- #1

Gill

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I have a math problem that does not seem to work out. It is a word problem about a suspension bridge. The bridge's main cable is in the shape of a parabola, and the entire bridge, the road, is 200 ft long. The ends of the bridge, the longest virtical cables on the bridge, are 50 ft high. the lowest point in the main cable is 10 ft off of the road with a virtical cable coming straight down. there are 4 additional cables inbetween the middle cable, and the longest cable on either side. i have to figure out the lengths of the virtical cables. i have put this diagram onto a graph, where the lowest part of the main cable is at the origin, the road is at (0,-10) as the directrix, and the focus would then be at the point (0,10). the 4 cables are evenly spaced out every 20 ft along the length of the bridge. since i know that the length from the focus to a point on the parabola is the same distance as that same point to the directrix, i used the distance formula and put the two distances equal to each other.

(A) sqrt ((X2-X1)^2 + (Y2-Y1)^2)

(X2,Y2)= (20,?) (X1, Y1)= (0,10) or the focus

sqrt ((20-0)^2 + (?-10)^2)

sqrt ((20)^2 + (?^2 -20? +100))

sqrt (400 + ?^2 -20? +100)

sqrt (?^2 -20? +500)

(B) sqrt ((X2-X1)^2 + (Y2-Y1)^2)

(X2,Y2)= (20,?) (X1, Y1)= (20,-10) or the directrix

sqrt ((20-20)^2 + (?+10)^2)

sqrt ((0)^2 + (?^2+20?+100))

sqrt (?^2 +20? +100)

(A=B) sqrt (?^2 -20? +500) = sqrt (?^2 +20? +100)

(?^2 -20? +500) = (?^2 +20? +100)

400 = 40?

10 = ? so the distance from the directrix to the top of the first virtical cable is 20 ft since the top is at point (20,10)

however, this does not work for the rest of the cables. the second cable, if im doing my math correctly, is 40 ft tall and the fifth cable that should b 50 ft tall is over 111 ft...

How can i solve this?