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Subject: Parabola Author: FB [ Next Thread | Previous Thread | Next Message | Previous Message ] Date Posted: 15:39:19 06/06/05 Mon Ok i am having a parabola problem i think i drew it right but im not sure. Anyways here is the question: A masonry bridge over a river has an arch in the form of a parabola with a vertical axis. The widths of the arch at water level are 5m above water level and 12m and 8m respectively. Find the height of the arch above the point at water level which is 2m from the middle. [ Next Thread | Previous Thread | Next Message | Previous Message ]

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[> Subject: Re: Parabola Author: QUITTNER [ Edit | View ] Date Posted: 08:17:07 06/10/05 Fri >>> A masonry bridge over a river has an arch in the form of a parabola with a vertical axis. The widths of the arch at water level are 5m above water level and 12m and 8m respectively. Find the height of the arch above the point at water level which is 2m from the middle. <<< ..... A shortened form of the equation with the origin (x=0 and y=0) at the top of the arch is y = p * x^2 with all y being negative. ..... Put y1 at x=2, put y2 at x=4, and y3 at x=6. The x of y2 and y3 are half the total widths. ..... We also know that (y2 - y3) = 5 (IS THAT RIGHT?) where y3 is the water level. ..... Insert into the equations y=p*x^2 where * means times, and ^2 means squared: y3=6*6*p=36p, y2=16p, and y2-y3=5. Solve for p and you then can get y1, y2 and y3. The required height above water level is y1-y3 where x=2. [ Post a Reply to This Message ]