Find Height of Elliptical Arch Spanning 118ft & 8ft High

[xsn53]

Homework Statement

A bridge is built in the shape of a semielliptical arch. It has a span of 118 feet. The height of the arch 25 feet from the center is to be 8 feet. Find the height of the arch at its center?

Homework Equations

not sure if the 25 feet from the center is the focal axis or not?

The Attempt at a Solution

with the given info i know that i have the variable a in the equation of an ellipse:

(x^2/a^2) + (y^2/b^2) = 1 (a>b)

and i know that i am looking for the minor axis or semiminor axis to be exact. The Foci:

(+ or - c,0) where c^2 = a^2 - b^2 I have, a, which is the span 118/2 = 59 for the semimajor axis. i believe i have, c, which is 25, but when i plug them in and solve for, b, i do not get the right answer. I believe i may need to do something with the 8 feet, but i have not seen it. If anyone could help point me in the right direction, I would be greatly appreciated.

Thanks.

 

[Deadleg]

I think the statement "The height of the arch 25 feet from the center is to be 8 feet" means that at $$x=\pm{25}$$, $$y=8$$ giving the coordinates $$(\pm{25},8)$$. So try putting those coodinates into the equation $${\frac{x^2}{25^2}}+{\frac{y^2}{b^2}}=1$$ and solve for $$b^2$$.

 

[HallsofIvy]

I think the statement "The height of the arch 25 feet from the center is to be 8 feet" means that at $$x=\pm{25}$$, $$y=8$$ giving the coordinates $$(\pm{25},8)$$. So try putting those coodinates into the equation $${\frac{x^2}{25^2}}+{\frac{y^2}{b^2}}=1$$ and solve for $$b^2$$.

Was that a typo? We are told that the span is 118 feet and you haven't used that. Put x= 25, y= 8 into
[tex]\frac{x^2}{118^2}+ \frac{y^2}{b^2}= 1[/itex]
and solve for b.

 

[Deadleg]

Oh yeah whoops :S But span=2a so a=59, so I believe the equation is $$\frac{x^2}{59^2}+ \frac{y^2}{b^2}= 1$$.

 

[xsn53]

yes, thank you for your quick replies, when i put the values in for x,y and a, (59) I get:

(118 * sqr root(714)) / 357 ---> which breaks into a cool +/- 8.83. which what do you

know, is exactly the right answer :) Thanks again for the help with the problem, I was thinking I had to use the 8 somewhere, once again thanks for all your help, and steering me in a right direction.