The dimensions of the supporting base of a bridge are as follows: 40 m height and 120 m wide. a) If the arch is an ellipse with the standard form (x-h)2/a2 + (y-k)2/b2 = 1, determine the equation of the arch. b) If the arch is a parabola with the standard form y-k=a(x-h)2, determine the equation of the arch. C)Determine the height of the arch at a point 30 m away from the centre using the equations found above. The actual height at this point is 30.2 m, determine whether the parabola or ellipse makes the closest approximation.
Assume the center of the support is at the origin.
(a) In the case of an ellipse where `((x-h)^2)/a^2+(y-k)^2/b^2=1` we have a=60 and b=40. (a is the length of the semi-major axis, or 1/2 the major axis which is 120 while b is the length of the semi-minor axis or 1/2 the minor axis). The center of the ellipse is at the origin so h=k=0.
Then the equation is `x^2/3600+y^2/1600=1`
(b) In the case of a parabola, assume the vertex lies on the y-axis. Then `y=a(x-h)^2+k` where h=0 and k=40. We also have the two intercepts, (-60,0) and (60,0). Using (60,0) to find a we get `0=a(60)^2+40==>a=-1/90`
(c) In the case of the ellipse if x=30 we get
For the parabola with x=30 we get
Since 30 is closer to 30.2 than 34.64 is, the parabola is a better model.
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