LAST REVISED December, 2008

The Parabola

Algebra Module A47

Copyright This publication © The Northern Alberta Institute of Technology 2002. All Rights Reserved.

A47 − The Parabola

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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.

A47 − The Parabola

OBJECTIVE ONE When you complete this objective you will be able to… will be able to… Determine the focus and the directrix given the equation of a parabola, with vertex at the origin. Determine the focus and the directrix given the equation of a parabola, with vertex at the origin.

Exploration Activity 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. 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 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:

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The distance to the directrix from point P is (x + p) and the distance from the focus to point p is ( ) ( )2 0x p y− + − 2 since these two distances are always the same in any parabola we get:

( ) ( )2 0x p y− + − 2 = x + p Squaring we get: (x − p)2 + y2 = x2 + 2xp + p2

After simplifying: y2 = 4px Equation 1

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y

F (p, 0)

= - px

p

x + p

P (x,y )

22 )0()( −+− ypx

V (0,0 )

Directrix

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.

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yx = -3

F (3,0)

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.

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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.

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y x = 2

F (-2, 0)

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 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.

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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 = -1

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V(0,0)

F (0, 1)

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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 5. 1.03y2 = −2.36x Show Me. 6. 2y2 = 18x 7. 2y2 = 40x

Experiential Activity One Answers 1. (6, 0); x = −6 2. (0, 2); y = −2 3. (0, 25); y = −25 4. (0, −1); y = 1 5. (−0.5728, 0); x = 0.5728 6. (2.25, 0); x = −2.25 7. (5, 0); x = −5

A47 − The Parabola

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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 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 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.

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Experiential Activity Two Find the equation of the parabola satisfying the given conditions: 1. Focus (−2, 0); directrix x = 2. Show Me. 2. Focus (0, 5); vertex (0,0). 3. Focus (0, −3); directrix y = 3. 4. Focus(10, 0); vertex (0,0)

Experiential Activity Two Answers 1. y2 = −8x 2. x2 = 20y 3. x2 = −12y 4. y2 = 40x

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.

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y x = 2

V(5,4) F (8,4)

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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 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) 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.

A47 − The Parabola

y = -0 5

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F (4,-3.5)

V(4,-2)

Try these online activities to expand your knowledge of parabolas: Parabola with horizontal directrix Parabola with vertical directrix

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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. y2 − 6y − 8x + 41 = 0 9. x2 − 6x − 14y − 5 = 0 10. x2 + 2x − 4y + 13 = 0 11. y2 −2y − 12x + 25 = 0 12. y2 + 6y − 4x + 17 = 0 13. x2 − 6x − 8y − 7 = 0 14. x2 − 6x + 10y + 9 = 0 15. y2 − 4y + 12x + 16 = 0

A47 − The Parabola

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Experiential Activity Three Answers 1. x2 − 10x − 8y + 33 = 0 2. y2 + 8y − 8x = 0 3. x2 + 4x + 16y − 44 = 0 4. y2 + 4y + 24x + 52 = 0 5. x2 + 2x − 4y − 7 = 0 6. y2 + 6y − 8x − 15 = 0 7. x2 − 4x + 16y − 12 = 0 8. V(4, 3); F(6, 3) 9. V(3, −1); F(3, 2.5) 10. V(−1, 3); F(−1, 4) 11. V(2, 1); F(5, 1) 12. V(2, −3); F(3, −3) 13. V(3, −2); F(3, 0) 14. V(3, 0); F(3, −2.5) 15. V(−1, 2); F(−4, 2)

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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.

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.

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V (0,400)

(150,0)

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Because the vertex is above the origin we use the equation:

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.

(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) 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.

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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. x2 = −357.1429y 2. x2 = −234.7222y 3. x2 = −10.6667y 4. 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.

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Appendix 1

The Parabola

2 4y p=

0,0

( ),0p

Vertex: ( )

Focus: Directrix: x p= −

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( ) (2 4y k p x h− = −

( ),h k)

Vertex:

Focus: ( ),p h k+

Directrix: x p h= − +

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x = -p+h

2 4x py= Vertex: ( )0,0

(

Focus: )0, p Directrix: y p= −

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