Conic Sections
    Parabola
      The parabola whose axis of symmetry is parallel to the y-axis
         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 y-axis
By translating the parabola  x2 = 2py  its vertex is moved  from the origin to the point A (x0, y0) so that its equation
transforms to (x - x0)2 = 2p(y - y0).                             
The axis of symmetry of this parabola is parallel to the       y-axis.
As we already mentioned, this parabola is a function that   we usually write
  y = ax2 + bx + c  or  y - y0 = a(x - x0)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 x-axis  b) the axis of the parabola parallel to the y-axis
    Ay2 + Bx + Cy + D = 0A and B not 0,      Ax2 + Bx + Cy + D = 0A and C not 0
      or   x = ay2 + by + ca not 0.         or   y = ax2 + bx + ca not 0.
Note that the parabola has equation that contains only one squared term.
Parametric equations of the parabola
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,    
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,     
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.
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 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.
Conic sections contents
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