Introduction to Parabolas
A conic section is a shape obtained when a cone is sliced. The study of conic sections began over two thousand years ago and we use their properties today. Planets in our solar system move around the sun in elliptical orbits. The crosssection of many reflecting surfaces is in the shape of a parabola. In fact, all of the conic sections have useful reflecting properties. There are three conic sections—parabolas, ellipses (including circles), and hyperbolas.
Parabolas
In Quadratic Functions Help, we learned how to graph parabolas when their equations were in the form y = a ( x − h ) ^{2} + k or y = ax ^{2} + bx + c . Now we will learn the formal definition for a parabola and another form for its equation.
DEFINITION: A parabola is the set of all points whose distance to a fixed point and a fixed line are the same.
The fixed point is the focus . The fixed line is the directrix . For example, the focus for the parabola , and the directrix is the horizontal line y = 7. The point (0, 2) is on the parabola. Its distance from the line y = 7 is 5. Its distance from the focus (−3, 6) is also 5.
The new form for a parabola that opens up or down is ( x − h ) ^{2} = 4 p ( y − k ). The vertex is still at ( h , k ), but p helps us to find the focus and the equation for the directrix. The focus is the point (h , k + p ), and the directrix is the horizontal line y = k − p . The form for the equation for a parabola that opens to the side is ( y − k ) ^{2} = 4 p ( x − h ). The focus for a parabola that opens to the right or to the left is the point ( h + p , k ), and the directrix is the vertical line x = h − p . This information is summarized in Table 12.1 and in Figures 12.1 and 12.2.
Table 12.1
( x − h ) ^{2} = 4 p ( y − k )

( y − k ) ^{2} = 4 p ( x − h )

The vertex is ( h, k ).

The vertex is ( h, k ).

The parabola opens up if p is positive and down if p is negative.

The parabola opens to the right if p is positive and to the left if p is negative.

The focus is ( h, k + p ).

The focus is ( h + p, k ).

The directrix is y = k − p .

The directrix is x = h − p .

The axis of symmetry is x = h .

The axis of symmetry is y = k .

Fig. 12.1
Fig. 12.2
Matching the Equation of a Parabola to its Graph Examples
In the following examples, we will be asked to match the equation to its graph. The vertex for each parabola will be at (0, 0). We can decide which graph goes to which equation either by finding the focus or the directrix in the equation and finding which graph has this focus or directrix.
Examples
Match the graphs in Figures 12.3 through 12.6 with their equations.
Fig. 12.3
Fig. 12.4
Fig. 12.5
Fig. 12.6
 x ^{2} = 6 y

The equation is in the form ( x − h ) ^{2} = 4 p ( y − k ), so the parabola will open up or down. We have . Now we know three things—that the parabola opens up (because p is positive), that the focus is ), and the directrix is (from The graph that behaves this way is in Figure 12.5 .

Fig. 12.5
 y ^{2} = 6 x

The equation is in the form ( y − k ) ^{2} = 4 p ( x − h ), so the parabola opens to the left or to the right. We have . Now we know that the parabola opens to the right, that the focus is , and that the directrix is . The graph for this equation is in Figure 12.3 .

Fig. 12.3
 y ^{2} = –6 x

The equation is in the form ( y − k ) ^{2} = 4 p ( x − h ), so the parabola opens to the left or to the right. We have . The parabola opens to the left, the focus is , and the directrix is . The graph for this equation is in Figure 12.4.

Fig. 12.4
 x ^{2} = –6 y

The equation is in the form ( x − h ) ^{2} = 4 p ( y − k ), so the parabola opens up or down. We have . The parabola opens down, the focus is . The directrix is (from . The graph for this equation is in Figure 12.6.

Fig. 12.6
Finding the Vertex, Focus, Directrix, and Direction of Parabola Examples
Using the information in Table 12.1, we can find the vertex, focus, directrix, and whether the parabola opens up, down, to the left, or to the right by looking at its equation.
Table 12.1
( x − h ) ^{2} = 4 p ( y − k )

( y − k ) ^{2} = 4 p ( x − h )

The vertex is ( h, k ).

The vertex is ( h, k ).

The parabola opens up if p is positive and down if p is negative.

The parabola opens to the right if p is positive and to the left if p is negative.

The focus is ( h, k + p ).

The focus is ( h + p, k ).

The directrix is y = k − p .

The directrix is x = h − p .

The axis of symmetry is x = h.

The axis of symmetry is y = k.

Examples
Find the vertex, focus, and directrix. Determine if the parabola opens up, down, to the left, or to the right.
 ( x − 3) ^{2} = 4( y − 2)

This equation is in the form ( x − h ) ^{2} = 4 p ( y − k ). The vertex is (3, 2). Once we have found p , we can find the focus and directrix and how the parabola opens. p = 1 (from 4 = 4 p ). The parabola opens up because p is positive; the focus is ( h , k + p ) = (3, 2 + 1) = (3, 3); and the directrix is y = 1 (from y = k − p = 2 − 1 = 1).
 ( y + 1) ^{2} = 8( x − 3)

The equation is in the form ( y − k ) ^{2} = 4 p ( x − h ). The vertex is (3, −1), p = 2 (from 8 = 4 p ); the parabola opens to the right; the focus is ( h + p , k ) = (3 + 2, −1) = (5, −1); and the directrix is x = 1 (from x = h − p = 3−2 = 1).
Finding the Equation of a Parabola
If we know any two of the vertex, focus, and directrix, we can find an equation of the parabola. From the information given, we first need to decide which form to use. Knowing the directrix is the fastest way to decide this. If the directrix is a horizontal line ( y = number), then the equation is ( x − h ) ^{2} = 4 p ( y − k ). If the directrix is a vertical line ( x = number), then the equation is ( y − k ) ^{2} = 4 p ( x − h ). If we do not have the directrix, we need to look at the coordinates of the vertex and focus. Either both the x coordinates will be the same or both y coordinates will be. If both x coordinates are the same, the parabola opens up or down. We need to use the form ( x − h ) ^{2} = 4 p ( y − k ). If both y coordinates are the same, the parabola opens to the side. We need to use the form ( y − k ) ^{2} = 4 p ( x − h ). Once we have decided which form to use, we might need to use algebra to find h , k , and p . For example, if we know the focus is (2, −1) and the directrix is x = 6, then we know h − p = 6 and h + p = 2 and k = −1. The equations h − p = 6 and h + p = 2 form a system of equations.
Now that we have all three numbers and the form, we are ready to write the equation: ( y + 1) ^{2} = −8( x − 4).
Examples
Find an equation for the parabola.
 The directrix is y = 2, and the vertex is (3, 1).

Because the directrix is a horizontal line, the equation we want is ( x − h ) ^{2} = 4 p ( y − k ). The vertex is (3, 1), giving us h = 3 and k = 1. From y = k − p and y = 2, we have 1 − p = 2, making p = −1. The equation is ( x − 3) ^{2} = −4( y − 1).
 The focus is (4, −2), and the vertex is (0, −2).

The y coordinates are the same, so this parabola opens to the side, and the equation we need is ( y − k ) ^{2} = 4 p ( x − h ). The vertex is ( h , k ) = (0, −2), giving us h = 0 and k = −2. The focus is ( h + p , k ) = (4, −2). From this we have h + p = 0 + p = 4, making p = 4. The equation is ( y + 2) ^{2} = 16 x .
Fig. 12.7
The directrix is the vertical line x = −1, and the focus is (3, 2). Because the parabola opens to the right, the form we need is ( y − k ) ^{2} = 4 p ( x − h ).
From the focus we have ( h + p , k ) = (3, 2), so h + p = 3 and k = 2. The directrix is x = −1 and x = h − p so h − p = −1.
Parabola Practice Problems
Practice
 Identify the vertex, focus, and directrix for ( y −5) ^{2} = 10( x − 1).
 Identify the vertex, focus, and directrix for .
 Find an equation for the parabola that has directrix y = − 2 and focus (4, 10).
For Problems 46, match the equation with its graph in Figures 12.812.10.
Fig. 12.8
Fig. 12.9
Fig. 12.10
 x ^{2} = −8( y + 1)
 ( x + 1) ^{2} = 4( y − 2)
 ( y − 2) ^{2} = −6( x + 3)
Solutions

h = 1, k = 5, and (from 4 p = 10). The vertex is (1, 5); the focus is and the directrix is (from .

h = −6, k = 4, and . The vertex is (−6, 4); the focus is ; and the directrix is .

We want to use the equation ( x − h ) ^{2} = 4 p ( y − k ). The focus is ( h , k + p ), so h = 4 and k + p = 10. The directrix is y = k − p , so k − p = −2.
4 + p = 10 Let k = 4 in k + p = 10
p = 6
The equation is ( x  4) ^{2} = 24( y − 4).

Figure 12.9

Figure 12.8

Figure 12.10
Practice problems for this concept can be found at: Conic Sections Practice Test.
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From PreCalculus Demystified: A SelfTeaching Guide. Copyright © 2005 by The McGrawHill Companies, Inc. All Rights Reserved.