Transcript Directrix

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
a
4p
1
x
( y  k )2  h
4p
Horizontal
Parabola
1
y
( x  h) 2  k
4p
Vertical
Parabola
Vertex: (h, k)
Vertex: (h, k)
If 4p > 0, opens right
If 4p > 0, opens up
If 4p < 0, opens left
If 4p < 0, opens down
The directrix is vertical
(x= )
The directrix is horizontal
(y= )
Remember: |p|
is the distance from the vertex to the focus
Find the focus and equation of the
directrix. Then sketch the graph.
1 2
y x
2
Focus :  0, p 
4p  2
p
1
2
Directrix : y   p
Opens up
 1
 0, 
 2
1
y
2
Vertex :  0, 0 
Find the focus and equation of the
directrix. Then sketch the graph.
1 2
x y
16
Focus :  p,0
4 p  16
Opens right
 4, 0 
p4
Directrix : x   p
x  4
Vertex (0,0)
Example:
Direction:
Vertex:
Focus:
Directrix:
2
x = -1/16 (y – 2) + 5 :
Example: Determine the focus and directrix of the
2
parabola y = 1/8 (x – 8) - 3 :
Direction:
Vertex:
Focus:
Directrix:
Convert the equation to standard form
Find the vertex, focus, and directrix
y2 – 2y + 12x – 35 = 0
1
y2 – 2y + 1___ = -12x + 35 + ___
(y – 1)2 = -12x + 36
(y – 1)2 = -12(x – 3)
x = -1/12 (y – 1)2 + 3
The parabola is horizontal and opens left
Vertex: (3, 1)
4p = -12
p = -3
Focus: (0, 1)
Directrix: x = 6
F
V
Write the equation in standard form by
completing the square. State the VERTEX,
focus, and directrix.
2
x  2 x  8y  17  0
Write the equation in standard form by
completing the square. State the VERTEX,
focus, and directrix.
2
y  6y  2x  9  0
Write the equation of a parabola with vertex
(-4, -1) that has a focus (-4, 2)
1
2
y
( x  h)  k
4p
Find p
3
1
2
y  ( x  4)  1
12
Write the equation of a parabola with vertex
(1, 2) that has a focus (5, 2)
1
2
x
( y  k)  h
4p
Find p
4
1
2
x  ( y  2)  1
16
Find the standard form of the equation of the parabola
given: the focus is (2, 4) and the directrix is x = - 4
The vertex is midway between the focus and
directrix, so the vertex is (-1, 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
Equation: x=1/4p (y – k)2 + h
|p| = 3
Equation: x= 1/12 (y – 4)2 - 1
V
F
A parabola has its focus at (1, -2) and its
directrix at y = 2. Does the point (5, -2)
lie on the parabola?
12
2
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?
(-6, 2)
(6, 2)
Consider a parabola cross-section of the dish
and create a coordinate system where the
origin is at the base of the dish.
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.