What Is the Equation of a Parabola for a Bridge Needing Specific Dimensions?

[SAC2009]

Homework Statement

1. It is a word problem I need done by tomorrow morning. so please help:
A new bridge is being constructed. The space between the support needs to be 1050 feet; the height at the center of the arch needs to be 350 feet. An empty tanker needs a 280 foot clearance to pass beneath it.
a) find the equation of a parabola with these characteristics. Place the vertex at the origin.
b) How wide is the channel that the tanker can pass through?
c) If the river were to flood and rise 10 feet, how would the clearance of the bridge be affected?

Homework Equations

The equation for a vertical parabola is y=a(x-h)squared + k

The Attempt at a Solution

I tried putting the vertex at (0,0) and plugging in (525,350) to get the a value. I don't know if this is anywhere close to what I was supposed to do.

 

[CompuChip]

Personally I find the equation
y = A (x - p)(x - q)
more useful, when you know where the parabola has to be zero.
Do you see why?

 

[HallsofIvy]

Homework Statement

1. It is a word problem I need done by tomorrow morning. so please help:
A new bridge is being constructed. The space between the support needs to be 1050 feet; the height at the center of the arch needs to be 350 feet. An empty tanker needs a 280 foot clearance to pass beneath it.
a) find the equation of a parabola with these characteristics. Place the vertex at the origin.
b) How wide is the channel that the tanker can pass through?
c) If the river were to flood and rise 10 feet, how would the clearance of the bridge be affected?

Homework Equations

The equation for a vertical parabola is y=a(x-h)squared + k

And if you Place the vertex at the origin, y(x)= ax²+ k. "The space between the support needs to be 1050 feet" so y(1050/2)= y(525)= 0 (NOT "526,350). "the height at the center of the arch needs to be 350 feet" so y(0)= 350. That is enough to find a and k.
In order for the tanker to pass through, the height must be at least 280 feet. What values of x make y= 280?

The Attempt at a Solution

I tried putting the vertex at (0,0) and plugging in (525,350) to get the a value. I don't know if this is anywhere close to what I was supposed to do.

 

[SAC2009]

I thought that if the vertex was at the origin (0,0) that the equation would be y=axsquared, because the vertex is (h,k). I keep getting an extremely small number for k, which wouldn't work because then the width at that point would be way too small.

 

[SAC2009]

sorry, i meant an extremely small number for a

 

[HallsofIvy]

You are right. I put the origin at (0, 350).

Now, either y= kx² or y= -x² depending on whether you want the parabola opening upward or downward. The second would better represent a support for a bridge but, mathematically, either way works.

If you want to use y= kx² then the bases of the support are at (525, 350) and (-525, 350) and you would need, to allow the tanker to pass, to solve y= x²= 280.