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Parabola (Horizontal) - finding vertex, focus and equation of line

  1. Oct 3, 2009 #1
    1. The problem statement, all variables and given/known data

    Use completing the square method to rewrite the equation of the parabola
    y^2 – 4y – 44 = 16x in the form (y-y0)^2 = 4A(x-x0)

    Hence find:

    a) the coordinates of the vertex
    b) the coordinates of the focus
    c) the equation of the line that passes through the focus and parallel to the y-axis.

    y^2 - 4y - 44 = 16x at this point I thought that the only way that I could get an equation whereby I could complete the square (and in the form required) was to add 48 to both sides of the equation.

    which would give me y^2 - 4y + 4 = 16x + 48 and I could complete the square and in the form required.

    (y - 2)^2 = 16(x+3)

    *Vertex therefore would be (2, -3)
    *Focal length A = 4
    *Focus S is 4 units to the right of the Vertex S(6, -3)
    *Equation of the line that passes through the focus and parallel to the y-axis x = -3


    2. Relevant equations

    As above


    3. The attempt at a solution

    As above - can someone confirm that I am on the right track with this?
     
  2. jcsd
  3. Oct 3, 2009 #2

    symbolipoint

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    The square piece which seems to be needed for adjustment purposes is (-4/2)^2, so you would add and subtract +4.

    One of your steps should include something of lefthand side as
    (y^2 - 4y + 4) - 4 - 44 which you should find by inspection contains a factorable expression, ultimately allowing you to find x as a function of y, in standard form ( I did not show all of the steps or full equations).
     
  4. Oct 3, 2009 #3
    Hi there - many thanks for the assistance.

    Just so that I have this right.

    y^2 - 4y - 44 = 16x

    y^2 - 4y + 4 - 44 = 16x + 4

    (y^2 - 4y + 4) - 44 = 16x + 4

    (y-2)^2 = 16x + 4 + 44

    (y-2)^2 = 16x + 48

    (y-2)^2 = 16(x + 3) I dont think that I need to simplify this any further?


    Cheers Petra


     
  5. Oct 3, 2009 #4
    Are these correct?

    *Vertex therefore would be (2, -3) - No (-3, 2)
    *Focal length A = 4
    *Focus S is 4 units to the right of the Vertex S( )
    *Equation of the line that passes through the focus and parallel to the y-axis x = -3

     
  6. Oct 3, 2009 #5
    Are these correct?

    *Vertex therefore would be (2, -3) - No (-3, 2)
    *Focal length A = 4
    *Focus S is 4 units to the right of the Vertex S(1,2)
    *Equation of the line that passes through the focus and parallel to the y-axis x = 1
     
  7. Oct 3, 2009 #6

    symbolipoint

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    Your algebra steps seem good, but you did not finish. Your first goal was determine how to complete the square. Your second goal is to find x as a function of y and have this be in standard form. Reading the vertex from the result would then be rendered easily.
     
  8. Oct 3, 2009 #7
    Hi there

    The question asks to "rewrite the equation" of the parabola in the form (y-yo)^2 = 4A(x-xo)
    and from there find the vertex etc.... From the point that I have reached I can Identify the vertex.

    So Im not sure that I understand you second statement? Just a bit confused.

    Cheers Petra d


     
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