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Graph y 2 = −16 x . Identify and label the focus, directrix, and endpoints of the latus rectum.

Focus: ( 4 , 0 ) ; Directrix: x = 4 ; Endpoints of the latus rectum: ( 4 , ± 8 )

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Graphing a parabola with vertex (0, 0) and the y -axis as the axis of symmetry

Graph x 2 = −6 y . Identify and label the focus , directrix    , and endpoints of the latus rectum    .

The standard form that applies to the given equation is x 2 = 4 p y . Thus, the axis of symmetry is the y -axis. It follows that:

  • 6 = 4 p , so p = 3 2 . Since p < 0 , the parabola opens down.
  • the coordinates of the focus are ( 0 , p ) = ( 0 , 3 2 )
  • the equation of the directrix is y = p = 3 2
  • the endpoints of the latus rectum can be found by substituting   y = 3 2   into the original equation, ( ± 3 , 3 2 )

Next we plot the focus, directrix, and latus rectum, and draw a smooth curve to form the parabola    .

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Graph x 2 = 8 y . Identify and label the focus, directrix, and endpoints of the latus rectum.

Focus: ( 0 , 2 ) ; Directrix: y = −2 ; Endpoints of the latus rectum: ( ± 4 , 2 ) .

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Writing equations of parabolas in standard form

In the previous examples, we used the standard form equation of a parabola to calculate the locations of its key features. We can also use the calculations in reverse to write an equation for a parabola when given its key features.

Given its focus and directrix, write the equation for a parabola in standard form.

  1. Determine whether the axis of symmetry is the x - or y -axis.
    1. If the given coordinates of the focus have the form ( p , 0 ) , then the axis of symmetry is the x -axis. Use the standard form y 2 = 4 p x .
    2. If the given coordinates of the focus have the form ( 0 , p ) , then the axis of symmetry is the y -axis. Use the standard form x 2 = 4 p y .
  2. Multiply 4 p .
  3. Substitute the value from Step 2 into the equation determined in Step 1.

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 ) and directrix     x = 1 2 ?

The focus has the form ( p , 0 ) , so the equation will have the form y 2 = 4 p x .

  • Multiplying 4 p , we have 4 p = 4 ( 1 2 ) = −2.
  • Substituting for 4 p , we have y 2 = 4 p x = −2 x .

Therefore, the equation for the parabola is y 2 = −2 x .

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What is the equation for the parabola with focus ( 0 , 7 2 ) and directrix y = 7 2 ?

x 2 = 14 y .

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Graphing parabolas with vertices not at the origin

Like other graphs we’ve worked with, the graph of a parabola can be translated. If a parabola is translated h units horizontally and k units vertically, the vertex will be ( h , k ) . This translation results in the standard form of the equation we saw previously with x replaced by ( x h ) and y replaced by ( y k ) .

To graph parabolas with a vertex ( h , k ) other than the origin, we use the standard form ( y k ) 2 = 4 p ( x h ) for parabolas that have an axis of symmetry parallel to the x -axis, and ( x h ) 2 = 4 p ( y k ) for parabolas that have an axis of symmetry parallel to the y -axis. These standard forms are given below, along with their general graphs and key features.

Standard forms of parabolas with vertex ( h , k )

[link] and [link] summarize the standard features of parabolas with a vertex at a point ( h , k ) .

Axis of Symmetry Equation Focus Directrix Endpoints of Latus Rectum
y = k ( y k ) 2 = 4 p ( x h ) ( h + p ,   k ) x = h p ( h + p ,   k ± 2 p )
x = h ( x h ) 2 = 4 p ( y k ) ( h ,   k + p ) y = k p ( h ± 2 p ,   k + p )
(a) When p > 0 , the parabola opens right. (b) When p < 0 , the parabola opens left. (c) When p > 0 , the parabola opens up. (d) When p < 0 , the parabola opens down.

Questions & Answers

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Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
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