A quadratic equation always represents a parabola.Every quadratic form of equation when graphically represented, forms a parabola. When a ball is thrown or kicked, It follows a parabola.
The equation of parabola is can be written as

Where,
a, b and c are constants.
(x, y) is an arbitrary point lying on the parabola.
a $\neq$ 0. If a equals zero, then equation will be no more a quadratic equation. Hence it will not represent a parabola.

## Standard Form of a Parabola

The equation of parabola is defined above, but more generally and specifically, the standard and general form of a parabola is given below.
Where,
A, B, C, D, E and F are constants.
x and y are coordinates of an arbitrary point (x, y) lying on the parabola.

## Vertex Form of a Parabola

Sometimes, the equation of parabola is given in vertex form. The general and standard form of parabola is deducible to the vertex form of parabola. The equations of parabola in vertex form are:
Where,
(h, k) is vertex of parabola.
(x, y) is any arbitrary point on parabola.
If vertex of parabola is origin (0, 0), then h = 0 and k = 0. So, following are the parabolic equations in this case are

## Equation of Parabola

Equation of parabola can be in any of the given forms:
• Standard Form
• Vertex Form
Standard form is referred as quadratic form. It can be in below two formats:
• y = ax2 + bx + c
• Ax2 + Bxy + Cy2 + Dx + Ey + F = 0
Vertex form can be as follows:
Vertex at origin:
• y2 = 4ax
• y2 = -4ax
• x2 = 4ay
• x2 = -4ay
Vertex at (h, k):
• (y - k)2 = 4a (x - h)
• (y - k)2 = - 4a (x - h)
• (x - h)2 = 4a (y - k)
• (x - h)2 = - 4a (y - k)

## Solving Parabola Equations

Here, let us see how to solve parabolic equations.
While solving parabola equation, following points should be kept in mind:
• Deduce given equation in standard parabolic form.
• Determine "a" by comparing it with standard form.
• Now, determine the coordinates of focus, vertex, directrix etc. according to given format.
Let us consider few examples:

### Solved Example

Question: Consider a parabola: y2 - 12x = 0. Determine its vertex and focus.
Solution:
y2 = 12x
Comparing it with standard parabola equation, y2 = 4ax:
Vertex = (0, 0)
4a = 12
a = 3
Focus = (a, 0) = (3, 0)

Example 2: Deduce the quadratic equation y2 - 4x - 3 = 0 in vertex form of parabola and find its vertex, focus and equation of directrix.
Solution: y2 - 4x - 3 = 0
y2 = 4x + 3

y2 = 4(x + $\frac{3}{4}$)

Comparing it with standard parabola equation, (y - k)2 = 4a(x - h):

Vertex = (-$\frac{3}{4}$, 0)

4a = 4
a = 1
Focus = (h + a, 0)

= (-$\frac{3}{4}$ + 1, 0)

= ($\frac{1}{4}$, 0)

Directrix:
x = -$\frac{3}{4}$ - 1

x = -$\frac{7}{4}$

## Graphing a Parabola Equation

Following steps should be followed while graphing a parabolic equation:
If equation is in quadratic form:
1. Find a, b and c by comparing it with standard quadratic form y = ax2 + bx + c.
2. Calculate x coordinate of vertex with the help of formula $x = -$$\frac{b}{2a}$. Plugging x in given equation, find y coordinate of vertex.
3. Find x intercept by putting y = 0 in the equation.
4. Find y intercept by substituting x = 0 in the equation.
5. Now plot the graph of parabola with the help of information so obtained.
If equation is in vertex form:
1. Compare it with vertex parabolic form and determine the vertex (h, k).
2. Find coordinates of focus.
3. Determine the equation of directrix.
4. Now plot vertex, focus, directrix, and other required points and graph the parabola.
Let us consider an example:

### Solved Example

Question: Graph the quadratic equation y2 - 2x - 4 = 0
Solution:
Deduce the given equation in vertex form:
y2 = 2(x + 2)
Comparing it with standard form (y - k)2 = 4a (x - h)
h = -2 and k = 0
So, vertex = (-2, 0)
4a = 2
a = $\frac{1}{2}$
Focus = (h + a, 0) = (-$\frac{3}{2}$, 0)

Equation of Directrix:
x = h - a
x = -$\frac{5}{2}$ = -2.5

Finding the value of Y intercept by substituting x = 0
y2 = 4
y = $\pm$ 2
By using above information, we get the following graph of a parabola: