Warm Up
1) Write the equation of a hyperbola with
center (1, -4), a = 5, b =2, and horizontal
transverse axis
2) Write the equation of the hyperbola below:
THE
PARAB
LA
A parabola is the collection of all points P in the plane that
are the same distance from a fixed point F as they are from
a fixed line D. The point F is called the focus of the
parabola, and the line D is its directrix. As a result, a
parabola is the set of points P for which d(F, P) = d(P, D)
Let's sort out this definition by looking at a graph:
directrix
focus
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8
by symmetry we
can get the other
half
Take a line segment
perpendicular to the
directrix and intersect
with a line segment
from the focus of the
same length. This will
be a point on the
parabola and will be
the same distance
from each.
Based on this definition and using the distance formula
we can get a formula for the equation of a parabola with
a vertex at the origin that opens left or right (see page
771 in book for derivation).
If the coefficient
If the coefficient
2
on x is negative the
y

4
ax
on x is positive the
parabola opens to
parabola opens to
the left
the right
a is the distance from the vertex to the focus
(or opposite way for directrix)
a
a
- - -5 - - - - 0 1 2 3 4 5 6 7 8
7 6
4 3 2 1
The equation for the parabola
shown is: 2
y  43x  12 x
The parabola opens to the
right and the vertex is 3
away from the focus.
Let's find the focus and directrix of the parabola:
y  4ax
2
y  16 x
2
This is 4a
 4a  16
a4
Since the coefficient is negative, this parabola opens to the
left. From the vertex count 4 in the negative direction to
get the focus.
The directrix
is a line
x=4
located the
a =4
same
focus a =4
distance
from the
(-4, 0)
vertex in the
other
direction.
We could make a line segment through the focus of the
parabola parallel to the directrix with endpoints on the
parabola. This segment is called the latus rectum.
The length of the latus rectum is 4a.
This is very helpful information when graphing a parabola
because we then know how wide the parabola is.
y  16 x
2
(-4, 8)
latus
rectum
focus
4a
(-4, 0)
(-4, -8)
x=4
a
a
The length of the
latus rectum is 16
so it is 8 each way
from the focus.
In college algebra you graphed parabolas that opened
up or down. The only difference with the equation is
the x and the y are in different places.
x  4ay
2
x  4ay
2
Let's look at the steps to graph the parabola. x  4 y
What
direction
Draw
Add
the
the
directrix
parabola
(not necessary
What
is
the
length
What
is a?
does
this
open?
containing
for
graphing
these
but we want to see
of the latus
rectum?
points.
how
it relates here).
-The
=directrix
- 4a
ais
= a1.
The
If4the
length
x isso
squared
of
the
The
focus
is
ais 4.
y=1
away
from
the
latus
it opens
rectum
up
or
away
the
vertex
in
the
Make
downfrom
a(depending
line
vertex
the
opposite
direction
segment
on thein
sign
4
units
of the
direction
the
as
the
long
coefficient).
(2focus.
each way)
If the
(-2, -1) (0, -1) (2, -1)
parabola
opens.
through
y is squared
the
focus.
it is
right or left.
2
Our parabola may have horizontal and/or vertical
transformations. This would translate the vertex from the
origin to some other place. The equations for these
parabolas are the same but h is the horizontal shift and k
the vertical shift:
The vertex will be at (h, k)
x  h 2  4a y  k 
opens down
opens up
x  h 2  4a y  k 
opens right
y  k 
2
 4a  x  h 
opens left
 y  k 2  4ax  h 
Let's try one:
 y  2
2
 8x  1
Opens? y is squared and 8
is positive so right.
Focus? 4a = 8 so a = 2. Focus is 2
away from vertex in
direction parabola opens.
(1, 6)
x = -3
(-1, 2)
(1, 2)
Vertex?
It is shifted to the left
one and up 2
(set (x + 1) = 0 and get
x = -1 and set
(y - 2) = 0 and get
y = 2).
Vertex is (-1, 2)
Length of latus rectum?
This is number in front of
parenthesis which is 8, so 4
each way from focus.
Directrix?
(1, -2)
"a" away from the vertex
so 2 away in opposite
direction of focus.
The secret to doing these is NOT to memorize a bunch
of formulas given in your book in Tables 1 and 2, but to
DRAW A PICTURE.
What if we were given the focus of a parabola was (-2, 2)
and the vertex was (-5, 2). If we draw a picture we can
figure out the equation and anything else we need to
know.
Just looking at this much
Focus is "a" away from graphed can you determine
vertex so a = 3
which way the parabola opens
3
(and therefore what the standard
form of the equation looks like)
(-5, 2)
(-2, 2)
The focus must be inside
the parabola so it must
open to the right.
(5h
yyyy22k244(412
3aa)xx
(55))
simplified:
2 222
The equation we are given may not be in standard form
and we'll have to do some algebraic manipulation to get it
that way. (you did this with circles in college algebra).
y  2y  x  0
2
factor
Since y is squared, we'll
complete the square on the y's
and get the x term to other side.
1  x  ___
1
y  2 y  ___
2
must add to this side too to keep equation =
middle coefficient divided by 2 and squared
 y  1
2
  x  1
Now we have it in standard form
we can find the vertex, focus,
directrix and graph.
 y  1
2
1 x  1
Opens?
Right (y squared & no negative)
opposites of these values
Vertex?
Focus? 4a = 1 so a = 1/4
Length of latus rectum? 1, so 1/2 each way
x = -5/4
(-3/4, -1/2)
(-3/4, -1)
(-1, -1)
(-3/4, -3/2)
Since the focus was at (-3/4, 1),
to get the ends of the latus
rectum, we'd need to increase
the y value of the focus by 1/2
and then decrease the y value
by 1/2. (look at the picture to
determine this).
Directrix? 1/4 away from vertex
There are many applications that involve parabolas.
One is paraboloids of revolution. This is taking a
parabola and revolving it to form "a dish".
The waves come in and hit the
surface and are reflected to the
focus of the parabola.
The receiver is
placed at the focus.
To work these problems, draw a picture with the
vertex at the origin.
Homework
Page 659 #13 – 35 odd