

Parabola

Definition and construction
of the parabola

Construction of the parabola

Vertex form of the equation of a parabola

Transformation of the equation of a parabola

Parabola examples






Definition and construction
of the parabola

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. 
A parabola is uniquely determined by the distance of
the focus from the directrix.

This distance is called the
focal parameter
p
and its
midpoint A is the vertex
(apex) of the parabola.

The parabola has an axis of symmetry which passes
through the
focus perpendicular to the directrix.

The distance of any point
P
of the parabola from the
directrix and from the focus is denoted r, so

d(P_{d}_{
}P) =
FP =
r.

Construction of the parabola

To a given parameter
p
of the parabola draw corresponding directrix d
and the focus F.

To the distances greater then
p/2
draw lines, of arbitrary dense, parallel to the
directrix.

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.

The distance
p/2
from the vertex A
to the directrix or focus is called focal
distance.





Vertex form of the equation of a parabola

If a parabola is placed so that its vertex coincides with the origin of the coordinate system and its axis lies
along the xaxis then for every point of the parabola 

According to
definition of the parabola FP =
d(P_{d}_{ }P) = r


after squaring 

or 

y^{2} =
2px 
the
vertex form of the equation of the parabola, 

or 
y^{2} =
4ax 
if the distance between the directrix and 

focus is given by
p = 2a.

The explicit form of the equation
y = ±
Ö2px

shows that to
every positive value of x
correspond two
opposite values of y
which are symmetric relative to the
xaxis.





The parabola
y^{2} =
2px,
p >
0

is not defined for
x <
0, it opens to
the right.

For x
= p/2
the
corresponding ordinate
y = ± p.

This parabola is not a function since the vertical line
crosses
the graph more than once.

A focal chord is a line segment passing through the focus
with endpoints on the curve.

The latus rectum is the focal chord
(P_{1}P_{2} =
2p)
perpendicular
to the axis of the parabola.

Therefore, we can easily sketch the graph of the parabola
using following points,

A(0,
0), P_{1,2}(p/2,
±
p) and P_{3,4}(2p,
±2p)





Transformation of the equation of a parabola

The equation
y^{2} = 2px,
p < 0
represents the parabola opens to the left since must be y^{2}
> 0. Its axis of
symmetry is the xaxis. 
If variables
x and
y change the role obtained is the parabola whose axis of
symmetry is yaxis.

For
p >
0
the parabola opens up, if
p <
0
the parabola opens down as shows
the below figure. 

This parabola we often write
y = ax^{2},
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.




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 or
general 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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