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Parabola
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Definition and construction
of the parabola
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Construction of the parabola
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Vertex form of the equation of a parabola
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Transformation of the equation of a parabola
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Parabola examples
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| Definition and construction
of the parabola
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| A parabola is the set of all points in a plane that are the same distance from a point F, called the focus, and a
line d, called the
directrix. |
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A parabola is uniquely determined by the distance of
the focus from the directrix.
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This distance is called the
focal parameter
p
and its
midpoint A is the vertex
(apex) of the parabola.
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| The parabola has an axis of symmetry which passes
through the
focus perpendicular to the directrix.
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The distance of any point
P
of the parabola from the
directrix and from the focus is denoted r, so
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d(Pd
P) =
FP =
r.
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Construction of the parabola
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To a given parameter
p
of the parabola draw corresponding directrix d
and the focus F.
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To the distances greater then
p/2
draw lines, of arbitrary dense, parallel to the
directrix.
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Then, intersect each line at two
symmetric points by arc centered at the focus with a radius
which equals the distance of that line from the
directrix.
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The distance
p/2
from the vertex A
to the directrix or focus is called focal
distance.
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| Vertex form of the equation of a parabola
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If a parabola is placed so that its vertex coincides with the origin of the coordinate system and its axis lies
along the x-axis then for every point of the parabola |
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According to
definition of the parabola FP =
d(Pd P) = r
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| after squaring |
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or |
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| y2 =
2px |
the
vertex form of the equation of the parabola, |
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| or |
y2 =
4ax |
if the distance between the directrix and |
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focus is given by
p = 2a.
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The explicit form of the equation
y = ±
Ö2px
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shows that to
every positive value of x
correspond two
opposite values of y
which are symmetric relative to the
x-axis.
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The parabola
y2 =
2px,
p >
0
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is not defined for
x <
0, it opens to
the right.
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For x
= p/2
the
corresponding ordinate
y = ± p.
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This parabola is not a function since the vertical line
crosses
the graph more than once.
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A focal chord is a line segment passing through the focus
with endpoints on the curve.
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The latus rectum is the focal chord
(P1P2 =
2p)
perpendicular
to the axis of the parabola.
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Therefore, we can easily sketch the graph of the parabola
using following points,
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A(0,
0), P1,2(p/2,
±
p) and P3,4(2p,
±2p)
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| Transformation of the equation of a parabola
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| The equation
y2 = 2px,
p < 0
represents the parabola opens to the left since must be y2
> 0. Its axis of
symmetry is the x-axis. |
| If variables
x and
y change the role obtained is the parabola whose axis of
symmetry is y-axis.
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| For
p >
0
the parabola opens up, if
p <
0
the parabola opens down as shows
the below figure. |
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This parabola we often write
y = ax2,
where
a = 1/(2p) , with the focus at
F(0,
1/(4a)) and the directrix |
| y =
-1/(4a).
This parabola is a function since a vertical line crosses its graph at only one point.
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| Example:
Write equation of the parabola
y2 = 2px
passing through the point P(-4,
4) and find the focus,
the equation of the directrix and draw its graph. |
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Solution:
The coordinates of the point P
must satisfy the equation of the parabola
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P(-4,
4)
=>
y2 =
2px
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42 = 2p(-4)
=> p =
-2
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thus, the
equation of the parabola y2 =
-4x.
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The coordinate
of the focus,
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since F(p/2,
0) then
F(-1,
0).
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The equation of the directrix,
as x =
- p/2,
x =
1.
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| Example:
Into a parabola y2 =
2px
inscribed is an equilateral triangle whose one vertex coincides with the
vertex of the parabola and whose area A =
243Ö3.
Determine equation of the parabola and remaining vertices of the triangle. |
Solution:
Let write coordinates of a point
P
of the
parabola as elements of the equilateral triangle
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As the point P
lies on the parabola then
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The area of the equilateral triangle we express by
coordinates of
P
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| Show the parameter of the parabola by the side of the triangle, |
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the vertices of the triangle P(6p,
2Ö3p) and
P'(6p,
-2Ö3p)
so that, P(27,
9Ö3) and
P'(27,
-9Ö3). |
| Therefore,
the equation of the parabola y2 =
2px or
y2 = 9x. |
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| Example:
Find the vertex, the focus and the equation of the directrix and draw the graph of the parabola |
| y =
-x2
+ 6x -
7. |
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Solution:
Rewrite
the equation of the parabola in the translatable or
general form
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(x
-
x0)2 = 2p(y
-
y0) or
y
-
y0 = a(x
-
x0)2
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so, y =
- x2
+ 6x -
7
=> y =
- (x2
-
6x) -
7
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y =
-[(x
-
3)2
-
9] -
7
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y
-
y0 = a(x
-
x0)2,
y
-
2 = -(x
-
3)2,
a
= -1.
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The vertex of the parabola
A(x0, y0),
or
A(3, 2).
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The focus
F(x0,
y0 + 1/(4a)),
or
F(3,
7/4).
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The equation of the directrix,
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y =
y0 -
1/(4a),
y = 2 +
1/4 or
y =
9/4.
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| Pre-calculus contents
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