8.2 Graph and Write
Equations of Parabolas
•Where is the focus and directrix compared to the vertex?
•How do you know what direction a parabola opens?
•How do you write the equation of a parabola given the
focus/directrix?
•What is the general equation for a parabola?
Parabolas
A parabola is defined in
terms of a fixed point, called
the focus, and a fixed line,
called the directrix.
A parabola is the set of all
points P(x,y) in the plane
whose distance to the focus
equals its distance to the
directrix.
directrix
focus
axis of
symmetry
Horizontal Directrix
Standard Equation of a parabola with its vertex
at the origin is
x2
= 4py
p > 0: opens upward
p < 0: opens downward
focus: (0, p)
directrix: y = –p
axis of symmetry: y-axis
y
P(x, y)
F(0, p)
O
D(x, –p)
x
y = –p
Vertical Directrix
Standard Equation of a parabola with its vertex
at the origin is
y
y2= 4px
p > 0: opens right
p < 0: opens left
focus: (p, 0)
directrix: x = –p
axis of symmetry: x-axis
D(x, –p)
P(x, y)
O
x = –p
F(p, 0)
x
Example 1
1 2
Graph x  y . Label the vertex, focus, and
directrix. 4 y2 = 4px
Identify p.
y2
= 4(1)x
4
So, p = 1
Since p > 0, the parabola
opens to the right.
Vertex: (0,0)
Focus: (1,0)
Directrix: x = -1
2
-4
-2
-2
-4
2
4
Example 1
1 2
Graph x  y . Label the vertex, focus, and
directrix. 4 Y2 = 4x
Use a table to sketch a graph
4
y
x
0
2
4
-2
-4
0
1
4
1
4
2
-4
-2
-2
-4
2
4
Graph x = –⅛y 2. Identify the focus, directrix, and axis
of symmetry.
SOLUTION
STEP 1
Rewrite the equation in standard form.
x = –1
Write original equation.
8
– 8x = y 2
Multiply each side by – 8.
STEP 2
Identify the focus, directrix, and axis of symmetry. The
equation has the form y 2 = 4px where p = – 2. The focus is
(p, 0), or (– 2, 0). The directrix is x = – p, or x = 2. Because
y is squared, the axis of symmetry is the x - axis.
STEP 3
Draw the parabola by making a table of values and
plotting points. Because p < 0, the parabola opens to
the left. So, use only negative x - values.
Graph the equation. Identify the focus, directrix, and
axis of symmetry of the parabola.
1.
Y 2 = –6x
SOLUTION
STEP 1
Rewrite the equation in standard form.
Y 2 = 4 (– 3 )x
2
STEP 2
Identify the focus, directrix, and axis of symmetry. The
equation has the form y 2 = 4px where p = – 3 . The focus
2
3
is (p, 0), or (– , 0). The directrix is x = – p, or x = 23 .
2
Because y is squared, the axis of symmetry is the
x - axis.
STEP 3
Draw the parabola by making a table of values and
plotting points. Because p < 0, the parabola opens to
the left. So, use only negative x - values.
2.45
2.45
4.24
4.90
5.48
Graph the equation. Identify the focus, directrix, and
axis of symmetry of the parabola.
3.
y = 1– x 2
4
SOLUTION
STEP 1 Rewrite the equation in standard form.
Write original equation.
y=– 1 x2
4
Multiply each side by – 4.
2
– 4y = x
STEP 2
equation
x2=–4
focus
0, –1
directrix
y=1
axis of symmetry
Vertical x = 0
STEP 3
Draw the parabola by making a table of values and
plotting points. Because p < 0, the parabola opens to
the left. So, use only negative y - values.
y
x
2
2.83
3.46
4
4.47
Graph the equation. Identify the focus, directrix, and
axis of symmetry of the parabola.
4.
x = 1– y 2
3
SOLUTION
STEP 1 Rewrite the equation in standard form.
x = 1Y2
Write original equation.
3
Multiply each side by 3.
3x = y 2
STEP 2
equation
focus
3 0,
Y 2 = 43 x
4
4
directrix
x=– 3
4
axis of symmetry
Horizontal y = 0
STEP 3
Draw the parabola by making a table of values and
plotting points. Because p < 0, the parabola opens to
the left. So, use only negative x - values.
y
x
1.73
2.45
3
3.46
3.87
Example 2
Write the standard equation of the parabola
with its vertex at the origin and the
directrix y = -6.
Since the directrix is
below the vertex, the
parabola opens up
Since y = -p and y = -6,
p=6
x2=4(6)y
x2 = 24y
Write an equation of the parabola shown.
SOLUTION
The graph shows that the vertex is (0, 0) and the
directrix is y = – p =– 3
for p in the standard form
2
of the equation of a parabola.
x2 = 4py
Standard form, vertical axis of symmetry
3 for p
x2 = 4• ( 3)y
Substitute
2
2
x 2 = 6y
Simplify.
Write the standard form of the equation of the parabola
with vertex at (0, 0) and the given directrix or focus.
5.
Directrix: y = 2
SOLUTION
x 2 = 4py
Standard form, vertical axis of symmetry
x 2 = 4 (–2)y
Substitute –2 for p
x 2 = – 8y
Simplify.
Write the standard form of the equation of the parabola
with vertex at (0, 0) and the given directrix or focus.
8.
Focus: (0, 3)
SOLUTION
x 2 = 4py
Standard form, vertical axis of symmetry
x 2 = 4 (3)y
Substitute 3 for p
x 2 = 12y
Simplify.
Solar Energy
The EuroDish, developed to provide electricity in
remote areas, uses a parabolic reflector to
concentrate sunlight onto a high-efficiency engine
located at the reflector’s focus. The sunlight heats
helium to 650°C to power the engine.
•
Write an equation for the
EuroDish’s cross section
with its vertex at (0, 0).
•
How deep is the dish?
Read more on page 498
• Where is the focus and directrix compared to
vertex?
The focus is a point on the line of symmetry and the
directrix is a line below the vertex. The focus and
directrix are equidistance from the vertex.
• How do you know what direction a parabola opens?
x2, graph opens up or down, y2, graph opens right or left
• How do you write the equation of a parabola given the
focus/directrix?
Find the distance from the focus/directrix to the
vertex (p value) and substitute into the equation.
• What is the general equation for a parabola?
x2= 4py (opens up [p>0] or down [p<0]), y2 = 4px (opens
right [p>0] or left [p<0])
8.2 Assignment
p. 499, 3-17 odd,
27-33 odd,
39-45 odd
8.2 Graph and Write
Equations of Parabolas
day 2
• What does it mean if a parabola has a
translated vertex?
• What general equations can you use
for a parabola when the vertex has
been translated?
Standard Equation of a Translated Parabola
Vertical axis:
(x
− h)2 = 4p(y − k)
vertex: (h, k)
focus: (h, k + p)
directrix: y = k – p
axis of symmetry: x = h
Standard Equation of a Translated Parabola
Horizontal axis:
(y − k)2 = 4p(x − h)
vertex: (h, k)
focus: (h + p, k)
directrix: x = h - p
axis of symmetry: y = k
Example 3
Write the standard equation of the parabola with a
focus at F(-3,2) and directrix y = 4. Sketch the info.
The parabola opens downward,
so the equation is of the form
(x − h)2 = 4p(y − k)
vertex: (-3,3)
h = -3, k = 3
p = -1
(x + 3)2 = 4(−1)(y − 3)
Graph (x – 2)2 = 8 (y + 3).
SOLUTION
STEP 1 Compare the given equation to the standard
form of an equation of a parabola . You can
see that the graph is a parabola with vertex
at (2, – 3) ,focus (2, – 1) and directrix y = – 5
Draw the parabola by making a table of value
and plot y point. Because p > 0, he parabola
open to the right. So use only points x- value
STEP 2
x
1
2
y
–2.875
–3
3
4
5
– 2.875 –2.5 – 1.875
STEP 3 Draw a curve through
the points.
Example 4
Write an equation of a parabola whose vertex is at
(−2,1) and whose focus is at (−3, 1).
Begin by sketching the parabola.
Because the parabola opens to the
left, it has the form
(y −k)2 = 4p(x − h)
Find h and k: The vertex is at
(−2,1) so h = −2 and k = 1
Find p: The distance between the
vertex (−2,1) and the focus (−3,1) by
using the distance formula.
p = −1
(y − 1)2 = −4(x + 2)
Write an equation of the parabola whose vertex is at
(– 2, 3) and whose focus is at (– 4, 3).
SOLUTION
STEP 1 Determine the form of the equation. Begin by
making a rough sketch of the parabola.
Because the focus is to the left of the vertex,
the parabola opens to the left, and its equation
has the form (y – k)2 = 4p(x – h) where p < 0.
STEP 2 Identify h and k. The vertex is at
(– 2, 3), so h = – 2 and k = 3.
STEP 3
Find p. The vertex (– 2, 3) and focus (4, 3)
both lie on the line y = 3, so the distance
between them is p | = | – 4 – (– 2) | = 2, and
thus p = +2. Because p < 0, it follows that p =
– 2, so 4p = – 8.
The standard form of the equation is (y – 3)2 = – 8(x + 2).
Write the standard form of a parabola with vertex at
(3, – 1) and focus at (3, 2).
SOLUTION
STEP 1 Determine the form of the equation. Begin by
making a rough sketch of the parabola.
Because the focus is to the left of the vertex,
the parabola opens to the left, and its
equation has the form (x – h)2 = 4p(y – k)
where p > 0.
STEP 2 Identify h and k. The vertex is at (3,– 1), so h = 3
and k = –1.
STEP 3
Find p. The vertex (3, – 1) and focus (3, 2) both
lie on the line x = 3, so the distance between
them is p | = | – 2 – (– 1) | = 3, and thus p = + 3.
Because p > 0, it follows that p = 3, so 4p = 12.
The standard form of the equation is (x – 3)2 = 12(y + 1)
• What does it mean if a parabola has a
translated vertex?
It means that the vertex of the parabola
has been moved from (0,0) to (h,k).
• What general equations can you use for a
parabola when the vertex has been
translated?
(y-k)2 =4p(x-h)
(x-h)2 =4p(y-k)
8.2 Assignment day 2
p. 499, 26-32 even, 40-46 even
p. 531, 3, 9, 15-16