The **general equation of the parabola** contains quadratic terms in *x* and in *and*, as well as linear terms in both variables plus an independent term. The axis of symmetry of the first is parallel to the vertical axis and that of the second is parallel to the horizontal axis.

In general, the quadratic equation that lacks the cross term *xy* is written as:

Ax2 + Cy2 +Dx + Ey + F = 0

The values of A, C, D, E, and F are real numbers. Imposing the conditions A∙C = 0 and A+C≠0, the curve that results from graphing the points that satisfy said equation is a parabola.

**Case 1**

For a vertical parabola, its general equation is:

Ax2 + Dx + Ey + F = 0

Where A and E are different from 0. In other words, when a term with x2 appears, the parabola is vertical.

**case 2**

On the other hand, for the horizontal parabola we have:

Cy2 + Dx + Ey + F = 0

Here C and D are also different from 0, therefore the quadratic term corresponds to y2.

In any case, the general equation of the parabola is quadratic in one of the variables and linear in the other.

**Elements of the parable**

The parabola, defined as a locus, consists of the set of points in a plane that are equidistant from another point called *focus* and also of a straight line, known as *directrix line*.

Starting from the general equation, it is possible to study the parabola by specifying its elements. Including the focus and the directrix line, said elements, briefly described are:

–**Axis**which refers to the axis of symmetry of the parabola, can be horizontal (parallel to the abscissa axis) or vertical (parallel to the ordinate axis).

–**Orientation**, which in turn corresponds to the orientation of the axis. The parabola is vertical if its axis of symmetry is vertical, and it is horizontal when the axis is also vertical.

–**Vertex**is the point at which the axis intersects the parabola.

–**Focus**point located on the axis, inside the parabola and at a distance *p *of the vertex. All the points of the parabola are equidistant from the focus and the directrix line.

–**Parameter**is the distance *p* between the focus and the vertex.

–**directrix line**which is perpendicular to the axis and is also a distance away *p* from the vertex of the parabola, but it does not intersect it, since it is outside.

–**straight side**is the chord that passes through the focus, intersecting the parabola in two points, perpendicular to its axis.

–**Eccentricity**which in the case of the parabola is always equal to 1.

–**Graphic representation**.

The information to determine all these elements is contained in the general equation.

**the canonical form**

To determine the elements of the parabola, it is sometimes convenient to go from the general form to the canonical form of the parabola, by means of the method of completing squares in the quadratic variable.

This canonical form is:

(xh)2 = 4p(y–k)

Where the point (h,k) is the vertex V of the parabola. Likewise, the canonical form can be converted to the general equation, developing the notable product and rearranging the terms.

**examples**

**Example 1**

The following are equations of the parabola in general form:

a) 4×2 + 5y – 3 = 0

b) 1 – 2y + 3x –y2 = 0

In a) the coefficients are identified: A = 4, C = 0, D = 0, E = 5, F = -3. It is a parabola whose axis of symmetry is vertical.

On the other hand, in b) the general equation is:

– y2 + 3x – 2y + 1= 0

And the coefficients are: C = –1, D = 3, E = -2 and F = 1.

**Example 2**

The following parable is in canonical form:

(y–1)2 = 6(x–3)

To find its general equation, first develop the notable product and make the parentheses on the right:

y2 –2y + 1 = 6x –18

Now all the terms are passed to the left and conveniently grouped:

y2 –2y + 1– 6x +18 = 0 → y2 – 6x –2y + 19 = 0

Since the quadratic term is y2, it is a horizontal parabola. The coefficients are:

C=1; D= –6; E = –2, F = 19.

**solved exercises**

**Exercise 1**

The following parable is given in general form:

x2 –10x–12y – 11 = 0

It is requested to write it in the canonical form.

**Solution**

Going to canonical form is achieved by completing squares, in this case, on the variable x. We begin by writing the terms in x between parentheses:

(x2 –10x) –12y – 11 = 0

You have to transform what is between parentheses into a perfect square trinomial, which is achieved by adding 52, which naturally has to be subtracted, because otherwise the expression is altered. It looks like this:

(x2 −10x+52) −12y – 11−52= 0

The three terms in parentheses make up the perfect square trinomial (x-5)2. You can check developing this remarkable product to corroborate. Now the parable remains:

(x–5)2 –12y –36 = 0

What follows is to factor the terms outside the parentheses:

(x–5)2 –12(y +3) = 0

Which finally becomes:

(x–5)2 =12(y +3)

**Example 2**

Find the elements of the previous parabola and build its graph.

**Solution **

**Vertex**

The vertex of the parabola has coordinates V (5, -3)

**Axis**

The line x = 5.

**Parameter**

Regarding the parameter value *p* which appears in the canonical form: (x–h)2 = 4p(y–k) is found by comparing both equations:

4p = 12

p = 12/4 = 3

**Orientation**

This parabola is vertical and opens up. Since the vertex is located at x = 5, y = -3, then the axis of symmetry is the vertical line x = 5.

**Focus**

The focus is on the line x = 5, therefore it has coordinate x = 5 as well.

the coordinate *and* of the focus must be p units above k, that is: p + k = 3 + (-3) = 0, then the focus is at the point (5,0).

**directrix line**

It is perpendicular to the axis, therefore it is of the form y = c, now, since it is a distance p from the vertex, but outside the parabola, it means that it is a distance p below k:

y = k – p = -3-3 = -6

**straight side**

This segment cuts the parabola, passes through the focus and is parallel to the directrix line, therefore it is contained in the line y = 0.

**Graphic representation**

It can be easily obtained from free online graphing software such as Geogebra. In the input box it is placed like this:

**References**

Baldor. 1977. Elementary Algebra. Venezuelan Cultural Editions.

Hoffman, J. Selection of Mathematics topics. Volume 2.

Jiménez, R. 2008. Algebra. Prentice Hall.

Stewart, J. 2006. Precalculus: Mathematics for Calculus. 5th. Edition. Cengage Learning.

Zill, D. 1984. Algebra and Trigonometry. McGraw Hill.