## find eqn for hyperbola to model sports arena's roof's arches

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king&i
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### find eqn for hyperbola to model sports arena's roof's arches

The roof of a sports arena is supported by hyperbolic arches anchored to the ground. These arches span a distance of sixty meters and have a maximum height of 20 meters.
a) Find an equation for a hyperbola to model one of these arches.
b) What is the height of the arch at a horizontal distance of 25 meters from the ceiling's peak?

stapel_eliz
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king&i wrote:The roof of a sports arena is supported by hyperbolic arches anchored to the ground. These arches span a distance of sixty meters and have a maximum height of 20 meters.
a) Find an equation for a hyperbola to model one of these arches.

Draw a picture of this situation. Well, okay; draw just the lower "half" of the hyperbola, representing the roof's arch, and the x-axis, representing the ground.

Since the arches are 60 units wide, then, if you (for simplicity's sake) center the drawing on the y-axis, the x-intercepts will be at x = -30 and x = +30. Label your picture with this information.

Since the peak is 20 units high, then the vertex (of the one half) is at y = 20. Label your picture with this information.

The center will obviously be on the y-axis, so (h, k) = (0, k). Since the vertex is "a" units from the center, then k = 20 + a. Label your picture with this information. Note that this relationship means that a = k - 20.

The conics equation of an hyperbola of this form is:

. . . . .$\frac{(y\, -\, k)^2}{a^2}\, -\, \frac{(x\, -\, h)^2}{b^2}\, =\, 1$

Since you already know that h = 0, then:

. . . . .$\frac{(y\, -\, k)^2}{a^2}\, -\, \frac{x^2}{b^2}\, =\, 1$

You have three points on the hyperbola: (-30, 0), (0, 20), and (30, 0). Since the first and third will work out pretty much the same, plug the second and the third into the above equation, and get equations with unknowns "a", "b", and "k". But a = k - 20, so:

. . . . .$\frac{(-k)^2}{(k\, -\, 20)^2}\, -\, \frac{30^2}{b^2}\, =\, 1$

. . . . .$\frac{(20\, -\, k)^2}{(k\, -\, 20)^2}\, -\, 0\, =\, 1$

For this second equation, note that 20 - k = -(k - 20), and the squaring takes care of the "minus" sign, so you really have:

. . . . .$1\, -\, 0\, =\, 1$

...which is true, but pretty pointless. This tells you that the reason you've been asked to find "an" equation, rather than "the" equation, is that there may be more than one equation possible! So work with what you've got, and pick something for "k". Then this will give you a value for "a", and then you can solve for the value of "b".

king&i wrote:b) What is the height of the arch at a horizontal distance of 25 meters from the ceiling's peak?

Twenty-five units from the peak is twenty-five units to the right (or the left) of the y-axis. So plug "x = 25" into your equation, and see what you get.