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12.3 The Parabola

The focus-directrix definition, equations for curves opening up, down, left, or right, and reflector and satellite-dish applications.

Algebra and Trigonometry 2e, 12.3 The Parabola

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28 exercises

Writing the Equation of a Parabola in Standard Form Given its Focus and Directrix

What is the equation for the parabola with focus ( − 1 2 ,0 ) \left(- \frac{1}{2} , 0\right) and directrix x= 1 2 ? x = \frac{1}{2} ?

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12.3 Exercise 12

What is the equation for the parabola with focus ( 0, 7 2 ) \left(0 , \frac{7}{2}\right) and directrix y=− 7 2 ? y = - \frac{7}{2} ?

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12.3 Exercise 18

Balcony-sized solar cookers have been designed for families living in India. The top of a dish has a diameter of 1600 mm. The sun’s rays reflect off the parabolic mirror toward the “cooker,” which is placed 320 mm from the base.

ⓐ Find an equation that models a cross-section of the solar cooker. Assume that the vertex of the parabolic mirror is the origin of the coordinate plane, and that the para…

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12.3 Exercise 19

Define a parabola in terms of its focus and directrix.

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12.3 Exercise 21

If the equation of a parabola is written in standard form and p p is negative and the directrix is a horizontal line, then what can we conclude about its graph?

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12.3 Exercise 23

As the graph of a parabola becomes wider, what will happen to the distance between the focus and directrix?

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12.3 Exercise 25

Determine whether the given equation is a parabola. If so, rewrite the equation in standard form.

y=4 x 2 y = 4 x^{2}

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12.3 Exercise 27

Determine whether the given equation is a parabola. If so, rewrite the equation in standard form.

( y−3 ) 2 =8( x−2 ) \left(y - 3\right)^{2} = 8 \left(x - 2\right)

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12.3 Exercise 29

Rewrite the given equation in standard form, and then determine the vertex (V), \left(V\right) , focus (F), \left(F\right) , and directrix (d) \left(d\right) of the parabola.

x=8 y 2 x = 8 y^{2}

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12.3 Exercise 31

Rewrite the given equation in standard form, and then determine the vertex (V), \left(V\right) , focus (F), \left(F\right) , and directrix (d) \left(d\right) of the parabola.

y=−4 x 2 y = −4 x^{2}

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12.3 Exercise 33

Rewrite the given equation in standard form, and then determine the vertex (V), \left(V\right) , focus (F), \left(F\right) , and directrix (d) \left(d\right) of the parabola.

x=36 y 2 x = 36 y^{2}

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12.3 Exercise 35

Rewrite the given equation in standard form, and then determine the vertex (V), \left(V\right) , focus (F), \left(F\right) , and directrix (d) \left(d\right) of the parabola.

( x−1 ) 2 =4( y−1 ) \left(x - 1\right)^{2} = 4 \left(y - 1\right)

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12.3 Exercise 37

Rewrite the given equation in standard form, and then determine the vertex (V), \left(V\right) , focus (F), \left(F\right) , and directrix (d) \left(d\right) of the parabola.

( y−4 ) 2 =2( x+3 ) \left(y - 4\right)^{2} = 2 \left(x + 3\right)

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12.3 Exercise 39

Rewrite the given equation in standard form, and then determine the vertex (V), \left(V\right) , focus (F), \left(F\right) , and directrix (d) \left(d\right) of the parabola.

( x+4 ) 2 =24( y+1 ) \left(x + 4\right)^{2} = 24 \left(y + 1\right)

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12.3 Exercise 41

Rewrite the given equation in standard form, and then determine the vertex (V), \left(V\right) , focus (F), \left(F\right) , and directrix (d) \left(d\right) of the parabola.

y 2 +12x−6y+21=0 y^{2} + 12 x - 6 y + 21 = 0

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12.3 Exercise 43

Rewrite the given equation in standard form, and then determine the vertex (V), \left(V\right) , focus (F), \left(F\right) , and directrix (d) \left(d\right) of the parabola.

5 x 2 −50x−4y+113=0 5 x^{2} - 50 x - 4 y + 113 = 0

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12.3 Exercise 45

Rewrite the given equation in standard form, and then determine the vertex (V), \left(V\right) , focus (F), \left(F\right) , and directrix (d) \left(d\right) of the parabola.

x 2 −4x+2y−6=0 x^{2} - 4 x + 2 y - 6 = 0

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12.3 Exercise 47

Rewrite the given equation in standard form, and then determine the vertex (V), \left(V\right) , focus (F), \left(F\right) , and directrix (d) \left(d\right) of the parabola.

3 y 2 −4x−6y+23=0 3 y^{2} - 4 x - 6 y + 23 = 0

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12.3 Exercise 63

Find the equation of the parabola given information about its graph.

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

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12.3 Exercise 65

Find the equation of the parabola given information about its graph.

Vertex is ( 2,2 ); \left(2 , 2\right) ; directrix is x=2− 2 , x = 2 - \sqrt{2} , focus is ( 2+ 2 ,2 ). \left(2 + \sqrt{2} , 2\right) .

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12.3 Exercise 67

Find the equation of the parabola given information about its graph.

Vertex is ( 2 ,− 3 ); \left(\sqrt{2} , - \sqrt{3}\right) ; directrix is x=2 2 , x = 2 \sqrt{2} , focus is ( 0,− 3 ). \left(0 , - \sqrt{3}\right) .

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12.3 Exercise 75

The vertex and endpoints of the latus rectum of a parabola are given. Find the equation.

V( 0,0 )V \left(0 , 0\right), Endpoints ( −2,4 )\left(−2 , 4\right), ( −2,−4 ) \left(−2 , −4\right)

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12.3 Exercise 77

The vertex and endpoints of the latus rectum of a parabola are given. Find the equation.

V( −3,−1 )V \left(−3 , −1\right), Endpoints ( 0,5 )\left(0 , 5\right), ( 0,−7 ) \left(0 , −7\right)

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12.3 Exercise 79

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 =4y. x^{2} = 4 y . At what coordinates should you place the light bulb?

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12.3 Exercise 81

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?

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12.3 Exercise 83

The reflector in 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.

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12.3 Exercise 85

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.

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12.3 Exercise 87

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

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Adapted from College Algebra © 2021 Rice University. Licensed under CC BY 4.0. Sources: College Algebra 2e, Algebra and Trigonometry 2e, Precalculus 2e, College Algebra with Corequisite Support 2e. Imported at revision 4922e46.