1. ## elliptical word problem

A bridge is built in the shape of a semielliptical arch. It has a span of 106 feet. The height of the arch 25 feet from the center is 5 feet. Find the height of the arch at its center.

Now I'm pretty sure what I can do is 25^2/11236 + 5^2/b^2 = 1
so.. 625/11236 + 25/b^2 = 1
then solve for b

is this correct??

2. Hello, NKS!

A bridge is built in the shape of a semielliptical arch.
It has a span of 106 feet.
The height of the arch 25 feet from the center is 5 feet.
Find the height of the arch at its center.
Code:
                  |
* o *
*       |       o
*          |       :  *
*           |b      :   *
|       :5
*            |       :    *
-*- - - - - - + - - - o - -o - -
-53            |      25   53

We have: . $\frac{x^2}{53^2} + \frac{y^2}{b^2} \:=\:1 \quad\Rightarrow\quad b^2 \:=\:\frac{53^2y^2}{2809 - x^2}$

Since $(25,5)$ is on the ellipse: . $b^2 \:=\:\frac{53^25^2}{2809 - 25^2} \;=\;\frac{53^25^2}{2184}$

. . $b \;=\;\frac{53\cdot5}{\sqrt{2184}} \:=\:\frac{265}{2\sqrt{546}} \:=\:\frac{265\sqrt{546}}{1092}$

The height is about 5.67 feet.

3. thanks so much! it seems I did set up the problem correctly. I also came to the same answer

4. I read that I wasn't suppose to add problems to a thread that is not mine; however, how about if it's similar?

My problem is the the one on here. Please do excuse me if I'm not suppose to do this, but I would really appreciate your help.

A bride is to be built in the shape of a semielliptical arch and is to have a span of 100 feet. The height of the arch, at a distace of 40 from the center, is to be 10 feet. Find the height of the arch at its center.

I have come to the conclusion that b=16.66

So, if I would like to go further, like search for its eccentricity, I would have to find e=c/a

Which in this case are a=50, and c=47.14280857

This gives me that e=.9428561716, which is in fact smaller than one, making it an elllipse. Please check if my answers are correct. I have been working on this problem for weeks now.