Difficult math question, please help! :]

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Ok here's the question. Please try and help me..

The viaduct is a large arch, which has the shape of a rectangular hyperbola. The arch is 120 metres wide at its base. The maximum height of the arch is 40 metres.

a) how high is the arch at a point 12m from the centre of its base.
b) How high would the arch be at this same point if the cross section of the arch was half an ellipse.

I just need help knowing what to use to figure these out. Any help would be much appreciated!
 
Are you sure that's all the information you have regarding the arch?
 
yes, thats all the information my teacher gave me. its really hard :p
 
Hello, breanne!

The viaduct is a large arch, which has the shape of a rectangular hyperbola.
The arch is 120 metres wide at its base. The maximum height of the arch is 40 metres.

a) How high is the arch at a point 12m from the centre of its base?
Click to expand...
I believe a "rectangular" hyperbola has an equation in which the denominators are equal.

For this problem, I'll use: \(\displaystyle \L\,\frac{y^2}{a^2}\,-\,\frac{x^2}{a^2}\:=\:1\;\) or \(\displaystyle \;y^2\,-x^2\:=\:a^2\)

The graph looks like this. \(\displaystyle \;\)We'll use the lower branch.
Code: 
      *     |     *
        *   |   *
           ***
            |
            |
    - - - - + - - - - - -
            |a
            |
           ***
        *   |   *
      *     |40   *
     *- - - + - - -*(60,-a-40)
            |
The lower region is the arch itself.
It is 40 m high and 120 m wide.

The lower-right point is located at: \(\displaystyle \,x\,=\,60,\:y\,=\,-a\,-\,40\)

Since this point is on the hyperbola, it satisifies its equation:
\(\displaystyle \;\;\;(-a\,-\,40)^2\,-\,60^2\:=\:a^2\)

\(\displaystyle \;\;\)and we have: \(\displaystyle \,a^2\,+\,80x\,+\,1600\,-\,3600\:=\:a^2\)

\(\displaystyle \;\;\)which simplifies to: \(\displaystyle \,80x\,=\,2000\;\;\Rightarrow\;\;x\,=\,25\)

Hence, the equation is: \(\displaystyle \,y^2\,-\,x^2\:=\:625\;\;\Rightarrow\;\;y\:=\:\pm\sqrt{625\,-\,x^2}\)


(a) When \(\displaystyle x\,=\,12,\:y =\,-\,\sqrt{625\,-\,12^2}\,=\,-\sqrt{481}\,\approx\,-21.9\)
The point is 21.9 m below the x-axis.

Therefore, it is \(\displaystyle 18.1\) m above the "floor" of the arch.
 
soroban said:
I believe a "rectangular" hyperbola has an equation in which the denominators are equal.
Click to expand...
So, "rectangular" means "square"? That's a little silly, isn't it?

Oh, I get it. The asymptotes (or axes) are perpendicular. That makes more sense.
 
Not (necessarily) disagreeing but the only rectangular hyperbola I remember is y=a/x which would have to be rotated for this problem. I don't seem to get a match when I do that. I haven't rotated in a long time but...
---------------
Gene
 
(a) When \(\displaystyle x\,=\,12,\:y =\,-\,\sqrt{625\,-\,12^2}\,=\,-\sqrt{481}\,\approx\,-21.9\)
The point is 21.9 m below the x-axis.

Therefore, it is \(\displaystyle 18.1\) m above the "floor" of the arch.
Click to expand...

Soroban, wouldn't that be \(\displaystyle {-}\sqrt{625+144}=-27.73\), therefore, 12.27 above the base of the hyperbola.

\(\displaystyle \frac{y^{2}}{625}-\frac{144}{625}=1\)

\(\displaystyle \frac{y^{2}-144}{625}=1\)

\(\displaystyle y^{2}-144=625\)

\(\displaystyle y=sqrt{625+144}=sqrt{769}=27.73\)
 
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