Parabola PowerPoint

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Section 9.1 Conics
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OBJECTIVE 1
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The Parabola
.
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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!
Focus (0,p)
p
2p
Vertex (h,k)
p
Directrix
y -p
And the equation is
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Directrix
p
Vertex (h,k)
p
Focus (0,-p)
And the equation is
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2p
Directrix
p
p
Focus (p,0)
Vertex (h,k)
And the equation is
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p
p
Directrix
Vertex (h,k)
Focus (-p,0)
And the equation is
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STANDARD FORMS
I like to call standard form Good Graphing Form
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A Couple More Things

• A parabola that opens up/down
• Has a vertex at (h , k)
• Has an Axis of Symmetry at x h
• 3) Coordinates of focus (h , kp)
• 4) Equation of Directrix y k-p

• A parabola that opens right/left
• Has a vertex at (h , k)
• Has an Axis of Symmetry at y k
• 3) Coordinates of focus (hp , k)
• 4) Equation of Directrix x h-p

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General Form of any Parabola
Where either A or B is zero! You will use the
Completing the Square method to go from the
General Form to Standard Form,
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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.
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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)
(6, 5)
Standard form
y2 - 10y 25 -12x 72 y2 12x -
10y - 47 0
General form
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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)
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)
Standard form
x2 4x 4 8y - 48 x2 4x 8y 52
0
General form
3.6.10
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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.
4p 8 p 2
y2 - 8x - 2y - 15 0 y2 - 2y _____ 8x
15 _____
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(y - 1)2 8x 16 (y - 1)2 8(x 2)
Standard 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
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Graphing a Parabola
y2 - 10x 6y - 11 0
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y2 6y _____ 10x 11 _____
(y 3)2 10x 20 (y 3)2 10(x 2)
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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 in form.
Since p 2 , the equation is
(Larson, Boswell, Kanold Stiff, 2005)
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10 Write the standard form of the equation of
the parabola with the given focus or directrix
with the vertex at (0,0). Focus
Since the focus has to be inside the parabola and
lie on the axis of symmetry, this parabola opens
up, and is the form
The distance p is the distance from the vertex to
the focus, or in this case 3.
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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