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Explore Deriving the Standard-Form Equation of a Parabola

A parabola is defined as a set of points equidistant from a line (called the directrix) and a point (called the focus). The focus will always lie on the axis of symmetry, and the directrix will always be perpendicular to the axis of symmetry. This definition can be used to derive the equation for a horizontal parabola opening to the right with its vertex at the origin using the distance formula. (The derivations of parabolas opening in other directions will be covered later.)

A The coordinates for the focus are given by

.

B Write down the expression for the distance from a point (x, y) to the coordinates of the focus:

d = ――――――――――――

( - ) 2 + ( - ) 2

C The distance from a point to a line is measured by drawing a perpendicular line segment from the point to the line. Find the point where a horizontal line from (x, y) intersects the directrix (defined by the line x = -p for a parabola with its vertex on the origin).

D Write down the expression for the distance from a point, (x, y) to the point from Step C:

d = ――――――――――――

( - ) 2 + ( - ) 2

E Setting the two distances the same and simplifying gives.

――――― (x - p) 2 + y 2 = ――― (x + p) 2

To continue solving the problem, square both sides of the equation and expand the squared binomials.

x 2 + xp + p 2 + y 2 = x 2 + xp + p 2

F Collect terms.

x 2 + px + p 2 + y 2 = 0

G Finally, simplify and arrange the equation into the standard form for a horizontal parabola (with vertex at (0, 0)):

y 2 =

Resource Locker

(x, y)

(p, 0)(-p, 0)

(-p, y)

Directrix

d

d

Focus

(p, 0)

(-p, y)

x

x

p

-p

y

y

0

y

1

0

4px

-2

-4

1

0

1 2 1

Module 4 175 Lesson 2

4.2 ParabolasEssential Question: How is the distance formula connected with deriving equations for both

vertical and horizontal parabolas?

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A2_MNLESE385894_U2M04L2.indd 175 22/05/14 5:00 PM

Common Core Math StandardsThe student is expected to:

A-CED.A.2

Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales. Also A-CED.A.3, G-GPE.A.2

Mathematical Practices

MP.7 Using Structure

Language ObjectiveExplain to a partner what the focus and directrix of a parabola are.

Fill in and label a graphic organizer describing different types of parabolas.

COMMONCORE

COMMONCORE

HARDCOVER PAGES 125134

Turn to these pages to find this lesson in the hardcover student edition.

Parabolas

ENGAGE Essential Question: How is the distance formula connected with deriving equations for both vertical and horizontal parabolas?Possible answer: When you use the distance formula to describe all the points that are equidistant from a given point and a horizontal line you get the equation of a vertical parabola. Similarly, when you use the distance formula to describe all the points that are equidistant from a given point and a vertical line, you get the equation of a horizontal parabola.

PREVIEW: LESSON PERFORMANCE TASKView the Engage section online. Discuss the photo and how the shape of a parabola can be used to build a microphone. Then preview the Lesson Performance Task.

175

HARDCOVER

Turn to these pages to find this lesson in the hardcover student edition.

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Explore Deriving the Standard-Form Equation

of a Parabola

A parabola is defined as a set of points equidistant from a line (called the directrix)

and a point (called the focus). The focus will always lie on the axis of symmetry, and

the directrix will always be perpendicular to the axis of symmetry. This definition can

be used to derive the equation for a horizontal parabola opening to the right with its

vertex at the origin using the distance formula. (The derivations of parabolas opening

in other directions will be covered later.)

The coordinates for the focus are given by

.

Write down the expression for the distance

from a point (x, y) to the coordinates of the

focus:

d = ―――――――――――― ( - ) 2 + ( - ) 2

The distance from a point to a line is measured

by drawing a perpendicular line segment from

the point to the line. Find the point where

a horizontal line from (x, y) intersects the

directrix (defined by the line x = -p for a

parabola with its vertex on the origin).

Write down the expression for the distance

from a point, (x, y) to the point from Step C:

d = ―――――――――――― ( - ) 2 + ( - ) 2

Setting the two distances the same and simplifying gives.

――――― (x - p) 2 + y 2 = ――― (x + p) 2

To continue solving the problem, square both sides of the equation and expand the squared binomials.

x 2 + xp + p 2 + y 2 = x 2 + xp + p 2

Collect terms.

x 2 + px + p 2 + y 2 = 0

Finally, simplify and arrange the equation into

the standard form for a horizontal parabola

(with vertex at (0, 0)):

y 2 =

Resource

Locker

A-CED.A.2 Create equations in two or more variables to represent relationships between quantities;

graph equations on coordinate axes with labels and scales. Also A-CED.A.3, G-GPE.A.2

COMMONCORE

(x, y)

(p, 0)(-p, 0)

(-p, y)

Directrix

d

d

Focus

(p, 0)

(-p, y)

x

x

p

-p

y

y

0

y

1

04px

-2

-4

1

0

1 21

Module 4

175

Lesson 2

4 . 2 Parabolas

Essential Question: How is the distance formula connected with deriving equations for both

vertical and horizontal parabolas?

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175 Lesson 4 . 2

L E S S O N 4 . 2

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Reflect

1. Why was the directrix placed on the line x = -p?

2. Discussion How can the result be generalized to arrive at the standard form for a horizontal parabola with a vertex at (h, k) : (y - k) 2 = 4p (x - h) ?

Explain 1 Writing the Equation of a Parabola with Vertex at (0, 0)

The equation for a horizontal parabola with vertex at (0, 0) is written in the standard form as y 2 = 4px. It has a vertical directrix along the line x = -p, a horizontal axis of symmetry along the line y = 0, and a focus at the point (p, 0) . The parabola opens toward the focus, whether it is on the right or left of the origin (p > 0 or p < 0) . Vertical parabolas are similar, but with horizontal directrices and vertical axes of symmetry:

Parabolas with Vertices at the Origin

Vertical Horizontal

Equation in standard form x 2 = 4py y 2 = 4px

p > 0 Opens upward Opens rightward

p < 0 Opens downward Opens leftward

Focus (0, p) (p, 0)

Directrix y = -p x = -p

Axis of Symmetry x = 0 y = 0

The directrix had to be as far from the vertex (at the origin) as the focus, but on the

opposite side. So if the focus is at (p, 0) , the directrix has to intersect the x-axis at

(-p, 0) . The line x = -p is perpendicular to the axis of symmetry (the line connecting the

focus and the origin) and contains the point (-p, 0) .

A parabola with a vertex at (h, k) can be described by a horizontal shift of h to the right

and a vertical shift of k upward, which can be achieved for any graph by substituting

(y - k) for y and (x - h) for x.


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Integrate Mathematical PracticesThis lesson provides an opportunity to address Mathematical Practice MP.7, which calls for students to “look for and make use of structure.” Students learn the relationships between quadratic equations and their graphs. Students learn that equations in the forms (y – k ) 2 = 4p(x – h) and (x – h ) 2 = 4p(y – k) have vertices (h, k), focus at either (h + p, k) or (h, k + p), and have the directrix y = k – p or x = h – p.

EXPLORE Deriving the Standard Form Equation of a Parabola

INTEGRATE MATHEMATICAL PRACTICESFocus on PatternsMP.8 Explain that if the equation of a parabola contains an x 2 term the parabola opens either up or down, while an equation that contains a y 2 term opens either right or left.

EXPLAIN 1 Writing the Equation of a Parabola with Vertex at (0, 0)

INTEGRATE MATHEMATICAL PRACTICESFocus on Math ConnectionsMP.1 Explain that for an equation in the form y = 1 __ 4p x 2 , the graph opens upward if 1 __ 4p is positive and downward if 1 __ 4p is negative. For an equation in theform x = 1 __ 4p y 2 , the graph opens to the right if 1 __ 4p is positive and to the left if 1 __ 4p is negative.

PROFESSIONAL DEVELOPMENT

Parabolas 176


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Example 1 Find the equation of the parabola from the description of the focus and directrix. Then make a sketch showing the parabola, the focus, and the directrix.

Focus (–8, 0) , directrix x = 8

A vertical directrix means a horizontal parabola.

Confirm that the vertex is at (0, 0) :a. The y-coordinate of the vertex is the same as the focus: 0.

b. The x-coordinate is halfway between the focus (-8) and the directrix (+8) : 0.

c. The vertex is at (0, 0) .

Use the equation for a horizontal parabola, y 2 = 4px, and replace p with the x coordinate of the focus: y 2 = 4 (-8) x

Simplify: y 2 = -32x

Plot the focus and directrix and sketch the parabola.

Focus (0, -2) , directrix y = 2

A [vertical/horizontal] directrix means a [vertical/horizontal] parabola.

Confirm that the vertex is at (0, 0) :a. The x-coordinate of the vertex is the same as the focus: 0.

b. The y-coordinate is halfway between the focus, and the

directrix, : 0

c. The vertex is at (0, 0) .

Use the equation for a vertical parabola, , and replace p with the y-coordinate of the focus: x 2 = 4 ⋅ ⋅ y

Simplify: x 2 =

Plot the focus, the directrix, and the parabola.

Your Turn

Find the equation of the parabola from the description of the focus and directrix. Then make a sketch showing the parabola, the focus, and the directrix.

3. Focus (2, 0) , directrix x = -2 4. Focus (0, - 1 _ 2 ) , directrix y = 1 _ 2

y

0-4-8

4

8

-44

x

y

0-4-8

4

8

-44

x

-8

8

y

0-4-8

4

8

-44

x

-8

8

y

0-4-8

4

8

-44

x

-8

8

x 2 = 4py

-2

-2

-8y

2

p = x-coordinate of the focus = 2

y 2 = 4 (2) x y 2 = 8x

p = y-coordinate of the focus

= (− 1 _ 2

)

x 2 = 4 (− 1 _ 2

) y

x 2 = −2y


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A2_MNLESE385894_U2M04L2 177 6/8/15 1:36 PMCOLLABORATIVE LEARNING

Peer-to-Peer ActivityHave students work in pairs. Instruct each student to create a design using graphs of parabolas. Students exchange designs and write the equations for the parabolas in the partner’s design, including the domain and range of each curve.

QUESTIONING STRATEGIESHow can you find the directrix of a parabola with an equation in the form y = 1 __ 4p x 2 or

x = 1 __ 4p y 2 ? The directrix is p units from the vertex. Remember that the parabola opens away from the directrix.

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Explain 2 Writing the Equation of a Parabola with Vertex at (h, k)

The standard equation for a parabola with a vertex (h, k) can be found by translating from (0, 0) to (h, k): substitute (x - h) for x and (y - k) for y. This also translates the focus and directrix each by the same amount.

Parabolas with Vertex (h, k)

Vertical Horizontal

Equation in standard form (x - h) 2 = 4p (y - k) (y - k) 2 = 4p (x - h) p > 0 Opens upward Opens rightward

p < 0 Opens downward Opens leftward

Focus (h, k + p) (h + p, k) Directrix y = k - p x = h - p

Axis of Symmetry x = h y = k

p is found halfway from the directrix to the focus:

• For vertical parabolas: p = (y value of focus) - (y value of directrix)

____ 2

• For horizontal parabolas: p = (x value of focus) - (x value of directrix) ____ 2

The vertex can be found from the focus by relating the coordinates of the focus to h, k, and p.

Example 2 Find the equation of the parabola from the description of the focus and directrix. Then make a sketch showing the parabola, the focus, and the directrix.

Focus (3, 2) , directrix y = 0

A horizontal directrix means a vertical parabola.

p = (y value of focus) - (y value of directrix)

____ 2 = 2 - 0 _ 2 = 1

h = the x-coordinate of the focus = 3

Solve for k: The y-value of the focus is k + p, so k + p = 2

k + 1 = 2

k = 1

Write the equation: (x - 3) 2 = 4 (y - 1) Plot the focus, the directrix, and the parabola.

y

0-4-8

4

8

-4

-8

84

x


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DIFFERENTIATE INSTRUCTION

ModelingStudents can write equations which model parabolic shapes that exist in the real world. These include bridges, arcs, and the paths traced in projectile motion.

Critical ThinkingHave students explain how to tell if the graph of a quadratic equation in standard form is a circle or parabola.

EXPLAIN 2 Writing the Equation of a Parabola with Vertex at (h, k)

INTEGRATE MATHEMATICAL PRACTICESFocus on Critical ThinkingMP.3 The focus of a parabola can be found at (h + p, k) or (h, k + p) . Alternatively, students can graph the vertex, find the focus by determining the opening direction of the parabola, then count p units in the appropriate direction.

QUESTIONING STRATEGIESGiven values of h, k, and p, describe the similarities and differences between the graph

of a parabola with an equation in the form (y - k) 2 = 4p (x - h) and an equation in the form (x - h) 2 = 4p (y - k) . Similarities: Both graphs have a vertex at (h, k) and the distance to the focus is the same. Differences: The graph of the equation in the form (y - k) 2 = 4p (x - h) opens to either the left or the right, while the graph of the equation in the form (x - h) 2 = 4p (y - k) opens either upward or downward.

CONNECT VOCABULARY Help students to understand the meanings of focus, directrix, and axis of symmetry by labeling these on the graph of a parabola.

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B Focus (-1, -1) , directrix x = 5

A vertical directrix means a parabola.

p = (x value of focus) - (x value of directrix) ____ 2 = -

__ 2 =

k = the y-coordinate of the focus =

Solve for h: The x-value of the focus is h + p, so

h + p =

h + (-3) =

h =

Write the equation: (y + 1) 2 = ( x - ) Your Turn

Find the equation of the parabola from the description of the focus and directrix. Then make a sketch showing the parabola, the focus, and the directrix.

5. Focus (5, -1) , directrix x = -3 6. Focus (-2, 0) , directrix y = 4

Explain 3 Rewriting the Equation of a Parabola to Graph the Parabola

A second-degree equation in two variables is an equation constructed by adding terms in two variables with powers no higher than 2. The general form looks like this:

a x 2 + b y 2 + cx + dy + e = 0

Expanding the standard form of a parabola and grouping like terms results in a second-degree equation with either a = 0 or b = 0, depending on whether the parabola is vertical or horizontal. To graph an equation in this form requires the opposite conversion, accomplished by completing the square of the squared variable.

y

0-4-8

4

8

-4

-8

84

x

y

0-4-8

4

8

-4

-8

84

x

y

0-4-8

4

8

-4

-8

84

x

-1-3

horizontal

-1

-1

-1

-12 2

5

2

p = 5 - (-3)

_ 2

= 4

k = -1 h + p = h + (4) = 5 → h = 1 (y + 1) 2 = 16 (x - 1)

p = 0 - 4 _ 2

= -2

h = -2 k + p = k + (-2) = 0 → k = 2 (x + 2) 2 = -8 (y - 2)




LANGUAGE SUPPORT

Connect VocabularyHave students work in pairs to fill in a graphic organizer. Write the word parabola in a circle in the middle of a sheet of paper. Fold the paper in fourths. Write opens upward in one corner of the paper, opens downward in another corner, opens to the right and opens to the left in the remaining corners. Have the students work together to sketch a parabola and write an equation for each kind of graph.

EXPLAIN 3 Rewriting the Equation of a Parabola to Graph the Parabola

INTEGRATE TECHNOLOGYStudents can solve equations of parabolas for y and graph the corresponding function(s) on their graphing calculators. If the equation is for a parabola that opens left or right, the parabola needs to be graphed using two functions.

AVOID COMMON ERRORSSome students may include both positive and negative values of 4p (x - h) when taking the square root of both sides of an equation in the form (y - k) 2 = 4p (x - h) . Remind them that when equations of this form are solved for y, the resulting equation should be in the form y = ± √

――― 4p (x - h) + k.

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Example 3 Convert the equation to the standard form of a parabola and graph the parabola, the focus, and the directrix.

x 2 - 4x - 4y + 12 = 0

Isolate the x terms and complete the square on x.

Isolate the x terms. x 2 - 4x = 4y - 12

Add ( -4 _ 2 ) 2 to both sides. x 2 - 4x + 4 = 4y - 8

Factor the perfect square trinomial on the left side. (x - 2) 2 = 4y - 8

Factor out 4 from the right side. (x - 2) 2 = 4 (y - 2) This is the standard form for a vertical parabola. Now find p, h, and

k from the standard form (x - h ) 2 = 4p(y - k) in order to graph the parabola, focus, and directrix.

4p = 4, so p = 1.

h = 2, k = 2.

Vertex = (h, k) = (2, 2).

Focus = (h, k + p) = (2, 2 + 1) = (2, 3).

Directrix: y = k - p = 2 - 1, or y = 1.

y 2 + 2x + 8y + 18 = 0

Isolate the terms. y 2 + 8y = −2x − 18

Add ( _ 2 ) 2

to both sides. y 2 + 8y + = −2x −

Factor the perfect square trinomial. (y + ) 2 = −2x −

Factor out on the right. (y + ) 2 = (x + ) Identify the features of the graph using the standard form of a

horizontal parabola, (y − k ) 2 = 4p(x − h):

4p = , so p = .

h = , k =

Vertex = (h, k) = ( , ) Focus = (h + p, k) = ( , ) Directrix: x = h - p or x =

y

0-4-8

4

8

-4

-8

84

x

y

0-4-8

4

8

-4

-8

84

x

y

816 2

4 2

4

-2

-1

- 1 __ 2

- 1 __ 2

- 3 __ 2

-2-2

-1

-4

-4

-4

1


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A2_MNLESE385894_U2M04L2 180 11/7/14 8:06 AM

QUESTIONING STRATEGIESHow would you solve an equation in the form (x - h) 2 = 4p (y - k) for y in order to

graph the equation on your graphing calculator? Divide both sides of the equation by 4p and then add k to both sides of the equation.

Parabolas 180


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Your Turn

Convert the equation to the standard form of a parabola and graph the parabola, the focus, and the directrix.

7. y 2 - 12x - 4y + 64 = 0 8. x 2 + 8x - 16y - 48 = 0

Explain 4 Solving a Real-World ProblemParabolic shapes occur in a variety of applications in science and engineering that take advantage of the concentrating property of reflections from the parabolic surface at the focus.

Parabolic microphones are so-named because they use a parabolic dish to bounce sound waves toward a microphone placed at the focus of the parabola in order to increase sensitivity. The dish shown has a cross section dictated by the equation x = 32 y 2 where x and y are in inches. How far from the center of the dish should the microphone be placed?

The cross section matches the standard form of a horizontal parabola with h = 0, k = 0, p = 8.

Therefore the vertex, which is the center of the dish, is at (0, 0) and the focus is at (8, 0) , 8 inches away.

y

0-4-8

4

8

-4

-8

84

x

y

0-4-8

4

8

-4

-8

84

x

y 2 - 4y = 12x - 64

y 2 - 4y + 4 = 12x - 64 + 4

(y - 2) 2 = 12x - 60

(y - 2) 2 = 12 (x - 5)

Vertex = (5, 2) , Focus = (8, 2) , Directrix: x = 2

x 2 + 8x = 16y + 48

x 2 + 8x + 16 = 16y + 48 + 16

(x + 4) 2 = 16y + 64

(x + 4) 2 = 16 (y + 4)

Vertex = (-4, -4) , Focus = (-4, 0) , Directrix: y = -8




EXPLAIN 4 Solving a Real-World Problem

CONNECT VOCABULARY Remind students that placing constraints on the values of x is equivalent to restricting the domain. Similarly, placing constraints on the values of y is equivalent to restricting the range.

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B A reflective telescope uses a parabolic mirror to focus light rays before creating an image with the eyepiece. If the focal length (the distance from the bottom of the mirror’s bowl to the focus) is 140 mm and the mirror has a 70 mm diameter (width), what is the depth of the bowl of the mirror?

The distance from the bottom of the mirror’s bowl to the focus is p. The vertex location is not specified (or needed), so use (0, 0) for simplicity. The equation for the mirror is a horizontal parabola (with x the distance along the telescope and y the position out from the center).

(y - ) 2

= 4p (x - ) y 2 = x

Since the diameter of the bowl of the mirror is 70 mm, the points at the rim of the mirror have y-values of 35 mm and -35 mm. The x-value of either point will be the same as the x-value of the point directly above the bottom of the bowl, which equals the depth of the bowl. Since the points on the rim lie on the parabola, use the equation of the parabola to solve for the x-value of either edge of the mirror.

2

= x

x ≈ mm

The bowl is approximately 2.19 mm deep.

Your Turn

9. A football team needs one more field goal to win the game. The goalpost that the ball must clear is 10 feet (~3.3 yd) off the ground. The path of the football after it is kicked for a 35-yard field goal is given by the equation y - 11 = -0.0125 (x - 20) 2 , in yards. Does the team win?

primefocus

eyepiece

parabolic mirror

plane mirror?

70 mm140 mm

0 0

560

35 560

2.19

y - 11 = -0.0125 (35 - 20) 2

y = 8.1875

Since 8.1875 is greater than 3.3, the ball goes over the goalpost and the team wins the game.





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