Solving a Parabola: Finding the Width at a Given Height

[rocomath]

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 :)

 

[rocomath]

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) ?

 

[tiny-tim]

Oh rocomath!

Forget all this detail.

Hint: for a standard parabola through the origin, y is proportional to what?

 

[rocomath]

OH! hehe, I had my x-intercepts but yet I wasn't using them. Thanks for your time tiny-tim ;)