

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.

Standard equation of a

PARABOLAThe equation of a parabola with vertex (0, 0) and focus on the x-axis is y2 = 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.

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 y2 = 4px.p is equal to the distance from the vertex to the focus, therefore p = -6.

The equation of the parabola is y2 = -24x.

b) The directrix is defined by x = 5.

The equation of the parabola is y2 = -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 y2 = 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 x2 = 4py.p is equal to the distance from the vertex to the focus, therefore p = 3.

The equation of the parabola is x2 = 12y.

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 y2 = 32x.

The equation of the parabola is y2 = -12x.

The equation of the parabola is x2 = -20y.

Finding the FOCUS DIRECTRIX

y = 4(4py)

y = 16py

1 = 16p

1/16 = p

FOCUS: (0, 1/16)

Directrix Y = - 1/16

y = 4x2 x = -3y2

x = -3(4px)

x = -12px

1 = -12p

-1/12 = pFOCUS: (-1/12, 0)Directrix x = 1/12

Practicey = 8x2 x = -4y2

FOCUS: (0, 1/32)

Directrix Y = - 1/32

FOCUS: (-1/16, 0)

Directrix x = 1/16

ParabolaConic section

WARM -UP

1. (try this 1. (try this one on your one on your own)own)

y = -6xy = -6x22

FOCUSFOCUS

(0, -(0, -11//2424))

DirectrixDirectrix

y = y = 11//2424

2. (try this 2. (try this one on your one on your own)own)

x = 8yx = 8y22

FOCUSFOCUS

(1/32, 0)(1/32, 0)

DirectrixDirectrix

x = -32x = -32

Find the focus and directrix of the following:

(y-k)2 =4p(x-h),p≠0Horizontal 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.

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

y2 - 10y + 25 = -12x + 72y2 + 12x - 10y - 47 = 0 General form

(y - 5)2 = 4(-3)(x - 6)(y - 5)2 = -12(x - 6)

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)

y2 + 8x -12y -12 General Form

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 equation of theaxis 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 Ax2 + Cy2 + Dx + Ey + F = 0where A = 0 or C = 0.

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 equationand 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

x2 + 4x + 4 = 8y - 48x2 + 4x - 8y + 52 = 0 General form

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)

x2 - 6x +36y -207 General Form

homework

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

X2 - 4x -12y + 16 = 0 General Form

• 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 theaxis of symmetry is x = h.

• The coordinates of the focus are (h, k + p).

• The equation of the directrix is y = k - p.

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.

4p = 8 p = 2

Standardform

Analyzing a Parabola

Finding the FOCUS DIRECTRIX

y = 4(4py)

y = 16py

1 = 16p

1/16 = p

FOCUS: (0, 1/16)

Directrix Y = - 1/16

y = 4x2 x = -3y2

x = -3(4px)

x = -12px

1 = -12p

-1/12 = pFOCUS: (-1/12, 0)Directrix x = 1/12

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 equationand 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

x2 + 4x + 4 = 8y - 48x2 + 4x - 8y + 52 = 0 General form

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

y2 - 10y + 25 = -12x + 72y2 + 12x - 10y - 47 = 0 General form

(y - 5)2 = 4(-3)(x - 6)(y - 5)2 = -12(x - 6)