# How Do You Solve This Word Problem. An Architect's Design For A Building Includes Some Large Pillars With Cross Sections In The Shape Of Hyperbolas. The Curves Can Be Modeled Be The Equation ((x^2)/.25)-((y^2/9))=1, Where The Units Are In Meters.?

you are trying to relate to a 3 dimensional object with a 2 dimensional equation, not impossible but a royal pain in the kazoo. I have just spent over an hour on your question and have determined that I dont have enough information to work with. To the best of my calculations the width of the pillar at the top in not 120 cm but actually 147.19114 cm. How I arrived at these numbers is quite simple.
First solve for y on your graphing calculator, : Y = + or - (4 * x^2 - 1)^(9/2)
graph the two functions in a -2.5 to 2.5 by -2.5 to 2.5 window and trace.
This step you do the hard way because you trace x not y.
Therefore y = o when x = .5 remember that is only half the diameter, double it and you get 1meter or 100cm
so when we find y = 2 half the height x = .7359557 doubled to 1.4719114 meters. Not 1.2
Is there a faster way? Probably but I havent started calculus yet.
A thought occurs, plug in your y value on the home menu and solve for x and double.
Remembering that the equation puts the center of the pillar at the origin of a Cartesian Graph plane. ( hence the top of the 4 meter pillar is at y = 2)

Now that I have thoroughly confused and intimidated you, did I shed some light.
thanked the writer.
As Andre21198 says, the center of the pillar is located at the origin of the coordinate system. As you say, the narrowest point is at the vertical center, where y=0. The half-width of the pillar is presumed to be the x-value, so the width is twice the x value.

At the center, y=0, so we have
(x^2)/.25 - (0^2)/9 = 1
x^2 = .25 (multiply both sides by .25)
x = √.25 = .5 (take the square root of both sides)
width at the center = 2*.5 meters = 1.00 meters = 100 cm

At the top of the pillar, y=2, so we have
(x^2)/.25 - (2^2)/9 = 1
x^2 = .25(1 + 4/9) (add 4/9 to both sides, then multiply by .25)
x = √(13/36) = (√13)/6 (take the square root of both sides)
width at the top = 2*(√13)/6 meters ≈ 1.202 meters = 120.2 cm

(Because the equation is symmetrical about x=0 and about y=0, for every point on the curve (x, y), there are 3 other points also on the curve, (x, -y), (-x, y), and (-x, -y). The distance between the point (x, y) and the point (-x, y)--the width of the pillar at y--is 2x.)
thanked the writer.
Oddman commented
Of course, 120.2 cm rounded to the nearest cm is 120 cm.