Math 143
Section 7.3
Parabolas
A parabola is a set of points in a plane that are equidistant from a fixed
line, the directrix, and a fixed point, the focus.
For any point Q that is on the
parabola, d2 = d1
Q
The latus rectum of a parabola is
a line segment that passes
through the focus, is parallel to
the directrix and has its
endpoints on the parabola.
The length of the latus rectum is
|4p| where p is the distance from
the vertex to the focus.
d1
Directrix
d2 Focus
Things you should already know
about a parabola.
Forms of equations
y = a(x – h)2 + k
opens up if a is positive
opens down if a is negative
vertex is (h, k)
y = ax2 + bx + c
V
opens up if a is positive
opens down if a is negative
vertex is
-b
2a
-b
, f(2a )
Thus far in this course we have studied parabolas that are vertical that is, they open up or down and the axis of symmetry is vertical
V
In this unit we will also study
parabolas that are horizontal –
that is, they open right or left
and the axis of symmetry is
horizontal
In these equations it is the
y-variable that is squared.
x = a(y – k)2 + h
or
x = ay2 + by + c
x = ay2 + by + c
y = ax2 + bx + c
Horizontal
Hyperbola
Vertical
Hyperbola
vertex:
y=
-b
2a
Vertex: x =
-b
2a
If a > 0, opens right
If a < 0, opens left
If a > 0, opens up
If a < 0, opens down
The directrix is vertical
The directrix is horizontal
a=
Remember:
1
4p
|p| is the distance from the vertex to the focus
the directrix is the same distance from the vertex as the focus is
(y – k)2 = 4p(x – h)
(x – h)2 = 4p(y – k)
Horizontal
Parabola
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
the vertex is midway
between the focus and
directrix
The directrix is horizontal
and the vertex is midway
between the focus and
directrix
Remember: |p|
is the distance from the vertex to the focus
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: (y – k)2 = 4p(x – h)
|p| = 3
Equation: (y – 4)2 = 12(x + 1)
V
F
Find the standard form of the equation of the parabola given:
the vertex is (2, -3) and focus is (2, -5)
Because of the location of the vertex
and focus this must be a vertical
parabola that opens down
Equation: (x – h)2 = 4p(y – k)
|p| = 2
V
Equation: (x – 2)2 = -8(y + 3)
The vertex is midway between the focus
and directrix, so the directrix for this
parabola is y = -1
F
(y + 3)2 = 4(x + 1)
Find the vertex, focus and directrix. Then graph the parabola
Vertex: (-1, -3)
The parabola is horizontal
and opens to the right
4p = 4
p=1
Focus: (0, -3)
Directrix: x = -2
V
F
x = ¼(y + 3)2 – 1
x 0
y -1
0
-5
3
1
3
-7
Convert the equation to standard form
Find the vertex, focus, and directrix
y2 – 2y + 12x – 35 = 0
y2 – 2y + ___
1 = -12x + 35 + ___
1
(y – 1)2 = -12x + 36
F
(y – 1)2 = -12(x – 3)
V
The parabola is horizontal and opens left
Vertex: (3, 1)
4p = -12
p = -3
Focus: (0, 1)
Directrix:
x=6
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:
x2 = 4py
At (6, 2), 36 = 4p(2)
so
p = 4.5
thus the focus is at the point (0, 4.5)
The receiver should be
placed 4.5 feet above the
base of the dish.
(400, 160)
(300, h)
300
100
The towers of a suspension bridge are
800 ft apart and rise 160 ft above the
road. The cable between them has the
shape of a parabola, and the cable just
touches the road midway between the
towers.
What is the height of the cable 100 ft
from a tower?
Since the parabola is vertical and has
its vertex at (0, 0) its equation must
be of the form:
x2
At (300, h),
= 4py
At (400, 160),
160,000 = 4p(160)
1000 = 4p
p = 250
thus the equation is x2 = 1000y
90,000 = 1000h
h = 90
The cable would be 90 ft long at
a point 100 ft from a tower.