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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

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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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


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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.

A 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.

B 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

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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.

A 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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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.

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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.

A 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.

B 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 =

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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.

A 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.

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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?

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eyepiece

parabolic mirror

plane mirror?

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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

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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?

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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. The bulb is located at the focus. How far is the bulb from 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. A cable 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 cable 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

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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 a 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?



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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



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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



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