Question about parabolas?

my math project asks to find the equation of the st louis arch....it is 615 ft high by 530 ft wide...how would i find an equation for this parabola?? plz helpQuestion about parabolas?
Assume a coordinate system with the origin on the ground under the center of the Arch.

The coordinates of one ground point is x = -530/2 and the other is x = +530/2. For both of these points y = 0.

The top of the arch is at x =0 and y = 615

so the equation is of the form

y = a(x +530/2)(x-530/2) = a(x^2 - 265^2)

at x = 0 y = 615

615 = a(0-70225)

a = -615/70225

y = -615/70225(x^2 - 265^2)

you can reduce -615/70225 if you wish
You might be disappointed. The Gateway Arch is not parabolic; it is based on a catenary curve. It is possible to fit a parabola to your specifications, but it will not overlay the arch.



Here, you may as well try. Start with a root form parabolic equation. Let the roots be 卤530/2.



y = a(x - 伪)(x - 尾)

y = a(x + 265)(x - 265)



Now substitute (0, 615).



615 = a(0 + 265)(0 - 265)

a = -615/70225 = -123/14045

y = (-123/14045)(x + 265)(x - 265)



Now plot that and see how well (or not) it fits the Gateway Arch.Question about parabolas?
Let the vertex be (0, 615) and the 2 intercepts be plus or minus (530/2)



530/2 %26lt;= x %26lt;= 530/2



Solve with y = a(x - h)^2 + k



When it's all done you should end up with:



f(x) = -(123/14045)x^2 + 615