Transcript Parabola PowerPoint
Chapter 3
Conics
3.6
MATHPOWER
TM
12, WESTERN EDITION
3.6.
1
The Parabola
The parabola is the locus of all points in a plane that are the same distance from a line in the plane, the directrix , as from a fixed point in the plane, the focus .
Point Focus = Point Directrix PF = PD |
p
| The parabola has one axis of symmetry , which intersects the parabola at its vertex .
The distance from the vertex to the focus is |
p
|.
The distance from the directrix to the vertex is also |
p
|.
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The Standard Form of the Equation of a Parabola with Vertex (0, 0) The equation of a parabola with vertex (0, 0) and focus on the
x
-axis is
y
2 = 4
px
.
The coordinates of the focus The equation of the directrix are (
p
,
0
).
is
x
= -
p
.
If
p
If
p
> 0 , the parabola opens right .
< 0 , the parabola opens left .
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The Standard Form of the Equation of a Parabola with Vertex (0, 0) The equation of a parabola with vertex (0, 0) and focus on the
y
-axis is
x
2 = 4
py
.
The coordinates of the focus The equation of the directrix are (0,
p
).
is
y
= -
p
.
If
p
If
p
> 0 , the parabola opens up .
< 0 , the parabola opens down .
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Sketching a Parabola A parabola has the equation the equation of the directrix.
y
2 = -8
x
. Sketch the parabola showing the coordinates of the focus and The vertex The focus of the parabola is ( is on the
x
-axis .
0, 0 ).
Therefore, the standard equation is
y
2 Hence, 4
p p
= -8 = -2 .
= 4
px
.
The coordinates of the focus are (-2, 0).
F(-2, 0) The equation of the directrix is
x
= -
p
, therefore,
x
= 2 .
x
= 2 3.6.
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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
y
2
p
= 4
px
is equal to the distance from the vertex to the focus, therefore .
p
= -6.
The equation of the parabola is
y
2 = -24
x
.
b) The directrix is defined by
x
= 5.
Since the focus is on the
x
-axis, the equation of the parabola is
y
2 = 4
px
.
The equation of the directrix is
x
The equation of the parabola is
y
2 = -
p
, therefore
-p =
5 or
p = -
5.
= -20
x
.
c) The focus is (0, 3).
Since the focus is (0, 3), the equation of the parabola is
x
2
p
= 4
py
.
is equal to the distance from the vertex to the focus, therefore
p
= 3.
The equation of the parabola is
x
2 = 12
y
.
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The Standard Form of the Equation with Vertex (
h, k
) For a parabola with the axis of symmetry parallel to the
y
-axis and vertex at (
h, k
):
• • • • • •
The equation of the axis of symmetry The coordinates of the focus are (
h
,
k
The equation of the directrix is
y
=
k
is
x
+
p
).
=
h
.
p
.
When
p
is positive , the parabola opens upward .
When
p
is negative , the parabola opens downward .
The standard form for parabolas parallel to the
y
-axis is: (
x
-
h
) 2 = 4
p
(
y
-
k
) The general form of the parabola is
Ax
2 +
Cy
2 +
Dx
+
Ey
+
F
= 0 where
A
= 0 or
C
= 0.
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The Standard Form of the Equation with Vertex (
h, k
) For a parabola with an axis of symmetry parallel to the
x
-axis and a vertex at (
h, k
):
• • • • •
The equation of the axis of symmetry The coordinates of the focus are (
h
+
p
is ,
k y
).
=
k
.
The equation of the directrix is
x
When
p
is positive , the parabola opens to the right .
When
p
is negative , the parabola opens to the left .
=
h
-
p
.
•
The standard form for parabolas parallel to the
x
-axis is: (
y
-
k
) 2 = 4
p
(
x
-
h
) 3.6.
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Finding the Equations of Parabolas Write the equation of the parabola with a focus at (3, 5) the directrix at
x
= 9 , in standard form and general form and The distance from the focus to the directrix is 6 units, therefore, 2
p
= -6,
p
= -3. Thus, the vertex is (6, 5).
The axis of symmetry is parallel to the
x
-axis: (
y
(
y
(
y
-
k
- 5) - 5) ) 2 2 2 = 4
p
( = 4(-3)( = -12(
x x
-
x h
) - 6) - 6)
h
= 6 and
k
= 5 Standard form
y
2
y
2 + 12
x
- 10
y
- 10
y
+ 25 = -12
x
- 47 = 0 + 72 General form (6, 5) 3.6.
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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 = 4
p
(
y
-
k
) (2 - (-2)) 2 = 4
p
(8 - 6) 16 = 8
p
2 =
p x
= 2 and
y
= 8
x
2
x
2 + 4
x
(
x
(
x
(
x
-
h
) 2 - (-2)) + 2) 2 2 = 4
p
(
y
= 4(2)(
y
= 8(
y
- 6)
k
) - 6) + 4 + 8
y x
+ 4 = 8 + 52 = 0
y
- 48 Standard form General form 3.6.
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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
y
2 - 8
x
- 2
y
- 15 = 0 .
y y
2 2 - 8
x
- 2
y
- 2
y
- 15 = 0 + _____ = 8
x
(
y
(
y
- 1) 2 - 1) 2 = 8
x
= 8(
x
+ 16 + 2) Standard form 4
p p
= 8 = 2 The vertex is (-2, 1).
The focus is (0, 1).
The equation of the directrix is
x
The axis of symmetry is
y
- 1 = 0 .
The parabola opens to the right .
+ 4 = 0 .
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Graphing a Parabola
y
2 - 10
x
+ 6
y
- 11 = 0
y
2 + 6
y
9
x
(
y
(
y
+ 3) 2 + 3) 2 = 10
x
= 10(
x
+ 20 + 2)
y
3
10(x
2)
y
10(x
2)
3
General Effects of the Parameters
A
and
C
When
A
x
C
= 0 , the resulting conic is an parabola .
When
A
If
C
is zero: is positive , the parabola opens to the left .
If
C
is negative , the parabola opens to the right .
When If
A C
is zero: is positive , the parabola opens up .
If
A
is negative , the parabola opens down .
When
A = D =
0, or when
C
a degenerate occurs.
=
E
= 0, E.g.,
x
2 + 5
x
+ 6 = 0
x
(
x x
2 + 5
x
+ 3)(
x
+ 6 = 0 + 2) = 0 + 3 = 0 or
x x
+ 2 = 0 = -3
x
= -2 The result is two vertical, parallel lines.
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