Parabolas - PowerPoint Presentation

Slide1

ParabolasSlide2
We already know A LOT about parabolas

2 forms (standard and vertex)

How to find Vertex (

h,k

) or (-b/2a)

Axis of Symmetry

Characteristics

Many ways to solve their equations

Solutions are x interceptsSlide3
We are going to add a couple things

Focus and

Directrix

Parabolas that are turned to left and rightSlide4
Focus and

Directrix

The Focus is a point inside the parabola

The

directrix

is a line outside the parabola

All points on the parabola are equidistant from the focus and

directrixThe vertex is midway between the focus and directrix Slide5

Here are some other applications of the focus...Slide6
The distance from the vertex to the focus (or the vertex to the

directrix

) is called pSlide7

Vertex

Form Equation of a Parabola

Horizontal Parabola

Vertical Parabola

Vertex: (h, k)

If 4p > 0, opens right

If 4p < 0, opens left

The

directrix

is

vertical (x= )

Vertex: (h, k)

If 4p > 0, opens up

If 4p < 0, opens down

The

directrix

is

horizontal (y= )

Remember:

|p|

is the distance from the vertex to the focusSlide8

Find the focus and

equation

of the directrix. Then sketch the graph.

Opens upSlide9

Find the focus and

equation

of the directrix. Then sketch the graph

.

Opens right

Vertex (0,0)Slide10

Example:

x = -1/16

(y

– 2)

2

+ 5

:

Direction:

Vertex:

Focus:

Directrix: Slide11

Example: Determine the focus and directrix of the parabola

y = 1/8

(x

– 8)

2

- 3

:

Direction:

Vertex:

Focus:

Directrix: Slide12

Converting an Equation

Directrix: x = 6

y

2

– 2y + 12x – 35 = 0

Convert the equation to standard form

Find the vertex, focus, and directrix

y

2

– 2y + ___ = -12x + 35 + ___

1

1

(y – 1)

2

= -12x + 36

(y – 1)

2

= -12(x – 3)

The parabola is horizontal and opens left

Vertex: (3, 1)

4p = -12

p = -3

F

V

Focus: (0, 1)

x

= -1/12 (y

– 1)

2

+ 3 Slide13

Write the equation in standard form by completing the square. State the VERTEX, focus, and

directrix

. Slide14

Write the equation in standard form by completing the square. State the VERTEX, focus, and

directrix

.Slide15
Write the equation of a parabola with vertex (-4, -1) that has a focus (-4, 2)

Find p

3Slide16
Write the equation of a parabola with vertex (1, 2) that has a focus (5, 2)

Find p

4Slide17

The vertex is midway between the focus and directrix, so the vertex is (-1, 4)

Equation:

x= 1/12 (y

– 4)

2

- 1

|p| = 3

Find the standard form of the equation of the parabola given:

the focus is (2, 4) and the

directrix

is x = - 4

The directrix is vertical so the parabola must be horizontal and since the focus is always inside the parabola, it must open to the right

F

Equation:

x=1/4p

(y – k)

2

+ h

VSlide18

A parabola has its focus at (1, -2) and its

directrix

at y = 2. Does the point (5, -2) lie on the parabola?Slide19

Applications

A satellite dish is in the shape of a parabolic surface. The dish is 12

ft

in diameter and 2

ft

deep. How far from the base should the receiver be placed?

Consider a parabola cross-section of the dish and create a coordinate system where the origin is at the base of the dish.

2

12

(-6, 2)

(6, 2)

Since the parabola is vertical and has its vertex at (0, 0) its equation must be of the form:

y = 1/4p x

2

At (6, 2),

2 = 1/4p (36)

so p = 4.5

The receiver should be placed 4.5 feet above the base of the dish.