# Homework Help: HINT Parabola

1. May 21, 2008

### 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.

2. May 21, 2008

### 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) ???

3. May 21, 2008

### tiny-tim

Oh rocomath!

Forget all this detail.

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

4. May 21, 2008

### rocomath

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