Parabolas

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 intercepts

We are going to add a couple things

• Focus and Directrix• Parabolas that are turned to left and right

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 directrix

• The vertex is midway between the focus and directrix

Here are some other applications of the focus...

• The distance from the vertex to the focus (or the vertex to the directrix) is called p

1

4a

p

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 focus

21( )

4x y k h

p 21

( )4

y x h kp

Find the focus and equation of the directrix. Then sketch the graph.

21

2y x

: 0,Focus p4 2p

10,

2

1

2p

:Directrix y p1

2y

Opens up : 0,0Vertex

Find the focus and equation of the directrix. Then sketch the graph.

21

16x y

: ,0Focus p

4 16p 4,04p

:Directrix x p

4x

Opens right Vertex (0,0)

Example: x = -1/16 (y – 2)2 + 5 :

Direction:

Vertex:

Focus:

Directrix:

Example: Determine the focus and directrix of the parabola y = 1/8 (x – 8)2 - 3 :

Direction:

Vertex:

Focus:

Directrix:

Directrix: x = 6

y2 – 2y + 12x – 35 = 0

Convert the equation to standard formFind the vertex, focus, and directrix

y2 – 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

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

2 2 8 17 0 x x y

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

2 6 2 9 0y y x

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

21( )

4y x h k

p

21( 4) 1

12y x

Find p 3

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

21( 2) 1

16x y

Find p

21( )

4x y k h

p

4

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

V

A parabola has its focus at (1, -2) and its directrix at y = 2. Does the point (5, -2) lie on the parabola?

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 x2

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.