# Homework Help: Pre calc word problem Parabolic archway

1. Apr 12, 2010

### unrealmatt3

1. The problem statement, all variables and given/known data

A Parabolic archway is 12 meters high at the vertex, at a height of 10 meters, the width of the archway is 8 meters. How wide is the archway at ground level?

2. Relevant equations

given in the picture it has (-4,10) and (4,10) along with vertex (0 ,12)

3. The attempt at a solution
(y- 12)^2 =4p(x - 0)

12 = 4p(0) p = 3 ?? I dont know if i need this

i have a picture but not sure what i need to do

2. Apr 13, 2010

### rl.bhat

Hi unrealmatt3, welcome to PF.
Consider vertex as the origin.
The equation of the parabola becomes
y^2 = 4*p*x.
In the first position x = (12 - 10) and y = 4. Find p.
In the second position x = 12. p is known. find y.
2y will give you the required result.

3. Apr 13, 2010

### unrealmatt3

hey thank rl.bhat for the welcome.
so i think i got it... first 16=4*P(x) where x=2 therefor P = 2
2nd y^2=4(2)(12)
Y^2 = 96
y=4 $$\sqrt{6}$$
then 2y = 19.6 meters

4. Apr 13, 2010

### rl.bhat

That is right.