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8.3 The parabola  (Page 2/11)

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d = ( x − 0 ) 2 + ( y − p ) 2       = x 2 + ( y − p ) 2

Set the two expressions for   d   equal to each other and solve for   y   to derive the equation of the parabola. We do this because the distance from   ( x , y )   to   ( 0 , p )   equals the distance from   ( x , y )   to   ( x , − p ) .

x 2 + ( y − p ) 2 = y + p

We then square both sides of the equation, expand the squared terms, and simplify by combining like terms.

x 2 + ( y − p ) 2 = ( y + p ) 2 x 2 + y 2 − 2 p y + p 2 = y 2 + 2 p y + p 2 x 2 − 2 p y = 2 p y            x 2 = 4 p y

The equations of parabolas with vertex   ( 0 , 0 )   are   y 2 = 4 p x   when the x -axis is the axis of symmetry and   x 2 = 4 p y   when the y -axis is the axis of symmetry. These standard forms are given below, along with their general graphs and key features.

Standard forms of parabolas with vertex (0, 0)

[link] and [link] summarize the standard features of parabolas with a vertex at the origin.

Axis of Symmetry Equation Focus Directrix Endpoints of Latus Rectum
x -axis y 2 = 4 p x ( p ,   0 ) x = − p ( p ,   ± 2 p )
y -axis x 2 = 4 p y ( 0 ,   p ) y = − p ( ± 2 p ,   p )
(a) When   p > 0   and the axis of symmetry is the x -axis, the parabola opens right. (b) When   p < 0   and the axis of symmetry is the x -axis, the parabola opens left. (c) When   p < 0   and the axis of symmetry is the y -axis, the parabola opens up. (d) When   p < 0   and the axis of symmetry is the y -axis, the parabola opens down.

The key features of a parabola are its vertex, axis of symmetry, focus, directrix, and latus rectum. See [link] . When given a standard equation for a parabola centered at the origin, we can easily identify the key features to graph the parabola.

A line is said to be tangent to a curve if it intersects the curve at exactly one point. If we sketch lines tangent to the parabola at the endpoints of the latus rectum, these lines intersect on the axis of symmetry, as shown in [link] .

Given a standard form equation for a parabola centered at (0, 0), sketch the graph.

  1. Determine which of the standard forms applies to the given equation:   y 2 = 4 p x   or   x 2 = 4 p y .
  2. Use the standard form identified in Step 1 to determine the axis of symmetry, focus, equation of the directrix, and endpoints of the latus rectum.
    1. If the equation is in the form   y 2 = 4 p x , then
      • the axis of symmetry is the x -axis,   y = 0
      • set   4 p   equal to the coefficient of x in the given equation to solve for   p .   If   p > 0 , the parabola opens right. If   p < 0 , the parabola opens left.
      • use   p   to find the coordinates of the focus,   ( p , 0 )
      • use   p   to find the equation of the directrix,   x = − p
      • use   p   to find the endpoints of the latus rectum,   ( p , ± 2 p ) .   Alternately, substitute   x = p   into the original equation.
    2. If the equation is in the form   x 2 = 4 p y , then
      • the axis of symmetry is the y -axis,   x = 0
      • set   4 p   equal to the coefficient of y in the given equation to solve for   p .   If   p > 0 , the parabola opens up. If   p < 0 , the parabola opens down.
      • use   p   to find the coordinates of the focus,   ( 0 , p )
      • use   p   to find equation of the directrix,   y = − p
      • use   p   to find the endpoints of the latus rectum,   ( ± 2 p , p )
  3. Plot the focus, directrix, and latus rectum, and draw a smooth curve to form the parabola.

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

Graph   y 2 = 24 x .   Identify and label the focus , directrix    , and endpoints of the latus rectum    .

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

  • 24 = 4 p , so   p = 6.   Since   p > 0 , the parabola opens right
  • the coordinates of the focus are   ( p , 0 ) = ( 6 , 0 )
  • the equation of the directrix is   x = − p = − 6
  • the endpoints of the latus rectum have the same x -coordinate at the focus. To find the endpoints, substitute   x = 6   into the original equation:   ( 6, ± 12 )

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

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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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