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HINT Parabola

  1. May 21, 2008 #1
    HINT!!! Parabola

    An arch is in the shape of a parabola with a vertical axis. The arch is 15 ft high at the center and the 40 ft wide at the base. At which height above the base is the width 20 ft?

    [tex]V(0,15)[/tex] so my equation becomes: [tex](x-0)^2=-4p(y-15)\rightarrow x^2=-4p(y-15)[/tex]

    [tex]F(0,15-p)[/tex] & [tex]D:y=15+p[/tex]

    I can substitute 20 into my equation, but I don't know how to find p.

    Just a hint please :)
  2. jcsd
  3. May 21, 2008 #2
    Oh crap, I'm misreading the problem.

    Base = 40 ft. So at the axis of symmetry, it's 20 ft to the left & right. So that means the x value we're at for a total of 20 ft is actually x = 10?

    My point is P(10,y) ???
  4. May 21, 2008 #3


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    Oh rocomath!

    Forget all this detail. :frown:

    Hint: for a standard parabola through the origin, y is proportional to what? :smile:
  5. May 21, 2008 #4
    OH!!! hehe, I had my x-intercepts but yet I wasn't using them. Thanks for your time tiny-tim ;)
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