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(Pd 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
x-axis then for every point of the parabola
According to definition of the parabola  FP = d(Pd P) = r
after squaring or
y2 = 2px the vertex form of the equation of the parabola,
or y2 = 4ax  if the distance between the directrix and 
focus is given by  p = 2a.
The explicit form of the equation  y = 2px
shows that to every positive value of x correspond two         opposite values of y which are symmetric relative to the      x-axis.
The parabola   y2 = 2pxp > 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 (P1P2 = 2p)               perpendicular to the axis of the parabola.
Therefore, we can easily sketch the graph of the parabola  
using following points, 
A(0, 0), P1,2(p/2, p) and P3,4(2p, 2p)
Transformation of the equation of a parabola
The equation  y2 = 2pxp < 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.
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 = 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.
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.
Solution:   The coordinates of the point P must satisfy the equation of the parabola
    P(-4, 4)   =>    y2 = 2px
                         42 = 2p(-4)   =>   p = -2
thus, the equation of the parabola   y2 = -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 y2 = 2px inscribed is an equilateral triangle whose one vertex coincides with the vertex of the parabola and whose area A = 2433. 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, 23p) and P'(6p, -23pso that,  P(27, 93) and P'(27, -93).
Therefore, the equation of the parabola  y2 = 2px  or  y2 = 9x.
Example:  Find the vertex, the focus and the equation of the directrix and draw the graph of the parabola 
y = -x2 + 6x - 7.
Solution:   Rewrite the equation of the parabola in the translatable or general form
    (x - x0)2 = 2p(y - y0or   y - y0 = a(x - x0)2
so,  y = - x2 + 6x - 7  =>   y = - (x2 - 6x) - 7
                                          y = -[(x - 3)2 - 9] - 7
 y - y0 = a(x - x0)2,    y - 2 = -(x - 3)2a = -1.
The vertex of the parabola  A(x0, y0),  or  A(3, 2).
The focus  F(x0, y0 + 1/(4a)),  or  F(3, 7/4).
The equation of the directrix,
y = y0 - 1/(4a),    y = 2 + 1/4  or  y = 9/4.
Pre-calculus contents H
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