What is the standard equation of a parabola that opens rightward and whose vertex is at (h, k)?

As we saw in Quadratic Functions , a parabola is the graph of a quadratic function. As part of our study of conics, we'll give it a new definition. A parabola is the set of all points equidistant from a line and a fixed point not on the line. The line is called the directrix, and the point is called the focus. The point on the parabola halfway between the focus and the directrix is the vertex. The line containing the focus and the vertex is the axis. A parabola is symmetric with respect to its axis. Below is a drawing of a parabola.

Figure %: In the parabola above, the distance d from the focus to a point on the parabola is the same as the distance d from that point to the directrix.

If a parabola has a vertical axis, the standard form of the equation of the parabola is this: (x - h)2 = 4p(y - k), where p≠ 0. The vertex of this parabola is at (h, k). The focus is at (h, k + p). The directrix is the line y = k - p. The axis is the line x = h. If p > 0, the parabola opens upward, and if p < 0, the parabola opens downward.

If a parabola has a horizontal axis, the standard form of the equation of the parabola is this: (y - k)2 = 4p(x - h), where p≠ 0. The vertex of this parabola is at (h, k). The focus is at (h + p, k). The directrix is the line x = h - p. The axis is the line y = k. If p > 0, the parabola opens to the right, and if p < 0, the parabola opens to the left. Note that this graph is not a function.

Let P = (x, y) be a point on a parabola. Let l be the tangent line to the parabola at the point P. Let

be a line segment whose endpoints are the focus of the parabola and P. Every parabola has the following property: the angle θ between the tangent line l and the segment

equal to the angle μ between the tangent line and the axis of the parabola. This means (in a physical interpretation) that a beam sent from the focus to any point on the parabola is reflected in a line parallel to the axis. Furthermore, if a beam traveling in a line parallel to the axis contacts the parabola, it will reflect to the focus. This is the principle on which satellite dishes are built.

Figure %: The reflective property of a parabola: θ = μ

Did you know you can highlight text to take a note? x

A parabola is a conic section. It is a slice of a right cone parallel to one side (a generating line) of the cone. Like the circle, the parabola is a quadratic relation, but unlike the circle, either x will be squared or y will be squared, but not both. You worked with parabolas in Algebra 1 when you graphed quadratic equations. We will now be investigating the conic form of the parabola equation to learn more about the parabola's graph.

Conic Equations of Parabolas:

You recognize the equation of a parabola as being y = x2 or
y = ax2 + bx + c from your study of quadratics. And, of course, these remain popular equation forms of a parabola. But, if we examine a parabola in relation to its focal point (focus) and directrix, we can determine more information about the parabola. We are now going to look more closely at the coefficient of the x2 term to see what additional information it can tell us about the graph of the parabola. Keep in mind that all information you already know about parabolas remains true!

Parabola with Vertex at Origin (0,0)
(axis of symmetry parallel to the y-axis)

Conic Forms of Parabola Equations:

with the vertex at (0,0),
focus at (0, p) and directrix y = -p

In the example at the right, the coefficient of x² is 1, so

, making p = ¼.
The vertex is (0,0), the focus is (0,¼), and the directrix is y = -¼.

The distance from the vertex (in this case the origin) to the

focus is traditionally labeled as "p".

The leading coefficient in y = ax2 + bx + c is labeled "a".
So when examining the coefficient of x2, we are examining a.

p is the distance from the vertex to the focus.

You remember the vertex form of a parabola as being y = a(x - h)2 + k where (h, k) is the vertex of the parabola. If we let the coefficient of x2 (or a) =

and perform some algebraic maneuvering, we can get the next equation.

Parabola with Vertex at (h, k)
(axis of symmetry parallel to the y-axis)
(Known as "standard form".)

Conic Form of Parabola Equation:
(x - h)2 = 4p(y - k)
with the vertex at (h, k), the focus
at (h, k+p) and the directrix
y = k - p

Since the example at the right is a translation of the previous graph, the relationship between the parabola and its focus and directrix remains the same (p = ¼). So with a vertex of (2,-3), we have:
(x - 2)2 = 4(¼) (y - (-3))
(x - 2)2 = y + 3
The focus is at (2,-3+¼) or (2,-2¾) and
the directrix is y = -3-¼ or y = -3¼

This "new" equation is just another form
of the old "vertex form" of a parabola.
OLD: y = (x - 2)² - 3
NEW: (x - 2)² = y + 3

S O M E T H I N G N E W ! ! !

Up to this point, all of your parabolas have been either opening upward or opening downward, depending upon whether the leading coefficient was positive or negative respectively. The axis of symmetry of those parabolas is parallel to the y-axis. We will now be looking at a parabola that opens to the right or to the left ("sideways"), with its axis of symmetry parallel to the x-axis.

"Sideways" Parabola Vertex (0,0)
(axis of symmetry parallel to x-axis)

Conic Forms of Parabola Equations:

with the vertex at (0,0),
focus at (p, 0), and directrix x = -p

We will now be examining the coefficient of y², instead of x².

In the example at the right, the coefficient of y² is 1, so

, making p = ¼.
The vertex is (0,0), the focus is (¼,0), and the directrix is x = -¼.

For parabolas opening to the right or
to the left, the y-variable is being squared (instead of the x² we are

used to seeing for parabolas).

"Sideways" Parabolas Vertex (h,k)
(axis of symmetry parallel to x-axis)

Conic Form of Parabola Equation:
(y - k)2 = 4p(x - h)
with the vertex at (h, k), the focus
at (h+p, k) and the directrix
x = h - p

Sideways Equation in Standard Vertex Form:
x = a(y - k)2 + h
with the vertex at (h, k).

Since the example at the right is a translation of the previous graph, the relationship between the parabola and its focus and directrix remains the same.
In the example at the right,
(y - 1)2 = 4(¼) (x - (-2))
(y - 1)2 = x + 2
The vertex is (-2,1), the focus is (-1¾,1)
and the directrix is y = -2¼.

Notice that parabolas that open right or left, are NOT functions. They fail the vertical line test for functions.

These parabolas are considered relations.

The Standard "Vertex Form" and the Conic Form are the same:

VERTEX FORM: x = (y - 1)² - 2

CONIC FORM: (y - 1)² = x + 2

S u m m i n g U p: (p is distance from vertex to focus)

Vertical Parabola (up/down)

Vertex (0,0):

Vertex (h,k):

Horizontal Parabola (left-right)

Vertex (0,0):

Vertex (h,k):

Deriving the Conic Parabola Equation:

Deriving the Parabola Equation

Start by placing the parabola's vertex at the origin, for ease of computation. By definition, the distance, p, from the origin to the focus will equal the distance from the origin to the directrix (which will be y = -p).

The focus is point F and FA = AB by definition.
Using the Distance Formula, we know

Since we know FA = AB, we have

The distance from the vertex (origin) to the focus is traditionally labeled as "p".
("p" is also the distance from the vertex
to the directrix.)

Given x2 = 16y, state whether the parabola opens upward, downward, right or left, and state the coordinates of the focus.

ANSWER:
Form: x2 = 4py
4p = 16
p = 4
The focal length is 4.

Since this "form" squares x, and the value of 4p is positive, the parabola opens upward. This form of parabola has its vertex at the origin, (0,0). The focal length (distance from vertex to focus) is 4 units.

The focus is located at (0,4).

Given the parabola, (x - 3)2 = -8(y - 2), state whether the parabola opens upward, downward, right or left, and state the coordinates of the vertex, the focus, and the equation of the directrix.

ANSWER:
Form: (x - h)2 = 4p(y - k)
Vertex: (h,k) = (3,2)
4p = 8
p = 2
The focal length is 2.

Since this "form" squares x, and the value of 4p is negative, the parabola opens downward.
This form of parabola has its vertex at (h,k) = (3,2).

The focal length (distance from vertex to focus) is 2 units.

The focus is located at (3,0).
The directrix is y = 4.

Write the equation of a parabola with a vertex at the origin and a focus of (0,-3).

ANSWER: Make a sketch.

Remember that the parabola opens "around" the focus. Vertex: (0,0) and Focus: (0,-3)

Focal length p = 3. Opening downward means negative.

Form of Equation: x2 = 4py

EQUATION:
x2 = 4(-3)y

x2 = -12y

Find the focus and directrix of the parabola whose equation
is x2 - 6x + 3y + 18 = 0.