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. Parabola
Directrix
Focus
Axis of
Symmetry
Vertex
Standard Forms of the Parabola
The 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 xaxis, 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.
y
y 2 = 4px
y
Focus (p, 0)
Directrix x = -p
x 2 = 4py
Focus (p, 0)
Vertex
x
x
Vertex
Directrix y = -p
1
Example
• Find the focus and directrix of the parabola given by:
x 2 = 16 y
Solution:
4p = 16
p=4
Focus (0,4) and directrix y=-4
Text Example
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 = -8
The focus, on the y-axis, is at (0, p)
and the directrix is given by y = - p.
p = -2
Text Example cont.
Find the focus and directrix of the parabola given by x2 = -8y. The graph
the parabola.
Solution
Because p < 0, the parabola opens downward. Using this value for p, we
obtain
5
Directrix: y = 2
Focus: (0, p) = (0, -2)
Vertex (0, 0)
4
3
Directrix: y = - p; y = 2.
2
1
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
(-4, -2)
-1
-2
-3
-4
-5
1 2 3 4 5
(4, -2)
Focus (0, -2)
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
7
x-axis. We use the standard form of the equation in
6
Directrix: x = -5
5
which x is not squared, namely y2 = 4px.
4
3
Focus (5, 0)
We need to determine the value of p. Recall
2
1
that the focus, located at (p, 0), is p units from the
-5 -4 -3 -2 -1
1 2 3 4 5 6 7
-1
vertex, (0, 0). Thus, if the focus is (5, 0), then p = 5.
-2
-3
We substitute 5 for p into y2 = 4px to obtain the
-4
standard form of the equation of the parabola. The
-5
-6
equation is
-7
2
2
y = 4 • 5x or y = 20x.
Text Example
Find the vertex, focus, and directrix of the parabola given by
y2 + 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.
y2 + 2y + 12x – 23 = 0
y2 + 2y = -12x + 23
y2 + 2y + 1 = -12x + 23 + 1
This is the given equation.
Isolate the terms involving y.
Complete the square by adding the square of
half the coefficient of y.
(y + 1)2 = -12x + 24
Text Example cont.
Solution
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)
We use this form to identify the vertex, (h, k), and the value for p needed to
locate the focus and the directrix.
The equation is in standard form.
(y – (-1))2 = -12(x – 2)
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) = 5
Thus, the focus is (-1, -1) and the directrix is x = 5.
3
Text Example cont.
Solution
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.
(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
Solve for y in each equation.
or
y = -7
Text Example cont.
Solution
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.
Directrix: x = 5
7
6
(-1, 5)
5
4
3
2
1
-5 -4 -3 -2
1
3 4
6
7
-2
Focus (-1, -1)
(-1, -7)
-3
-4
-5
-6
Vertex (2, -1)
-7
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.
4
The Parabola
5