The standard form equation of a general quadratic (polynomial
functions of degree 2) function is
f(x) = ax^{2} + bx + c where a
≠
0.
If b = 0, the quadratic function has the form f(x) = ax^{2}
+ c.
Since f(-x) = a(-x)^{2} + c =
ax^{2} +c = f(x),
Such quadratic functions are even functions, which means that the y-axis is a line of symmetry of the graph of f.
The graph of a quadratic function is a parabola, a line-symmetric curve whose shape is like the graph of y = x^{2 }shown in figure. The point of intersection of the parabola and its line of symmetry is the vertex of the parabola and is the lowest or highest point of the graph. The graph of a parabola either opens upward like y=x^{2 }or opens downward like the graph of y = -x^{2 }.
In the figure, the vertex of the graph of y=x^{2 } is (0,0) and the line of symmetry is x = 0.
Definition: Parabola
From the geometric point of view, the given point is the focus of the parabola and the given line is its directrix. It can be shown that the line of symmetry of the parabola is the line perpendicular to the directrix through the focus. The vertex of the parabola is the point of the parabola that is closet to both the focus and directrix. |
Show that an equation for the parabola with the focus (o, p) and directex y = -p is y = 1/4p x^{2}
Proof:
We must show that a point (x, y) that is equidistant from (o, p) and the line y = -p also satisfies the equation y = 1/4p x^{2}
Conversely, we must also show that a point satisfying the equation y = 1/4p x^{2 } is equidistance from (o,p) and the line y = -p.
We assume that p>0. The argument is similar for p<0.
First, if (x, y) is equidistance from (0, p) and the line y = -p, then
Consequently, we can drive an equation for the parabola as follows:
|y + p| = [(x^{2} + (y-p)^{2}]^{1/2}
squaring both sides.
(y + p)^{2} = (x^{2} + (y-p)^{2}
y^{2} + 2py + p^{2} = x^{2} + y^{2} -2py + p^{2}
4py = x^{2}
y = 1/4p x^{2}
By reversing the above steps, we see that a solution (x, y) of y = 1/4p x^{2 }is equidistance from (0, p) and the line y =-p.
This completes the proof.
The standard form of a parabola with vertex (0,0) are as follows:
Algebra |
Geometry |
y = ax^{2} | Focus: (0, 1/4a) Directrix: y= -1/4a |
x = ay^{2} | Focus (1/4a, 0) Directrix: x = -1/4a |
The line of symmetry for y =ax^{2} is the y-axis. Similarly, the line of symmetry for x=ay^{2} is the x-axis.
Compare with the form y = ax2
=> a = -1/2
Therefore, the focus is (0, 1/4a) = (0, 1/[4(-1/2)])
= (0, -1/2) and the directrix is the line y = -1/4a = -1/[4(-1/2)] = 1/2
Because x = 2 = -1/4a
=> 8a = -1
=> a = -1/8
The standard form equation for the parabola is
x = ay^{2} = -1/8 y^{2}