Algebra
Module A47
The Parabola
Copyright
This publication © The Northern
Alberta Institute of Technology
2002. All Rights Reserved.
LAST REVISED December, 2008
The Parabola
Statement of Prerequisite Skills
Complete all previous TLM modules before beginning this module.
Required Supporting Materials
Access to the World Wide Web.
Internet Explorer 5.5 or greater.
Macromedia Flash Player.
Rationale
Why is it important for you to learn this material?
Many scientific and technical applications use the parabola. Satellite dish technology, the
path of projectiles, and telescopic mirrors are a few examples.
Pre Self-assessment
Pre-Test
Learning Outcome
When you complete this module you will be able to…
Solve problems using parabolas.
Learning Objectives
1. Determine the focus and the directrix given the equation of a parabola, with vertex at
the origin.
2. Determine the equation of a parabola, with vertex at the origin, given the focus and
the equation of the directrix, or given the focus and the directrix.
3. Determine the critical parts of translated parabolas.
4. Solve applied parabola problems.
Connection Activity
A satellite television dish is in the shape of a paraboloid of revolution, the surface formed
when rotating a parabola around its axis of symmetry. Signals from a satellite strike the
surface of the dish and are reflected back to a focal point where the receiver is located.
The shape of the parabola gathers many signals and sends them to the focal point
resulting in a stronger signal.
1
A47 − The Parabola
OBJECTIVE ONE
When you complete this objective you will be able to…
Determine the focus and the directrix given the equation of a parabola, with vertex at the
origin.
Exploration Activity
DEFINITION: A parabola is defined as the path of a point that moves so that it is always
equidistant from a given line and a given point.
The given line is called the directrix and the given point is called the focus. The principal
axis of the parabola is the line through the focus that is perpendicular to the directrix. The
point on the parabola that is midway between the directrix and the focus is called the
vertex.
Using the definition we shall find the standard equation of the parabola for which the
focus is the point (p, 0), the directrix is the line x = −p, and the vertex is at the origin.
According to the definition of the parabola the distance from a point P(x, y) on the
parabola to the focus (p, 0) must equal the distance from P(x, y) to the directrix x = −p.
Thus, as indicated in the diagram:
y
x =- p
10
x + 8p
p 6
4
P (x,y )
2
( x − p ) + ( y − 0)
2
-10 -8
-6
0
-4V (0,0
-2 ) 0
-2
Directrix
2
x
2
F
0)
4 (p, 6
8
10
-4
-6
-8
-10
The distance to the directrix from point P is (x + p) and the distance from the focus to
point p is
( x − p)
2
+ ( y − 0 ) since these two distances are always the same in any
2
parabola we get:
( x − p)
2
+ ( y − 0)
2
=x+p
Squaring we get: (x − p)2 + y2 = x2 + 2xp + p2
After simplifying: y2 = 4px Å Equation 1
2
A47 − The Parabola
Equation 1 is called the standard form of the equation of a parabola with its (principal)
axis along the x-axis and the vertex at the origin. If 4p is positive the curve opens to the
right, if 4p is negative the curve opens to the left.
EXAMPLE 1
Find the coordinates of the focus and the equation of the directrix of the parabola
y2 = 12x. Make a sketch.
SOLUTION:
From the form of y2 = 12x we know it has its vertex at the origin and opens to the right.
Comparing this equation to the standard equation (1) we see that:
12 = 4p, hence p = 3 and the focus is on the x axis with coordinates (3, 0); and the
directrix is a vertical line 3 units to the left of the origin so its equation is: x = −3.
y
x = -3
10
8
6
4
2
F (3,0)
x
0
-10 -8
-6
-4
-2 0
-2
2
4
6
8
10
-4
-6
-8
-10
AGAIN: If the parabola has its vertex at the origin BUT opens to the left then its equation
has a negative coefficient in front of the x term, i.e. y2 = −5x opens to the left. See the
next example.
3
A47 − The Parabola
EXAMPLE 2
Find the focus and the directrix for: y2 = −8x. Make a sketch.
SOLUTION:
From the form of the given equation we know it has its vertex at the origin and opens to
the left. Comparing this equation to equation (1) we see that:
−8= 4p, hence p = −2 so its focus is at (−2, 0); and the directrix is a vertical line 2 units to
the right of the origin so its equation is x = 2.
y
x = 2
10
8
6
4
F (-2, 0)
-10 -8
-6
-4
2
0
-2 0
-2
x
2
4
6
8
10
-4
-6
-8
-10
NOTE: If we were to choose the focus as the point (0, p) and the directrix as the line
y = −p, we would find that the resulting equation is:
x2 = 4py
Å Equation 2
Equation 2 is the standard form of the equation of a parabola with the y axis as its
principal axis and the vertex at the origin. If 4p is positive the curve opens up, if 4p is
negative the curve opens down.
Remember with all parabolas:
When y is squared the curve opens horizontally
When x is squared the curve opens vertically.
**Also make sure the coefficient of the squared term is +1, this guarantees that the
coefficient of the linear term is equal to 4p.
4
A47 − The Parabola
EXAMPLE 3
Sketch the graph of 2x2 = 8y, and label its focus, vertex, and directrix.
SOLUTION:
Divide both sides of the equation by 2 so it will be in standard form. From x2 = 4y, we
know it has its vertex at the origin and it opens vertically. Comparing this equation to
equation (2) above we see that:
4 = 4p, hence p = 1, the focus is on the y axis, and its coordinates are (0, 1); the directrix
is a horizontal line 1 unit below the origin so its equation is: y = −1. The coordinates of
the vertex are (0,0). The graph follows:
y
10
8
6
4
F (0, 1)
2
V(0,0)
x
0
-10 -8
-6
-4
-2
-2
0
2
4
6
8
10
y = -1
-4
-6
-8
-10
5
A47 − The Parabola
Experiential Activity One
Find the coordinates of the focus and the equation of the directrix for each of the
following parabolas. Make a sketch to illustrate your work.
1. y2 = 24x
2. x2 = 8y
3. x2 = 100y
4. −3x2 = 12y
Show Me.
5. 1.03y2 = −2.36x
2
6. 2y = 18x
7. 2y2 = 40x
Experiential Activity One Answers
1.
2.
3.
4.
5.
6.
7.
(6, 0); x = −6
(0, 2); y = −2
(0, 25); y = −25
(0, −1); y = 1
(−0.5728, 0); x = 0.5728
(2.25, 0); x = −2.25
(5, 0); x = −5
6
A47 − The Parabola
OBJECTIVE TWO
When you complete this objective you will be able to…
Determine the equation of a parabola, with vertex at the origin, given the focus and the
equation of the directrix, or given the focus and the directrix.
Exploration Activity
In objective 1 we saw the definition of a parabola, what it looks like, and how to extract
information from a given parabolic equation. In this objective we shall see how to get the
equation from given information, specifically when given the focus (focal point) and the
equation of the directrix. The following examples show how this is done.
EXAMPLE 1
A parabola with vertex at the origin has its focal point at (3,0). Determine its equation.
SOLUTION:
Because this parabola has its vertex at the origin, its standard form is either y2 = 4px or
x2 = 4py. Its focus is on the x-axis at (3,0), so it must open "around" the focus to the right.
Thus we use the equation y2 = 4px with 4p as positive. In this case p = 3 (distance from
origin to focus) so the equation becomes:
y2 = 4px
y2 = 4(3)x
y2 = 12x Å the required equation.
EXAMPLE 2
Determine the equation of the parabola that has its focal point at (0,2) and the equation of
its directrix is y = −2.
SOLUTION:
The focal point is on the y axis, two units above the origin. Thus the curve opens up and
the standard equation we use is: x2 = 4py. In our case p = 2, so substituting this for p we
get:
x2 = 4(2)y
x2 = 8y Å required equation.
Note in this solution we did not have to use the equation of the directrix. The equation
y = −2 confirms that the parabola opens up (away from y = −2) and that the vertex is at
the origin, i.e. the vertex is 2 units from the focus and 2 units from the directrix.
7
A47 − The Parabola
Experiential Activity Two
Find the equation of the parabola satisfying the given conditions:
1.
2.
3.
4.
Focus (−2, 0); directrix x = 2.
Focus (0, 5); vertex (0,0).
Focus (0, −3); directrix y = 3.
Focus(10, 0); vertex (0,0)
Show Me.
Experiential Activity Two Answers
1.
2.
3.
4.
y2 = −8x
x2 = 20y
x2 = −12y
y2 = 40x
8
A47 − The Parabola
OBJECTIVE THREE
When you complete this objective you will be able to…
Determine the critical parts of translated parabolas.
Exploration Activity
Until now we have dealt with parabolas that had their vertices at the origin. These curves
are said to be in standard position.
If a parabola is not in standard position, its equation is more complicated. If the vertex is
not at the origin (0,0), but the principal axis is still vertical or horizontal the parabola is
said to be translated. Translation of any curve may be considered a dislocation without
any rotation.
The translated forms of the equation of the parabola are:
(y − k)2 = 4p(x − h) and (x − h)2 = 4p(y − k)
where h is the horizontal coordinate of the vertex and k is the vertical coordinate of the
vertex.
The vertex is the point (h,k).
NOTE: (y − k)2 = 4p(x − h) is the equation for a parabola opening horizontally.
(x − h)2 = 4p(y − k) is the equation for a parabola opening vertically.
EXAMPLE 1
Write the equation of the parabola whose focus is at (8,4) and whose directrix is
the line x = 2.
SOLUTION:
First we make a graph.
y
x =2
10
8
V(5,4 )
F (8,4)
6
4
2
-10 -8
-6
-4
0
-2 0
-2
x
2
4
6
8
10
-4
-6
-8
-10
9
A47 − The Parabola
Since the vertex is midway between the focus and the directrix, its coordinates must be
(5,4), i.e. (h, k) = (5,4). Also note that p = +3 since p is the distance from the vertex to the
focus.
Because the curve opens horizontally we use:
(y − k)2 = 4p(x − h)
Upon substitution our equation becomes: (y − 4)2 = 4(3)(x − 5),
y2 − 8y + 16 = 12x − 60
y2 − 8y − 12x + 76 = 0 Å the required equation.
EXAMPLE 2
Determine h, k, and p for the following parabola and make a sketch:
x2 − 8x + 6y + 28 = 0.
SOLUTION:
The student may wish to review the quadratic equations module if you need help in how
to complete the square.
We first put the equation into the form for a translated parabola, (x − h)2 = 4p(y − k), in
order to locate h, k, and p. To do this we complete the square on the x terms and move the
other terms to the right.
x2 − 8x = −6y − 28
x2 − 8x + 16 = −6y − 28 + 16 (completing the square)
(x − 4)2 = −6y − 12
(x − 4)2 = −6(y + 2) Å in this step we factored −6 from the right hand
side to make the coefficient of our y "+1", as this is the standard form.
Thus (h, k) = (4, −2) and 4p = −6, so p = −3/2.
Because the square term is x, this tells us the curve opens vertically, 4p being negative
indicates the curve opens down. Locate V(4, −2) and focus (4, −3.5) and make a sketch as
follows.
10
A47 − The Parabola
y
10
8
6
4
2
x
0
-10 -8
-6
-4
-2
-2
0
2
4
6
8 10
V (4,-2)
y= 05
-4
F (4,-3.5)
-6
-8
-10
Try these online activities to expand your knowledge of parabolas:
Parabola with horizontal directrix
Parabola with vertical directrix
11
A47 − The Parabola
Experiential Activity Three
EXERCISE
Write the equations of each of the parabolas:
1. Vertex at (5,1), focus at (5,3).
2. Vertex at (−2, −4), focus at (0, −4).
3. Vertex at (−2, 3), focus at (−2, −1). Show Me.
4. Vertex at (−2, −2), directrix at x = 4.
5. Focus at (−1, −1), directrix at y = −3.
6. Focus at (−1, −3), directrix at x = −5.
7. Focus at (2, −3), directrix at y = 5.
Locate the vertex and focus of each of the following translated parabolas and sketch:
8.
9.
10.
11.
12.
13.
14.
15.
y2 − 6y − 8x + 41 = 0
x2 − 6x − 14y − 5 = 0
x2 + 2x − 4y + 13 = 0
y2 −2y − 12x + 25 = 0
y2 + 6y − 4x + 17 = 0
x2 − 6x − 8y − 7 = 0
x2 − 6x + 10y + 9 = 0
y2 − 4y + 12x + 16 = 0
12
A47 − The Parabola
Experiential Activity Three Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
x2 − 10x − 8y + 33 = 0
y2 + 8y − 8x = 0
x2 + 4x + 16y − 44 = 0
y2 + 4y + 24x + 52 = 0
x2 + 2x − 4y − 7 = 0
y2 + 6y − 8x − 15 = 0
x2 − 4x + 16y − 12 = 0
V(4, 3); F(6, 3)
V(3, −1); F(3, 2.5)
V(−1, 3); F(−1, 4)
V(2, 1); F(5, 1)
V(2, −3); F(3, −3)
V(3, −2); F(3, 0)
V(3, 0); F(3, −2.5)
V(−1, 2); F(−4, 2)
13
A47 − The Parabola
OBJECTIVE FOUR
When you complete this objective you will be able to…
Solve applied parabola problems.
Exploration Activity
There are many applications of parabolic curves in the world around us. In this objective
we shall endeavor to illustrate a few of them.
EXAMPLE 1
Under certain conditions a cable that hangs between two supports can be closely
approximated as being parabolic. Assuming that this cable hangs in the shape of a
parabola, find its equation if a point 6 meters horizontally from its lowest point is 1 meter
above its lowest point. Choose the lowest point as the origin of the coordinate system.
SOLUTION:
Because this is a parabola with its vertex at the origin and opening up we use the standard
equation: x2 = 4py.
We know the point (6,1) is on the curve so we can substitute these values for x and y into
the standard equation to find p.
62 = 4(p)1, so p = 9, substitute now
and the required equation is: x2 = 36y.
EXAMPLE 2
A parabolic mirror, 8 cm in diameter, has a maximum depression of 0.30 mm. What is
the focal length of the mirror?
SOLUTION:
We are asked to find p, the focal length. We must change units to a common unit, so
change 8 cm to 80 mm. Also the radius = 40 mm. Assume the curve opens up, so the
point (40,0.3) is on the curve.
Substituting (40,0.3) into the equation x2 = 4py we get:
1600 = 4(p)(0.3) from which p = 1333.3 mm.
14
A47 − The Parabola
EXAMPLE 3
A bullet, when shot into the air at a certain angle, follows a path that is a parabolic curve.
The bullet reaches a height of 400 m and strikes the ground at a distance of 300 m from
the firing point. What is the equation of the parabola?
SOLUTION:
This is a parabola that opens downward. We choose to place the vertex above the origin
and on the y axis. This puts the curve in a translated position.
y
500
V (0,400)
400
This is one point we know on the parabola
that we determined from the given
information. We will need at least one point
that is not the vertex to determine the
equation of the parabola.
300
200
100
(150,0)
x
0
-300
-200
-100
0
100
200
300
Because the vertex is above the origin we use the equation:
(x − h)2 = 4p(y − k)
substitute h = 0, k = 400, and x = 150, y = 0 to get:
(150 − 0)2 = 4p(0 − 400)
p = −14.1
Thus the equation is: (x − 0)2 = 4(−14.1)(y − 400)
x2 = −56.3(y − 400) Å the required equation.
NOTE: now that we have the equation that describes the curve we can find any height we
wish simply by substituting the values for x.
Also when placing the curve on the coordinate axes pick a convenient place for the
vertex. You may place the vertex at the origin or translate it away from the origin. It does
not make any difference. The resulting equations will look different, but each is correct
for the positioning you chose.
15
A47 − The Parabola
Experiential Activity Four
1. When Frank Lovejoy hit a chip shot, the ball travels along a path that is parabolic. If
a ball, which travels 200 m horizontally, reaches a maximum height of 28 m, what is
the equation of the parabolic flight path? Place the vertex at the origin.
2. A baseball, when thrown into the air at a certain angle, follows a path that is a
parabolic curve. If the ball reaches a height of 18 m and strikes the ground 130 m
from the throwing point, what is the equation of the parabola? Place the vertex at the
origin.
Show Me.
3. A bridge across a river is built on a supporting arch that is in the shape of a parabola
that opens downward. The vertex is 6 m above the water. If the width of the water
surface between the two pillars is 16 m, what is the equation of the parabolic arch?
Place the vertex at the origin.
4. A parabolic mirror, 10 cm in diameter, has a maximum depression of 0.40 mm. What
is the focal length of the mirror?
Experiential Activity Four Answers
1.
2.
3.
4.
x2 = −357.1429y
x2 = −234.7222y
x2 = −10.6667y
1562.5 mm
Practical Application Activity
Complete the parabola module assignment in TLM.
Summary
The purpose of this module was to give the student some basic information about the
parabolic curve and some of its uses. This familiarity with the parabola will be extra
useful to the student when he/she is doing area problems in calculus.
16
A47 − The Parabola
Appendix 1
The Parabola
y
y
10
10
P>0
y 2 = 4 px
Vertex:
Focus:
( 0,0 )
( p,0 )
Directrix: x = − p
P<0
8
8
6
6
4
4
2
2
V(0,0)
F(p,0)
x
0
-10
-8
-6
-4
-2
0
2
4
6
8
x
0
-10
10
V(0,0)
F(p,0)
-8
-6
-4
-2
0
-2
-2
-4
-4
-6
-6
-8
-8
-10
-10
2
4
6
y
y
10
10
P<0
P>0
8
8
6
6
F(p+h,k)
V(h,k)
2
F(p+h,k)
2
2
( h, k )
Focus: ( p + h, k )
V(h,k)
4
4
= 4 p ( x − h)
Vertex:
10
x = -p
x = -p
(y − k)
8
x
0
-10
-8
-6
-4
Directrix: x = − p + h
-2
0
2
4
6
8
10
x
0
-10
-8
-6
-4
-2
0
-2
-2
-4
-4
-6
-6
-8
-8
-10
-10
x = -p+h
2
4
6
8
10
x = -p+h
y
y
P>0
10
P<0
10
8
8
6
6
F(0,p)
4
x = 4 py
4
y = -p
2
Vertex:
Focus:
( 0,0 )
( 0, p )
Directrix: y = − p
2
2
V(0,0)
-10
-8
-6
-4
-2
V(0,0)
x
0
0
2
4
6
8
-10
10
x
0
-8
-6
-4
-2
0
2
4
6
8
10
-2
-2
-4
-4
y = -p
F(0,p)
-6
-6
-8
-8
-10
-10
y
y
P>0
10
P<0
10
8
8
F(h,p+k)
6
6
y = -p+k
4
( x − h)
2
V(h,k)
= 4 p( y − k )
Vertex:
Directrix: y = − p + k
V(h,k)
2
( h, k )
Focus: ( h, p + k )
4
2
x
0
-10
-8
-6
-4
-2
0
2
4
6
8
10
x
0
-10
-8
-6
-4
-2
0
2
4
6
-2
-2
y = -p+k
F(h,p+k)
-4
-4
-6
-6
-8
-8
-10
-10
8
10