9.2 Graph & Write
Equations of
Parabolas
Algebra II
Parabolas
• We already know the graph of
y=ax2 is a parabola w/ vertex (0,0)
and AOS x=0
• Every parabola has the property
that any point on it is equidistant
from a point called the Focus and
a line called the directrix.
Focus
Lies on AOS
Directrix
The focus and directrix each lie IpI
units from the vertex. (the vertex is
½ way between the focus and
directrix)
•
x2=4py, p>0
Focus (0,p)
Directrix
y=-p
x2=4py, p<0
Directrix
y=-p
Focus (0,p)
y2=4px, p>0
Directrix
x=-p
Focus (p,0)
y2=4px, p<0
Focus (p,0)
Directrix
x=-p
Standard equation of Parabola
(vertex @ origin)
Equation
x2=4py
y2=4px
Focus
(0,p)
(p,0)
Directrix
AOS
y=-p
Vertical
(x=0)
x=-p
Horizontal
(y=0)
Steps to graphing
• 1.) isolate variable squared
• 2.) Set coefficient of the variable to the first
power equal to 4p; solve for p.
• 3.) Identify focus & directrix (using table or
pictures)
• 4.) Make a T-chart accordingly.
Ex. 1) Identify the focus and
directrix of the parabola
x = -1/6y2
EX. 1
• Since y is squared, AOS is horizontal
• Isolate the y2 → y2 = -6x
• Since 4p = -6
•
p = -6/4 = -3/2
• Focus : (-3/2,0) Directrix : x=-p, x= -(3/2), p=3/2
• To draw: make a table of values & plot
• p<0 so opens left so only choose neg
Ex. 2
• Find the focus and directrix, then graph
x=
•
•
•
•
2
3/4y
y2 so AOS is Horizontal
Isolate y2 → y2 = 4/3 x
4p = 4/3 p = 1/3
Focus (1/3,0) Directrix x=-p=-1/3
Standard equation of Parabola
(vertex @ origin)
Equation
x2=4py
y2=4px
Focus
(0,p)
(p,0)
Directrix
AOS
y=-p
Vertical
(x=0)
x=-p
Horizontal
(y=0)
Ex. 3) Writing the equation of a
parabola.
• The
graph
shows
V=(0,0)
• Directrex
y=-p=-2
• So
substitute
2 for p
• = 4py
2
• x = 4(2)y
• x2 = 8y
2
x
• y = 1/8 and check in your
calculator
2
x
Standard equation of Parabola
(vertex @ origin)
Equation
x2=4py
y2=4px
Focus
(0,p)
(p,0)
Directrix
AOS
y=-p
Vertical
(x=0)
x=-p
Horizontal
(y=0)
Ex. 4 Write equation of
parabola.
• Focus =
(0,-3)
• X2 = 4py
• X2 = 4(-3)y
• X2 = -12y
• y=-1/12x2
to check
Assignment