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
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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