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Conics: Parabolas:
  Finding the Equation from Information
(page 3 of 4)

Sections: Introduction, Finding information from the equation, Finding the equation from information, Word problems & Calculators


You will also need to work the other way, going from the properties of the parabola to its equation.

  • Write an equation for the parabola with focus at (0, 2) and directrix x = 2.
  • The vertex is always halfway between the focus and the directrix, and the parabola always curves away from the directrix, so I'll do a quick graph showing the focus, the directrix, and a rough idea of where the parabola will go:

      focus in green, directrix in purple, with a rough sketch of where the graph will go

     

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    So the vertex, exactly between the focus and directrix, must be at (h, k) = (1,
    2)
    . The absolute value of p is the distance between the vertex and the focus and the distance between the vertex and the directrix. (The sign on p tells me which way the parabola faces.) Since the focus and directrix are two units apart, then this distance has to be one unit, so | p | = 1.

    Since the focus is to the left of the vertex and directrix, then the parabola faces left (as I'd shown in my picture) and I get a negative value for p: p = 1. Since this is a "sideway" parabola, then the y part gets squared, rather than the x part. So the conics form of the equation must be:

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

  • Write an equation of the parabola with vertex (3, 1) and focus (3, 5).
  • Since the x-coordinates of the vertex and focus are the same, they are one of top of the other, so this is a regular vertical parabola, where the x part is squared. Since the vertex is below the focus, this is a right-side up parabola and p is positive. Since the vertex and focus are 5 1 = 4 units apart, then p = 4. Copyright Elizabeth Stapel 2010-2011 All Rights Reserved

    And that's all I need for my equation, since they already gave me the vertex.

      (x h)2 = 4p(y k)
      (x 3)2 = 4(4)(y 1)

      (x 3)2 = 16(y 1)

  • Write an equation for the parabola with vertex (5, 2) and directrix y = 5.
  • The directrix is an horizontal line; since this line is perpendicular to the axis of symmetry, then this must be a regular parabola, where the x part is squared.

    The distance between the vertex and the directrix is |5 (2)| = |5 + 2| = 3. Since the directrix is below the vertex, then this is a right-side up parabola, so p is positive: p = 3. And that's all I need to find my equation:

      (x h)2 = 4p(y k)
      (x 5)2 = 4(3)(y (2))

      (x 5)2 = 12(y + 2)

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Cite this article as:

Stapel, Elizabeth. "Conics: Parabolas: Finding the Equation From Information." Purplemath.
    Available from
http://www.purplemath.com/modules/parabola3.htm.
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