The Parabola

Definition of a Parabola• A Parabola is the set of all points in a plane

that are equidistant from a fixed line (the directrix) and a fixed point (the focus) that is not on the line.

Directrix

Parabola

Vertex

Focus

Axis of Symmetry

x

Directrix x = -p

y 2 = 4px

Vertex

Focus (p, 0)

y

x

Directrix y = -p

x 2 = 4py

Vertex

Focus (p, 0)y

Standard Forms of the ParabolaThe standard form of the equation of a parabola with vertex at the

origin is = 4px or x2 = 4py.

The graph illustrates that for the equation on the left, the focus is on the x-axis, which is the axis of symmetry. For the equation of the right, the focus is on the y-axis, which is the axis of symmetry.

Example

yx 162 =

• Find the focus and directrix of the parabola given by:

Solution:4p = 16p = 4Focus (0,4) and directrix y=-4

Find the focus and directrix of the parabola given by x2 = -8y. The graph the parabola.Solution The given equation is in the standard form x2 = 4py, so 4p = -8.

x2 = -8y

This is 4p.

We can find both the focus and the directrix by finding p.4p = -8p = -2

The focus, on the y-axis, is at (0, p) and the directrix is given by y = - p.

Text Example

Find the focus and directrix of the parabola given by x2 = -8y. The graph the parabola.SolutionBecause p < 0, the parabola opens downward. Using this value for p, we obtain

Focus: (0, p) = (0, -2)Directrix: y = - p; y = 2.

To graph x2 = -8y, we assign y a value that makes the right side a perfect square. If y = -2, then x2 = -8(-2) = 16, so x is 4 and –4. The parabola passes through the points (4, -2) and (-4, -2).

-5 -4 -3 -2 -1 1 2 3 4 5

54321

-1-2-3-4-5

(4, -2)(-4, -2)

Vertex (0, 0)Directrix: y = 2

Focus (0, -2)

Text Example cont.

Find the standard form of the equation of a parabola with focus (5, 0) and directrix x = -5.Solution The focus is (5, 0). Thus, the focus is on the x-axis. We use the standard form of the equation in which x is not squared, namely y2 = 4px.

We need to determine the value of p. Recall that the focus, located at (p, 0), is p units from the vertex, (0, 0). Thus, if the focus is (5, 0), then p = 5. We substitute 5 for p into y2 = 4px to obtain the standard form of the equation of the parabola. The equation is

y2 = 4 • 5x or y2 = 20x.

-1-5 -4 -3 -2 1 2 3 4 5 6 7

54

321

76

-3-4

-5-6-7

-1-2

Focus (5, 0)

Directrix: x = -5

Text Example cont.

Find the vertex, focus, and directrix of the parabola given byy2 + 2y + 12x – 23 = 0.

Then graph the parabola.Solution We convert the given equation to standard form by completing the square on the variable y. We isolate the terms involving y on the left side.

(y + 1)2 = -12x + 24

y2 + 2y + 12x – 23 = 0 This is the given equation.

y2 + 2y = -12x + 23 Isolate the terms involving y.

y2 + 2y + 1 = -12x + 23 + 1 Complete the square by adding the square of half the coefficient of y.

Text Example

To express this equation in the standard form (y – k)2 = 4p(x – h), we factor –12 on the right. The standard form of the parabola’s equation is

(y + 1)2 = -12(x – 2)

Solution

We use this form to identify the vertex, (h, k), and the value for p needed to locate the focus and the directrix.

(y – (-1))2 = -12(x – 2) The equation is in standard form.

We see that h = 2 and k = -1. Thus, the vertex of the parabola is (h, k) = (2, -1). Because 4p = -12, p = -3. Based on the standard form of the equation, the axis of symmetry is horizontal. With a negative value for p and a horizontal axis of symmetry, the parabola opens to the left. We locate the focus and the directrix as follows.

Focus: (h + p, k) = (2 + (-3), -1) = (-1, -1)Directrix: x = h – p

x = 2 – (-3) = 5Thus, the focus is (-1, -1) and the directrix is x = 5.

Text Example cont.

To graph (y + 1)2 = -12(x – 2), we assign x a value that makes the right side of the equation a perfect square. If x = -1, the right side is 36. We will let x = -1 and solve for y to obtain points on the parabola.

Solution

(y + 1)2 = -12(-1 – 2) Substitute –1 for x.

(y + 1)2 = 36 Simplify.

y + 1 = 6 or y + 1 = -6 Write as two separate equations.

y = 5 or y = -7 Solve for y in each equation.

Text Example cont.

Because we obtained these values of y for x = -1, the parabola passes through the points (-1, 5) and (-1, -7). Passing a smooth curve through the vertex and these two points, we sketch the parabola below.

Solution

-5 -4 -3 -2 1 3 4 6 7

54

321

76

-3-4

-5-6-7

-2

Focus (-1, -1)

Directrix: x = 5

Vertex (2, -1)

(-1, -7)

(-1, 5)

Text Example cont.

The Latus Rectum and Graphing Parabolas

• The latus rectum of a parabola is a line segment that passes through its focus, is parallel to its directrix, and has its endpoints on the parabola.

The Parabola