Section 9.1
Conics
OBJECTIVE 1
.
Geometric Definition of a Parabola: The collection of all the points
P(x,y) in a a plane that are the same distance from a fixed point, the
focus, as they are from a fixed line called the directrix.
y
Focus
(0,p)
P
Vertex
(h,k)
x
As you can plainly see
the distance from the
focus to the vertex is a
and is the same distance
from the vertex to the
directrix! Neato!
y
Focus
(0,p)
p
2p
Directrix y = -p
And the
equation
is…
Vertex
(h,k)
x
p
1 2
x  4 py or y 
x
4p
2
y
Directrix
ya
Vertex
(h,k)
p
x
Focus (0,-p)
And the
equation is…
p
1 2
x  4 py or y  
x
4p
2
y
2p
Directrix
x  a
p
Vertex
(h,k)
p
Focus
(p,0)
x
And the
equation is…
1 2
y  4 px or x 
y
4p
2
y
p
Vertex
Focus (-p,0)
p
Directrix
xa
(h,k)
x
1 2
y  4 px or x  
y
4p
2
And the
equation is…
STANDARD FORMS
1
1) (x  h)  4 p(y  k) or y 
(x  h) 2  k
4p
Vertex at ( h, k )
Opens up
1
2) (x  h)  4 p(y  k) or y  
(x  h) 2  k
4p
Vertex at ( h, k )
Opens down
1
2
3) (y  k)  4 p(x  h) or x 
(y  k) 2  h
4p
Vertex at ( h, k )
Opens right
1
4) (y  k)  4 p(x  h) or x  
(y  k) 2  h
4p
Vertex at ( h, k )
Opens left
2
2
2
I like to call standard form “Good Graphing Form”
A Couple More Things…………
A parabola that opens up/down:
1) Has a vertex at (h , k)
2) Has an Axis of Symmetry at x = h
3) Coordinates of focus (h , k+p)
4) Equation of Directrix y = k-p
A parabola that opens right/left:
1) Has a vertex at (h , k)
2) Has an Axis of Symmetry at y = k
3) Coordinates of focus (h+p , k)
4) Equation of Directrix x = h-p
General Form of any Parabola
Ax  By  Cx  Dy  E  0
2
2
*Where either A or B is zero!
* You will use the “Completing the Square”
method to go from the General Form to
Standard Form,
Finding the Equation of a Parabola with Vertex (0, 0)
A parabola has vertex (0, 0) and the focus on an axis.
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.
Finding the Equations of Parabolas
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
(6, 5)
Finding the Equations of Parabolas
Find the equation of the parabola that has a minimum 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)
(2 - (-2))2 = 4p(8 - 6)
16 = 8p
2=p
x = 2 and y = 8
(x - h)2 = 4p(y - k)
(x - (-2))2 = 4(2)(y - 6)
(x + 2)2 = 8(y - 6) Standard form
x2 + 4x + 4 = 8y - 48
x2 + 4x + 8y + 52 = 0
General form
3.6.10
Analyzing a Parabola
Find the coordinates of the vertex and focus,
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
4p = 8
p=2
form
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.
3.6.11
Graphing a Parabola
y2 - 10x + 6y - 11 = 0
y2 + 6y + _____
9 = 10x + 11 + _____
9
(y + 3)2 = 10x + 20
(y + 3)2 = 10(x + 2)
y  3   10(x  2)
y   10(x  2)  3
Example #2 Writing the equation
of a parabola
From the graph, the vertex is at the origin,
(0,0), and the directrix is 2 units away
from the vertex.
The parabola opens up, so the equation is
2
in x  4 py form.
Since p = 2 , the equation is
p2
x 2  4 py

x  4( 2) y
2
x  8y
2
(Larson, Boswell, Kanold & Stiff, 2005)
#10 Write the standard form of the
equation of the parabola with the
given focus or directrix with the
vertex at (0,0). Focus
(0,3)
Since the focus has to be inside the
parabola and lie on the axis
of symmetry, this parabola opens up,
and is the form
x 2  4 py
The distance p is the distance from
the vertex to the focus, or in this
case 3.
 x  4(3) y
2
x  12 y
2
General Effects of the Parameters A and C
When A x C = 0, the resulting
conic is an parabola.
When A is zero:
If C is positive,
the parabola opens to the left.
If C is negative,
the parabola opens to the right.
When C is zero:
If A is positive,
the parabola opens up.
If A is negative,
the parabola opens down.
When A = D = 0, or when C = E = 0,
a degenerate occurs.
E.g., x2 + 5x + 6 = 0 x2 + 5x + 6 = 0
(x + 3)(x + 2) = 0
x + 3 = 0 or x + 2 = 0
x = -3
x = -2
The result is two vertical,
parallel lines.
3.6.13