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Axis of a Parabola

  1. Dec 4, 2006 #1
    If someone could tell me how to go about it, I'd really appreciate it. Just point me in the right direction please. Kinda stuck.

    The vertex of a parabola is (-3,1) and it touches the x axis at the origin. Find the equation of the axis of the parabola.
     
  2. jcsd
  3. Dec 4, 2006 #2

    radou

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    Did you mean, the parabola intersects the x axis at the origin? Or am I missing something?
     
  4. Dec 4, 2006 #3
    No, the x-axis is the tangent to the parabola at the origin. I dont know what to do with it.
     
  5. Dec 4, 2006 #4
    I dont think that possible.
     
  6. Dec 4, 2006 #5
    Couldnt it be a tilted parabola of the type [tex]y^2=4ax[/tex] so that its lower arm touches the x-axis at the origin?
     
  7. Dec 4, 2006 #6
    yes it could. In that case, it would be opening to the right.
     
  8. Dec 4, 2006 #7
    So how would you find its axis?
     
  9. Dec 4, 2006 #8
    it would simply be y = 1.
     
  10. Dec 4, 2006 #9
    Why would it be y=1?
     
  11. Dec 4, 2006 #10

    arildno

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    Since the axis has to go through the parabola's vertex, and must be parallell to the x-axis.
     
  12. Dec 4, 2006 #11

    HallsofIvy

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    Why would the axis of the parabola have to parallel to the x-axis? We are told that the x-axis is tangent to the parabola at (0,0). No line parallel to the axis of a parabola is ever tangent to it.
     
  13. Dec 5, 2006 #12

    arildno

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    THe x-axis is tangent to the parabola at the origin.

    The symmetry axis of the parabola is parallell to the x-axis, because the the tangent to the vertex is parallell to the y-axis.
    The symmetry axis of a parabola is perpendicular to the tangent at the vertex.
     
  14. Dec 5, 2006 #13
    It doesnt say the tangent at the vertex is paralell to the y-axis. You dont know the equation so you cannot assume that. the parabola could be tiled so that the axis of the parabola makes some angle with the x-axis and one of the arms just touches the x-axis at the origin.
     
  15. Dec 5, 2006 #14

    HallsofIvy

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    Yes, that is given.

    Where was that said?

     
  16. Dec 5, 2006 #15
    Where do you begin on something like this?
     
  17. Dec 5, 2006 #16

    arildno

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    I'm sorry, I was wrong.
     
  18. Dec 5, 2006 #17
    Any ideas on how to go about this?
     
  19. Dec 5, 2006 #18
    Well, if it is a parabola, the equation is of the form y = f(x) = ax^2+bx+c

    Now, f(-3)=1 and

    f(0)=0

    Finally, f'(0)=0

    Also, since f(x)= ax^2+bx+c and f(0)=0 c must also equal zero.

    I'm not sure how any of this will help you though.

    EDIT: I think I'm wrong again (as usual). I guess the equation of a parabola does not have to be of the form ax^2+bx+c=f(x)
     
    Last edited: Dec 5, 2006
  20. Dec 6, 2006 #19
    Could it be that the question is incomplete?
     
  21. Dec 6, 2006 #20
    I think the equation should be y=a(x-h)^2+k
    Since (-3,1) is the vertex h=-3 and k=1
    we get y=a(x+3)^2+1
    If x=0 and y=0
    9a+1=0 a=-1/9 y=1-((x+3)^2)/9
     
  22. Dec 7, 2006 #21
    Yes, that would be true if the axis were parallel to the x-axis, but it isnt so. The parabola is tilted.
     
  23. Dec 7, 2006 #22

    uart

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    accidental double post.:eek:
    please remove. :)
     
    Last edited: Dec 7, 2006
  24. Dec 7, 2006 #23

    uart

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    You can construct a geometric solution by using the property of a parabola that the gradient (tan of angle) of the "vertex to point" chord is equal to one half of the gradient of the tangent at the point.

    See if you can follow through on this and after a bit of basic geometry you should be able to get an equation (based on the tan of sum/difference formula) of something like :

    [tex]\frac{\tan(\theta)}{1 + 2 \tan^2(\theta)} = \frac{1}{3}[/tex]

    Actually that is the solution for the angle whose tan is one half that of the required angle of rotation, but it's going to be pretty easy if you can get to this point.

    Note that the above is a quadratic so there are actually two solutions. One solution has smaller rotation (measured from vertical) and a longer focal length whereas the second solution has greater rotation (axis is closer to the x-axis) and a shorter focal length.
     
    Last edited: Dec 7, 2006
  25. Dec 9, 2006 #24
    Sorry, you lost me. The slope of the chord from a point to the vertex is equal to half the slope of the tangent at that point?
     
  26. Dec 9, 2006 #25

    uart

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    Yeah that's correct for a parabola of the usual form (x^2=4ay). Say you denote the vertex point "O" and some general point on the parabola as point "P", then the gradient of the straight line joining "O" and "P" is one half of the gradient of the tangent at "P". This a well known property of a parabola that's very easily proven.

    Take a look at the attachment. This diagram shows the parabola after its been rotated back to a vertical axis. Your goal is to determine the angle PRQ though which we have rotated it to make it vertical. This of course will be the original angle of the axis (measured from vertical) of the original given parabola.

    In this diagram we know the following facts.
    1. tan(PRQ) = 2 tan(POR)
    2. PS = 3
    3. OS = 1
    4. OSP is a right angle.

    I used the above information and some geometry to solve for tan(POR) and hence to find angle PRQ.
     

    Attached Files:

    Last edited: Dec 9, 2006
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