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A2.4.B Write the equation of a parabola using given attributes, including … focus, directrix …

(x, y)

(p, 0)(-p, 0) x

y(-p, y)

Directrix

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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 =

(p, 0)

(-p, y)

x

x

p

-p

y

y

0

y

1

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4px

-2

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Module 5 241 Lesson 1

5 . 1 ParabolasEssential Question: How is the distance formula connected with deriving equations for both

vertical and horizontal parabolas?

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Texas Math StandardsThe student is expected to:

A2.4.B

Write the equation of a parabola using given attributes, including vertex, focus, directrix, axis of symmetry, and direction of opening.

Mathematical Processes

A2.1.F

Analyze mathematical relationships to connect and communicate mathematical ideas.

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

1.E.2, 4.F.3, 5.B.1, 5.B.2

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

The student is expected to:

HARDCOVER PAGES 171180

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 design a microphone. Then preview the Lesson Performance Task.

241

HARDCOVER PAGES

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

Resource

Locker

A2.4.B Write the equation of a parabola using given attributes, including … focus, directrix …

(x, y)

(p, 0)(-p, 0)

x

y

(-p, y)

Directrix

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Focus

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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 =

(p, 0)

(-p, y)

x

x

p

-p

y

y

0

y

1

04px

-2

-4

1

0

1 21

Module 5

241

Lesson 1

5 . 1 Parabolas

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

vertical and horizontal parabolas?

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241 Lesson 5 . 1

L E S S O N 5 . 1

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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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A2_MTXESE353930_U2M05L1 242 20/02/14 4:05 AM

Integrate Mathematical ProcessesThis lesson provides an opportunity to address Mathematical Process TEKS A2.1.F, which calls for students to “analyze mathematical relationships to connect and communicate mathematical ideas.” Students learn the relationships between quadratic equations and their graphs. Students learn that equations in the forms (y - k)2 = 4 p (x - h) and (x - h)2 = 4 p (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 PROCESSESFocus on PatternsExplain 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 PROCESSESFocus on Math ConnectionsExplain 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

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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 expression 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 expression for a vertical parabola, , and

replace p with the x 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

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



A2_MTXESE353930_U2M05L1 243 1/27/15 12:14 PM

COLLABORATIVE 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.




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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 PROCESSESFocus on Critical ThinkingThe focus of a parabola is (h + p, k) for a horizontal parabola and (h, k + p) for a vertical parabola. Alternatively, students can graph the vertex and 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.

-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)



A2_MTXESE353930_U2M05L1 245 20/02/14 4:05 AM

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

816 2

4 2

4

- 1 __ 2

- 1 __ 2

- 3 __ 2

-2-2

-1

-2

-1

-4

-4

-4

1




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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.

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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 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?

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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• Online Homework• Hints and Help• Extra Practice

Elaborate

10. Examine the graphs in this lesson and determine a relationship between the separation of the focus and the vertex, and the shape of the parabola. Demonstrate this by finding the relationship between p for a vertical parabola with vertex of (0, 0) and a, the coefficient of the quadratic parent function y = a x 2 .

11. Essential Question Check-In How can you use the distance formula to derive an equation relating x and y from the definition of a parabola based on focus and directrix?

Find the equation of the parabola with vertex at (0, 0) from the description of the focus and directrix and plot the parabola, the focus, and the directrix.

1. Focus at (3, 0) , directrix: x = -3 2. Focus at (0, -5) , directrix: y = 5

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

Evaluate: Homework and Practice

The parabola gets wider as the focus moves away from the vertex. To convert from the

standard form of a parabola to the standard form of a quadratic, isolate y:

x 2 = 4py

y = 1 _ 4p

x 2 = a x 2

⇒ a = 1 _ 4p

Write the expressions for the distance from a point on the parabola to each of the focus

and the directrix. Then equate the two distances per the definition of a parabola.

y 2 = 4px

y 2 = 4 (3) x

y 2 = 12x

x 2 = 4py

x 2 = 4 (-5) y

x 2 = -20y

y 2 = 4px

y 2 = 4 (-1) x

y 2 = -4x

x 2 = 4py

x 2 = 4 (2) y

x 2 = 8y




ELABORATE INTEGRATE MATHEMATICAL PROCESSESFocus on Math ConnectionsExplain that there are alternate forms for the equations of parabolas. Parabolas with vertices at the origin may be written in the forms y = a x 2 (for vertical parabolas) and x = a y 2 (for horizontal parabolas). Parabolas with vertices at points other than the origin may be written in the forms (x - h) 2 = 4p (y - k ) (vertical parabolas) and (y - k) 2 = 4p (x - h) (horizontal parabolas). In these forms, a = 1 __ 4p .

SUMMARIZE THE LESSONExplain how a parabola can be graphed given its equation. Use the equation to graph the vertex of

the parabola. Then find the value of p and

determine the direction in which the parabola

opens. Graph the focus and directrix accordingly.

Substitute an x-value to find a point on the

parabola. Graph the reflection of the point over the

axis of symmetry. Then complete the graph.

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Find the equation of the parabola with the given information.

5. Vertex: (-3, 6) ; Directrix: x = -1.75 6. Vertex: (6, 20) ; Focus: (6, 11)

Find the equation of the parabola with vertex at (h, k) from the description of the focus and directrix and plot the parabola, the focus, and the directrix.

7. Focus at (5, 3) , directrix: y = 7 8. Focus at (-3, 3) , directrix: x = 3

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

9. y2 - 20x - 6y - 51 = 0 10. x2 - 14x - 12y + 73 = 0

11. Communications The equation for the cross section of a parabolic satellite television dish is y = 1 __ 50 x 2 , measured in inches. How far is the focus from the vertex of the cross section?

(y -k) 2 = 4p (x - h)

(y - 6) 2 = 4 (-1.25) (x + 3)

(y - 6) 2 = -5 (x + 3)

(x - h) 2 = 4p (y - k)

(x - 6) 2 = 4 (-9) (y - 20)

(x - 6) 2 = -36 (y - 20)

p = 3 - 7 _ 2

= -2 h = 5

k + p = k + (- 2) = 3

k = 5

(x - 5) 2 = - 8 (y - 5)

p = -3 - 3 _ 2

= -3 k = 3

h + p = h + (- 3) = -3

h = 0

(y - 3) 2 = -12x

y 2 - 6y = 20x + 51

y 2 - 6y + 9 = 20x + 51 + 9

(y - 3) 2 = 20x + 60

(y - 3) 2 = 20 (x + 3)

Vertex = (-3, 3) , Focus = (2, 3) , Directrix : x = -8

x 2 - 14x = 12y - 73

x 2 - 14x + 49 = 12y - 73 + 49

(x - 7) 2 = 12y - 24

(x - 7) 2 = 12 (y - 2)

Vertex = (7, 2) , Focus = (7, 5) , Directrix : x = -1

y = 1 _ 50

x 2 = 1 _ 4·12.5 x 2

The focus is 12.5 in. from the vertex of the cross section.


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A2_MTXESE353930_U2M05L1.indd 250 1/11/15 2:09 AMExercise Depth of Knowledge (D.O.K.) Mathematical Process

1–4 1 Recall of Information 1.D Multiple representations

5–6 1 Recall of Information 1.F Analyze relationships

7–8 1 Recall of Information 1.D Multiple representations

9–10 2 Skills/Concepts 1.D Multiple representations

11–14 2 Skills/Concepts 1.A Everyday life

15 2 Skills/Concepts 1.D Multiple representations

EVALUATE

ASSIGNMENT GUIDE

Concepts and Skills Practice

Explore Deriving the Standard Form Equation of a Parabola

Exercises 16–17

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

Exercises 1–4

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

Exercises 5–8, 15

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

Exercises 9–10

Example 4Solving a Real-World Problem

Exercises 11–14

INTEGRATE MATHEMATICAL PROCESSESFocus on ReasoningStudents can check that their parabolas have the correct widths by verifying that the distance across the parabola at the focus equals 4p.

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12. Engineering The equation for the cross section of a spotlight is y + 5 = 1 __ 12 x 2 , measured in inches. Where is the bulb located with respect to the vertex of the cross

section?

13. When a ball is thrown into the air, the path that the ball travels is modeled by the parabola y - 7 = -0.0175 (x - 20) 2 , measured in feet. What is the maximum height the ball reaches? How far does the ball travel before it hits the ground?

14. The equation of the cables for a suspension bridge is modeled by y - 55 = 0.0025 x 2 where x is the horizontal distance in feet from the support tower and y is the height in feet above the bridge. How far is the lowest point of the cables above the bridge?

15. Match each equation to its graph.

A.

y + 1 = 1 _ 16 (x - 2) 2

B.

y - 1 = 1 _ 16 (x + 2) 2

C.

x + 1 = - 1 _ 16 (y - 2) 2

y + 5 = 1 _ 12

x 2

y + 5 = 1 _ 4·3

x 2

The bulb is 3 in. from the vertex of the cross section.

The vertex of the parabola is (20, 7) , so the maximum height of the ball is 7 feet.

0 - 7 = -0.0175 (x - 20) 2

±20 = x - 20

x = 0, 40

The ball travels 40 feet before it hits the ground.

The vertex of the parabola is (0, 55) , so the lowest point of the cables is 55 feet above the bridge.

B C A



A2_MTXESE353930_U2M05L1 251 1/12/15 8:37 PM

Exercise Depth of Knowledge (D.O.K.) Mathematical Process

16–17 3 Strategic Thinking 1.F Analyze relationships

18 2 Skills/Concepts 1.A Everyday life

19 3 Strategic Thinking 1.F Analyze relationships

20 2 Skills/Concepts 1.G Explain and justify arguments

AVOID COMMON ERRORSSome students may confuse the equations of horizontal and vertical parabolas. It may help them to make a chart listing the general equations for both horizontal and vertical parabolas.

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Derive the equation of the parabolas with the given information.

16. An upward-opening parabola with a focus at (0, p) and directrix y = -p.

17. A leftward-opening parabola with a focus at (-p, 0) and directrix x = p.

H.O.T. Focus on Higher Order Thinking

18. Multi-Step A tennis player hits a tennis ball just as it hits one end line of the court. The path of the ball is modeled by the equation y - 4 = - 4 ____ 1521 (x - 39) 2 where x = 0 is at the end line. The tennis net is 3 feet high, and the total length of the court is 78 feet.

a. How far is the net located from the player?

b. Explain why the ball will go over the net.

c. Will the ball land “in,” that is, inside the court or on the opposite endline?

distance from (x, y) to focus = distance from (x, y) to directrix

√ ――――――― (x - 0) 2 + (y - p) 2 = √ ――――――― (x - x) 2 + (y + p) 2

√ ――――― x 2 + (y - p) 2 = √ ――― (y + p) 2

x 2 + (y - p) 2 = (y + p) 2

x 2 + y 2 - 2py + p 2 = y 2 + 2py + p 2

x 2 = 4py

distance from (x, y) to focus = distance from (x, y) to directrix

√ ――――――― (x + p) 2 + (y - 0) 2 = √ ――――――― (x - p) 2 + (y - y) 2

(x + p) 2 + y 2 = (x - p) 2

x 2 + 2px + p 2 + y 2 = x 2 - 2px + p 2

y 2 = -4px

Find where the ball's height is 0.

0 − 4 = − 4 _ 1521

(x − 39) 2

−4 = − 4 _ 1521

(x − 39) 2

1521 = (x − 39) 2

±39 = x − 39

x = 0, 7 8 The ball will travel 78 feet before it hits the ground. Since the ball’s path and the

court are 78 feet long, the ball will land on the opposite endline, or “in.”

The vertex of the parabola is (39, 4) , so the maximum height the ball reaches is 4 feet. This maximum height occurs at the midline, where the net is located. Since the net is only 3 feet high, the ball will go over the net.

Because the court is 78 feet long and the net is at the court's midline, the net is 39 feet from the player.




A2_MTXESE353930_U2M05L1.indd 252 1/14/15 9:28 PM

CONNECT VOCABULARY Have students find and label the focus and directrix of different parabolas and describe their locations in relation to the axis of symmetry and the vertex of each parabola.

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19. Critical Thinking The latus rectum of a parabola is the line segment perpendicular to the axis of symmetry through the focus, with endpoints on the parabola. Find the length of the latus rectum of a parabola. Justify your answer. Hint: Set the coordinate system such that the vertex is at the origin and the parabola opens rightward with the focus at (p, 0) .

20. Explain the Error Lois is finding the focus and directrix of the parabola y - 8 = - 1 __ 2 (x + 2) 2 . Her work is shown. Explain what Lois did wrong, and then find the correct answer.

h = -2, k = 8

4p = - 1 _ 2 , so p = - 1 _ 8 , or p = -0.125

Focus = (h, k + p) = (-2, 7.875)

Directrix: y = k - p, or y = 8.125

The parabola has the equation: y 2 = 4px.

The axis of symmetry of this parabola is the x-axis. The line containing the latus rectum is perpendicular to the x-axis and goes through the focus so it has an equation of x = p. Setting x = p in the equation above and solving for y we obtain the coordinates of the endpoints of the latus rectum. Their coordinates are (p, 2p) and (p, -2p) . The length of this segment is 2p - (-2p) = 4p as expected for a vertical segment with those endpoints.

y 2 =4px

y 2 =4p ⋅ p y 2 =4 p 2

y = ± √ ―― 4 p 2

y = ±2p

Lois did not find p correctly. The equation for a vertical parabola is (x - h)

2 = 4p(y - k), so Lois should have begun by finding an equivalent

equation with the coefficient in front of y - 8 instead of in front of (x + 2) 2 :

y - 8 = - 1 __ 2

(x + 2) 2

-2(y - 8) = (x + 2) 2 .

h = –2, k = 8

4p = –2, so p = – 1 __ 2

, or p = –0.5

Focus = (h, k + p) = (–2, 7.5)

Directrix: y = k – p, or y = 8.5




JOURNALHave students compare and contrast the methods they have learned for graphing parabolas and writing equations for parabolas.

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Parabolic microphones are used for field audio during sports events. The microphones are manufactured such that the equation of their cross section is x = 1 __ 34 y 2 , in inches. The feedhorn part of the microphone is located at the focus.

a. How far is the feedhorn from the edge of the parabolic surface of the microphone?

b. What is the diameter of the microphone? Explain your reasoning.

c. If the diameter is increased by 5 inches, what is the new equation of the cross section of the microphone?

Lesson Performance Task

a. x = 1 _ 34

y 2 = 1 _ 4 ∙ 8.5

y 2

b. The point directly above the focus is at (p, y) . Since p = 8.5, we can plug (8.5, y) into the equation of the parabola and find y = 17. The radius is 17 inches, so the diameter is 34 inches.

c. The new diameter is 39 inches, so the new radius is 19.5 inches. So the point (p, 19.5) directly above the focus is on the parabola.

x = 1 _ 4p

y 2 ⎯→ p = 1 _ 4p

(19.5) 2 ⎯→ p = 9.75

x = 1 _ 4 (9.75)

y 2 ⎯→ x = 1 _ 39

y 2

The new equation is x = 1 _ 39

y 2 .




EXTENSION ACTIVITY

Sound rays parallel to the axis of a parabolic microphone are reflected off its inner surface and pass through the focal point. To explore this phenomenon, have students graph the parabola y = x 2 . The slope of any line tangent to a point on this parabola is m = 2x. Have students pick different points on the parabola (except the origin), draw tangent lines through these points, then draw a line parallel to the y-axis that ends at the point, forming an acute angle a, which students can measure using a protractor. Then have students draw a second line from the point, forming an angle with the tangent line congruent to a. Encourage the students to draw many lines, to observe that the lines intersect at the focus.

INTEGRATE TECHNOLOGYStudents can plot parabolas which open to the left or right using a calculator or a graphing program on a computer by first solving the equation x 2 = 4py for y, and plotting two functions, y = ± √ ―― 4px .

QUESTIONING STRATEGIESAsk students to derive x 2 = 4py by assuming that the distance from the focus to a point on

a parabola is equal to the shortest distance from that point to the directrix, and using the distance formula. P(x, y) is on the parabola. The focus is

(0, p), the distance from P to (0, p) is

――

(x - 0) 2 + (y - p) 2 , and the distance from P to

the directrix is y + p. Setting the two equal:

―― x 2 + (y - p) 2 = y + p

x 2 + (y - p) 2 = (y + p) 2

x 2 = 4py

Scoring Rubric2 points: Student correctly solves the problem and explains his/her reasoning.1 point: Student shows good understanding of the problem but does not fully solve or explain his/her reasoning.0 points: Student does not demonstrate understanding of the problem.

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