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HINT Parabola

  • Thread starter rocomath
  • Start date
  • #1
1,752
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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 :)
 

Answers and Replies

  • #2
1,752
1
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) ???
 
  • #3
tiny-tim
Science Advisor
Homework Helper
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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:
 
  • #4
1,752
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OH!!! hehe, I had my x-intercepts but yet I wasn't using them. Thanks for your time tiny-tim ;)
 

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