
Conic
Sections 


Parabola

The parabola whose axis
of symmetry is parallel to the yaxis 
Equations of the parabola
written in the general form 
Parametric equations of the parabola

Equations
of the parabola examples 





The parabola whose axis of symmetry is parallel to the yaxis 
By translating the parabola
x^{2} = 2py
its vertex is moved from the origin to the point A
(x_{0}, y_{0})
so that its equation

transforms to
(x

x_{0})^{2} = 2p(y

y_{0}).

The axis of symmetry of this parabola is parallel to the
yaxis.

As we already
mentioned, this parabola is a function that we usually
write


y = ax^{2}
+ bx + c
or y

y_{0} =
a(x

x_{0})^{2}, 


where 



This parabola opens up if
a >
0 and opens down if
a < 0.





Equations of the parabola written in the general form 
a) the axis of the parabola parallel to the
xaxis


b) the axis of the parabola parallel to the
yaxis 
Ay^{2}
+ Bx + Cy +
D =
0, A and
B not
0, 

Ax^{2}
+ Bx + Cy +
D =
0, A and
C not 0 
or x =
ay^{2}
+ by
+ c,
a not 0. 

or
y =
ax^{2}
+ bx
+ c,
a not 0. 

Note
that the parabola has equation that contains only one squared term. 

Parametric equations of the parabola

Parametric equations of the parabola
y^{2} =
4ax
with the vertex A
at the origin and the focus F(a,
0), and of its translation (y

y_{0})^{2} =
4a(x

x_{0})
with the vertex A(x_{0}, y_{0})
and the focus F(x_{0}
+ a, y_{0})
written 
respectively
are, 





Parametric equations of the parabola
x^{2} =
4ay
with the vertex A
at the origin and the focus F(0,
a), and of its translation
(x

x_{0})^{2} =
4a(y

y_{0})
with the vertex A(x_{0}, y_{0})
and the focus F(y_{0},
y_{0}
+ a) written 
respectively
are, 






Equations
of the parabola examples 
Example:
Write equation of the parabola
y^{2} = 2px
passing through the point P
(4,
4) and find the focus,
the equation of the directrix and draw its graph. 

Solution:
The coordinates of the point P
must satisfy the equation of the parabola

P(4,
4)
=>
y^{2} =
2px

4^{2} = 2p(4)
=> p =
2

thus, the
equation of the parabola y^{2} =
4x.

The coordinate
of the focus,

since F(p/2,
0) then
F(1,
0).

The equation of the directrix,
as x =
 p/2,
x =
1.





Example:
Into a parabola y^{2} =
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


As the point P
lies on the parabola then


The area of the equilateral triangle we express by
coordinates of
P





Show
the parameter of the parabola by the side of the triangle, 


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 y^{2} =
2px or
y^{2} = 9x. 

Example:
Find the vertex, the focus and the equation of the directrix and draw the graph of the parabola 
y =
x^{2}
+ 6x 
7. 
Solution:
Rewrite
the equation of the parabola in the translatable form

(x

x_{0})^{2} = 2p(y

y_{0}) or
y

y_{0} = a(x

x_{0})^{2}

so, y =
x^{2}
+ 6x 
7
=> y =
(x^{2}

6x) 
7

y =
[(x

3)^{2}

9] 
7

y

y_{0} = a(x

x_{0})^{2},
y

2 = (x

3)^{2},
a
= 1.

The vertex of the parabola
A(x_{0}, y_{0}),
or
A(3, 2).

The focus
F(x_{0},
y_{0 }+ 1/(4a)),
or
F(3,
7/4).

The equation of the directrix,

y =
y_{0 }
1/(4a),
y = 2 +
1/4 or
y =
9/4.












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