Parabola  Parabola examples
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 = 0,  A and B not 0, Ax2 + Bx + Cy + D = 0,  A and C not 0 or   x = ay2 + by + c,  a not 0. or   y = ax2 + bx + c,  a not 0.
Note that the parabola has equation that contains only one squared term.
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 = 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, -2Ö3pso that,  P(27, 9Ö3) and P'(27, -9Ö3).
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 - y0)  or   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)2,  a = -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 