Parabola
Conic section
Warm-up
Graph the following parabola using:
I Finding the solution of the equations (Factoring)
1. y =
2
x
- 6x + 8
II Finding the VERTEX (Using formula)
2. y =
2
–x
+ 4x – 4
III Graphing on y-axis (using vertex)
3. y =
2
x
-5
•
Parabola: the set of points in a
plane that are the same distance
from a given point called the focus
and a given line called the directrix.
•
Note the line through the
focus, perpendicular to the
directrix
 Axis of symmetry
• Note
the point
midway between the
directrix and the focus

Vertex
4
The equation of a parabola
with vertex (0, 0) and focus on
the y-axis is x2 = 4py.
The coordinates of the focus are (0, p).
The equation of the directrix is y = -p.
If p > 0, the parabola opens up.
If p < 0, the parabola opens
down.
Standard
equation of a
PARABOLA
The equation of a parabola
with vertex (0, 0) and focus on
the x-axis is y2 = 4px.
The coordinates of the focus are (p, 0).
The equation of the directrix is x = -p.
If p > 0, the parabola opens right.
If p < 0, the parabola opens left.
Finding the Equation of a
A parabola has
vertex Vertex
(0, 0) and the(0,
focus 0)
on an axis.
Parabola
with
Write the equation of each parabola.
a) The focus is (-6, 0).
Since the focus is (-6, 0), the equation of the parabola is y2 = 4px.
p is equal to the distance from the vertex to the focus, therefore p = -6.
The equation of the parabola is y2
= -24x.
b) The directrix is defined by x = 5.
Since the focus is on the x-axis, the equation of the parabola is y2 = 4px.
The equation of the directrix is x = -p, therefore -p = 5 or p = -5.
The equation of the parabola is y2 = -20x.
c) The focus is (0, 3).
Since the focus is (0, 3), the equation of the parabola is x2 = 4py.
p is equal to the distance from the vertex to the focus, therefore p = 3.
The equation of the parabola is x2 = 12y.
Practice
Finding the Equation of a
A parabola has
vertex Vertex
(0, 0) and the(0,
focus 0)
on an axis.
Parabola
with
Write the equation of each parabola.
a) The focus is (8, 0).
The equation of the parabola is y2
= 32x.
b) The directrix is defined by x = 3.
The equation of the parabola is y2
= -12x.
c) The focus is (0, -5).
The equation of the parabola is x2
= -20y.
Finding the FOCUS
y=
DIRECTRIX
2
4x
x=
2
-3y
y = 4(4py)
x = -3(4px)
y = 16py
x = -12px
1 = 16p
1 = -12p
1/16 = p
-1/12 = p
FOCUS:
(0, 1/16)
FOCUS:
Directrix
Y = - 1/16
Directrix
(-1/12, 0)
x = 1/12
Practice
y=
2
8x
x=
FOCUS:
(0, 1/32)
FOCUS:
Directrix
Y = - 1/32
Directrix
2
-4y
(-1/16, 0)
x = 1/16
Parabola
Conic section
WARM -UP
Find the focus and directrix of the following:
1. (try this one
on your own)
y=
-6x2
FOCUS
Directrix
(0, -1/24)
y = 1/24
2. (try this one
on your own)
x = 8y2
FOCUS
Directrix
(1/32, 0)
x = -32
2
(y-k) =4p(x-h),p≠0
Horizontal Axis, directrix: x = h-p
• The equation of the
axis of symmetry is y = k.
• The coordinates of the
focus are (h + p, k).
• The equation of the
directrix is x = h - p.
Ex: Write the equation of the parabola with a focus at (3, 5) and
the directrix at x = 9, in standard form and general form.
The distance from the focus to the directrix is 6 units, therefore, 2p
= -6, p = -3. Thus, the vertex is (6, 5).
The axis of symmetry is parallel to the x-axis:
(y - k)2 = 4p(x - h)
h = 6 and k = 5
(y - 5)2 = 4(-3)(x - 6)
(y - 5)2 = -12(x - 6)
Standard form
y2 - 10y + 25 = -12x + 72
y2 + 12x - 10y - 47 = 0
General form
Practice
Write the equation of the parabola with a focus at (4, 6) and
the directrix at x = 8, in standard form and general form.
Vertex: (6,6)
(y-6)2 = -8(x-6) Standard Form
2
y
+ 8x -12y -12 General Form
Standard Equation of a
Parabola with vertex at (h,k)
2
(x-h) =4p(y-k),p≠0
Vertical Axis, directrix: y = k-p
• The equation of the
axis of symmetry is x = h.
• The coordinates of the focus
are (h, k + p).
• The equation of the
directrix is y = k - p.
The general form of the parabola
is Ax2 + Cy2 + Dx + Ey + F = 0
where A = 0 or C = 0.
Find the equation of the parabola that has a
min at (-2, 6) and passes through the point (2, 8).
The axis of symmetry is parallel to the y-axis.
The vertex is (-2, 6), therefore, h = -2 and k = 6.
Substitute into the standard form of the equation
and solve for p:
(x - h)2 = 4p(y - k)
x = 2 and y = 8
(2 - (-2))2 = 4p(8 - 6)
16 = 8p
2=p
(x - h)2 = 4p(y - k)
(x - (-2))2 = 4(2)(y - 6)
(x + 2)2 = 8(y - 6)
x2 + 4x + 4 = 8y - 48
x2 + 4x - 8y + 52 = 0
Standard form
General form
homework
Find the equation of the parabola that has a
maximum at (3, 6) and passes through the point (9, 5).
Vertex: (3,6)
(x-3)2 = -36(y-6) Standard Form
2
x
- 6x +36y -207 General Form
Find the equation of the parabola that has a
vertex at (2,1) and focus (2,4).
2
(x-h)
h=2,
= 4p (y-k)
k=1,
p= 4-1 = 3
(x-2)2 = 4(3) (y-1)
2
(x-2)
2
X
= 12 (y-1)
- 4x -12y + 16 = 0
Standard Form
General Form
• The equation of the
axis of symmetry is y = k.
• The coordinates of the focus are (h
+ p, k).
• The equation of the directrix is
x = h - p.
• The equation of the
axis of symmetry is x = h.
• The coordinates of the focus
are (h, k + p).
• The equation of the directrix is
y = k - p.
Find the coordinates of the vertex and focus,Analyzing a Parabola
the equation of the directrix, the axis of symmetry,
and the direction of opening of y2 - 8x - 2y - 15 = 0.
y2 - 8x - 2y - 15 = 0
y2 - 2y + _____
1 = 8x + 15 + _____
1
(y - 1)2 = 8x + 16
(y - 1)2 = 8(x + 2) Standard
form
4p = 8
p=2
The vertex is (-2, 1).
The focus is (0, 1).
The equation of the directrix is x + 4 = 0.
The axis of symmetry is y - 1 = 0.
The parabola opens to the right.
Finding the FOCUS
y=
DIRECTRIX
2
4x
x=
2
-3y
y = 4(4py)
x = -3(4px)
y = 16py
x = -12px
1 = 16p
1 = -12p
1/16 = p
-1/12 = p
FOCUS:
(0, 1/16)
FOCUS:
Directrix
Y = - 1/16
Directrix
(-1/12, 0)
x = 1/12
Find the equation of the parabola that has a
min at (-2, 6) and passes through the point (2, 8).
The axis of symmetry is parallel to the y-axis.
The vertex is (-2, 6), therefore, h = -2 and k = 6.
Substitute into the standard form of the equation
and solve for p:
(x - h)2 = 4p(y - k)
x = 2 and y = 8
(2 - (-2))2 = 4p(8 - 6)
16 = 8p
2=p
(x - h)2 = 4p(y - k)
(x - (-2))2 = 4(2)(y - 6)
(x + 2)2 = 8(y - 6)
x2 + 4x + 4 = 8y - 48
x2 + 4x - 8y + 52 = 0
Standard form
General form
Ex: Write the equation of the parabola with a focus at (3, 5) and
the directrix at x = 9, in standard form and general form.
The distance from the focus to the directrix is 6 units, therefore, 2p
= -6, p = -3. Thus, the vertex is (6, 5).
The axis of symmetry is parallel to the x-axis:
(y - k)2 = 4p(x - h)
h = 6 and k = 5
(y - 5)2 = 4(-3)(x - 6)
(y - 5)2 = -12(x - 6)
Standard form
y2 - 10y + 25 = -12x + 72
y2 + 12x - 10y - 47 = 0
General form