 ## Parabola The standard form equation of a general quadratic (polynomial functions of degree 2) function is
f(x) = ax2 + bx + c where a ≠ 0.

If b = 0, the quadratic function has the form f(x) = ax2 + c.
Since f(-x) = a(-x)2 + c = ax2 +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 = x2 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=x2 or opens downward like the graph of y = -x2 .

In the figure, the vertex of the graph of y=x is (0,0) and the line of symmetry is x = 0.

Definition:  Parabola

1.Algebric:
A Parabola is the graph of a quadratic relation of either form where a ≠ 0;
y = ax2 + bx + c or x = ay2 + by + c
2. Geometric:
A parabola is the set of all points in a plane and a given line.
 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. #### Connection between Algebra and Geometry of Parabola

Show that an equation for the parabola with the focus (o, p) and directex y = -p is  y = 1/4p x2 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 x2

Conversely, we must also show that a point satisfying the equation y = 1/4p x2 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

• distance from (x, y) to y = -p is |x + y|
• distance from (x, y) to (0, p) is [(x2 + (y-p)2]1/2

Consequently, we can drive an equation for the parabola as follows:

|y + p| =  [(x2 + (y-p)2]1/2

squaring both sides.

(y + p)2 =  (x2 + (y-p)2

y2 + 2py + p2 = x2 + y2 -2py + p2

4py = x2

y = 1/4p x2

By  reversing the above steps, we see that a solution (x, y) of  y = 1/4p x2 is equidistance from (0, p) and the line y =-p.

This completes the proof.

#### Characteristics of a Parabola

The standard form of a parabola with vertex (0,0) are as follows:

 Algebra Geometry y = ax2 Focus: (0, 1/4a) Directrix: y= -1/4a x = ay2 Focus (1/4a, 0) Directrix: x = -1/4a

The line of symmetry for y =ax2 is the y-axis. Similarly, the line of symmetry for x=ay2 is the x-axis.

• Find the focus and Directrix for the parabola y = -1/2x2

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

• Find an equation in standard form for the parabola whose directrix is the line x = 2 and focus is the point (-2, 0).
The directrix, x = 2 is to the left. Therefore, the parabola is one with a horizontal line of symmetry and opens  to the left.

Because x = 2 = -1/4a

=> 8a = -1
=> a = -1/8

The standard form equation for the parabola is

x = ay2 = -1/8 y2

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