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?

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

F(0,15-p) & D:y=15+p

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 ;)