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| Conic
Sections |
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Parabola
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The parabola whose axis
of symmetry is parallel to the y-axis |
| Equations of the parabola
written in the general form |
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Parametric equations of the parabola
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Equations
of the parabola examples |
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| The parabola whose axis of symmetry is parallel to the y-axis |
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By translating the parabola
x2 = 2py
its vertex is moved from the origin to the point A
(x0, y0)
so that its equation
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transforms to
(x
-
x0)2 = 2p(y
-
y0).
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The axis of symmetry of this parabola is parallel to the
y-axis.
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As we already
mentioned, this parabola is a function that we usually
write
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y = ax2
+ bx + c
or y
-
y0 =
a(x
-
x0)2, |
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| where |
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This parabola opens up if
a >
0 and opens down if
a < 0.
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| Equations of the parabola written in the general form |
| a) the axis of the parabola parallel to the
x-axis
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b) the axis of the parabola parallel to the
y-axis |
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Ay2
+ Bx + Cy +
D =
0, A and
B not
0, |
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Ax2
+ Bx + Cy +
D =
0, A and
C not 0 |
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or x =
ay2
+ by
+ c,
a not 0. |
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or
y =
ax2
+ bx
+ c,
a not 0. |
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| Note
that the parabola has equation that contains only one squared term. |
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| Parametric equations of the parabola
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| Parametric equations of the parabola
y2 =
4ax
with the vertex A
at the origin and the focus F(a,
0), and of its translation (y
-
y0)2 =
4a(x
-
x0)
with the vertex A(x0, y0)
and the focus F(x0
+ a, y0)
written |
| respectively
are, |
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| Parametric equations of the parabola
x2 =
4ay
with the vertex A
at the origin and the focus F(0,
a), and of its translation
(x
-
x0)2 =
4a(y
-
y0)
with the vertex A(x0, y0)
and the focus F(y0,
y0
+ a) written |
| respectively
are, |
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| Equations
of the parabola examples |
| 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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| and
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 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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| Conic
sections contents |
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