1. Suspension Bridge

The cables of a suspension bridge are in the shape of a parabola. The towers supporting the cable are 400 feet apart and 100 feet high. If the cables are at a height of 10 feet midway between the towers, what is the height of the cable at a point 50 feet from the center of the bridge?

2. Hello, magentarita!

The cables of a suspension bridge are in the shape of a parabola.
The towers supporting the cable are 400 feet apart and 100 feet high.
If the cable is at a height of 10 feet midway between the towers,
what is the height of the cable at a point 50 feet from the center of the bridge?
Place the parabola on a graph.
Code:
                  |
(-200,100)*       |       *(200,100)
:       |       :
:*      |      *:
: *     |     * :
:   *   |   *   :
:       *       :
:       |(0,10) :
:       |       :
- - + - - - + - - - + --
-200      |      200

The parabola opens upward and is symmetric to the y-axis.
. . Its general form is: .$\displaystyle y \;=\;ax^2 + c$

Its y-intercept is (0, 10) . . . Hence: .$\displaystyle y \;=\;ax^2 + 10$

It passes through (200, 100).
. . We have: .$\displaystyle 100 \:=\:a\cdot200^2 + 10 \quad\Rightarrow\quad a \:=\:\frac{9}{4000}$
Hence: .$\displaystyle y \;=\;\tfrac{9}{4000}x^2 + 10$

When $\displaystyle x = \pm50,\;\;y \:=\:\tfrac{9}{4000}(50^2) + 10 \:=\:\frac{125}{8}$

Therefore, 50 feet from the center, the cable is $\displaystyle 15\tfrac{5}{8}$ feet high.

3. wow...

Originally Posted by Soroban
Hello, magentarita!

Place the parabola on a graph.
Code:
                  |
(-200,100)*       |       *(200,100)
:       |       :
:*      |      *:
: *     |     * :
:   *   |   *   :
:       *       :
:       |(0,10) :
:       |       :
- - + - - - + - - - + --
-200      |      200
The parabola opens upward and is symmetric to the y-axis.
. . Its general form is: .$\displaystyle y \;=\;ax^2 + c$

Its y-intercept is (0, 10) . . . Hence: .$\displaystyle y \;=\;ax^2 + 10$

It passes through (200, 100).
. . We have: .$\displaystyle 100 \:=\:a\cdot200^2 + 10 \quad\Rightarrow\quad a \:=\:\frac{9}{4000}$
Hence: .$\displaystyle y \;=\;\tfrac{9}{4000}x^2 + 10$

When $\displaystyle x = \pm50,\;\;y \:=\:\tfrac{9}{4000}(50^2) + 10 \:=\:\frac{125}{8}$

Therefore, 50 feet from the center, the cable is $\displaystyle 15\tfrac{5}{8}$ feet high.

What a marvelous explanation. How do you make your diagrams?

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parabolic suspension bridge problem

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