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

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y 2 + 12 x 6 y + 21 = 0

( y 3 ) 2 = −12 ( x + 1 ) , V : ( 1 , 3 ) ; F : ( 4 , 3 ) ; d : x = 2

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x 2 4 x 24 y + 28 = 0

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5 x 2 50 x 4 y + 113 = 0

( x 5 ) 2 = 4 5 ( y + 3 ) , V : ( 5 , 3 ) ; F : ( 5 , 14 5 ) ; d : y = 16 5

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y 2 24 x + 4 y 68 = 0

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x 2 4 x + 2 y 6 = 0

( x 2 ) 2 = −2 ( y 5 ) , V : ( 2 , 5 ) ; F : ( 2 , 9 2 ) ; d : y = 11 2

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y 2 6 y + 12 x 3 = 0

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3 y 2 4 x 6 y + 23 = 0

( y 1 ) 2 = 4 3 ( x 5 ) , V : ( 5 , 1 ) ; F : ( 16 3 , 1 ) ; d : x = 14 3

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x 2 + 4 x + 8 y 4 = 0

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Graphical

For the following exercises, graph the parabola, labeling the focus and the directrix.

( y 2 ) 2 = 4 3 ( x + 2 )

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−5 ( x + 5 ) 2 = 4 ( y + 5 )

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−6 ( y + 5 ) 2 = 4 ( x 4 )

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y 2 6 y 8 x + 1 = 0

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x 2 + 8 x + 4 y + 20 = 0

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3 x 2 + 30 x 4 y + 95 = 0

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y 2 8 x + 10 y + 9 = 0

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y 2 + 2 y 12 x + 61 = 0

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2 x 2 + 8 x 4 y 24 = 0

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For the following exercises, find the equation of the parabola given information about its graph.

Vertex is ( 0 , 0 ) ; directrix is y = 4 , focus is ( 0 , −4 ) .

x 2 = −16 y

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Vertex is ( 0 , 0 ) ; directrix is x = 4 , focus is ( −4 , 0 ) .

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Vertex is ( 2 , 2 ) ; directrix is x = 2 2 , focus is ( 2 + 2 , 2 ) .

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

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Vertex is ( −2 , 3 ) ; directrix is x = 7 2 , focus is ( 1 2 , 3 ) .

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Vertex is ( 2 , 3 ) ; directrix is x = 2 2 , focus is ( 0 , 3 ) .

( y + 3 ) 2 = −4 2 ( x 2 )

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Vertex is ( 1 , 2 ) ; directrix is y = 11 3 , focus is ( 1 , 1 3 ) .

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For the following exercises, determine the equation for the parabola from its graph.

( y 2 ) 2 = 1 4 ( x + 2 )

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( y 3 ) 2 = 4 5 ( x + 2 )

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Extensions

For the following exercises, the vertex and endpoints of the latus rectum of a parabola are given. Find the equation.

V ( 0 , 0 ) , Endpoints  ( 2 , 1 ) , ( −2 , 1 )

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V ( 0 , 0 ) , Endpoints  ( −2 , 4 ) , ( −2 , −4 )

y 2 = −8 x

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V ( 1 , 2 ) , Endpoints  ( −5 , 5 ) , ( 7 , 5 )

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V ( −3 , −1 ) , Endpoints  ( 0 , 5 ) , ( 0 , −7 )

( y + 1 ) 2 = 12 ( x + 3 )

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V ( 4 , −3 ) , Endpoints  ( 5 , 7 2 ) , ( 3 , 7 2 )

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Real-world applications

The mirror in an automobile headlight has a parabolic cross-section with the light bulb at the focus. On a schematic, the equation of the parabola is given as x 2 = 4 y . At what coordinates should you place the light bulb?

( 0 , 1 )

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If we want to construct the mirror from the previous exercise such that the focus is located at ( 0 , 0.25 ) , what should the equation of the parabola be?

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A satellite dish is shaped like a paraboloid of revolution. This means that it can be formed by rotating a parabola around its axis of symmetry. The receiver is to be located at the focus. If the dish is 12 feet across at its opening and 4 feet deep at its center, where should the receiver be placed?

At the point 2.25 feet above the vertex.

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Consider the satellite dish from the previous exercise. If the dish is 8 feet across at the opening and 2 feet deep, where should we place the receiver?

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A searchlight is shaped like a paraboloid of revolution. A light source is located 1 foot from the base along the axis of symmetry. If the opening of the searchlight is 3 feet across, find the depth.

0.5625 feet

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If the searchlight from the previous exercise has the light source located 6 inches from the base along the axis of symmetry and the opening is 4 feet, find the depth.

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An arch is in the shape of a parabola. It has a span of 100 feet and a maximum height of 20 feet. Find the equation of the parabola, and determine the height of the arch 40 feet from the center.

x 2 = −125 ( y 20 ) , height is 7.2 feet

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If the arch from the previous exercise has a span of 160 feet and a maximum height of 40 feet, find the equation of the parabola, and determine the distance from the center at which the height is 20 feet.

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An object is projected so as to follow a parabolic path given by y = x 2 + 96 x , where x is the horizontal distance traveled in feet and y is the height. Determine the maximum height the object reaches.

2304 feet

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For the object from the previous exercise, assume the path followed is given by y = −0.5 x 2 + 80 x . Determine how far along the horizontal the object traveled to reach maximum height.

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Practice Key Terms 4

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Source:  OpenStax, College algebra. OpenStax CNX. Feb 06, 2015 Download for free at https://legacy.cnx.org/content/col11759/1.3
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