There are two vertical poles, one of height 100 feet and the other 70 feet, positioned a horizontal distance of 80 feet apart on level ground. A rope of length 100 feet connects the tops of the two poles. A weight, placed on the rope so that it can slide freely along it, is allowed to come to rest.
Find the horizontal distance of the weight from the 100 feet pole and the vertical distance from the ground.
Let h be the horizontal distance sought and similarly let v be the vertical distance. Set everything up on an h-v coordinate plane. The ground is the h-axis, the top of the first pole is at (0,100), the top of the second pole is (80,70), and the position of the weight is some point (h,v).
The possible points the weight can be while keeping the rope taut is an arc of an ellipse. Then the lowest point the weight can be is the bottom horizontal tangent point of the ellipse. This point is where the weight will be at when it comes to rest as at that point it has the lowest potential energy.
Using the definition of an ellipse this equation can be formed:
sqrt(h^2+(100-v)^2) + sqrt((80-h)^2+(70-v)^2) = 100
Squaring and rearranging gives an intermediate result:
sqrt(h^2+(100-v)^2) * sqrt((80-h)^2+(70-v)^2) = -h^2 + 80h - v^2 + 170v - 5650
Squaring and rearranging again gives the standard form of the ellipse:
36h^2 + 48hv + 91v^2 - 6960h - 17390v + 810775 = 0
To find the horizontal tangents first complete the square with respect to h.
(6h + 4v - 580)^2 + (75v^2 - 12750v + 474375) = 0
Then the roots of the quadratic 75v^2 - 12750v + 474375 are the points where there are horizontal tangents to the ellipse. Those values are v=55 and v=115, which imply the points (60,55) and (20,115) are the locations of the tangent points. (20,115) is the upper tangent and (60,55) is the lower tangent. Then the answer to the problem is a horizontal distance of 60 feet and a vertical distance of 55 feet.
Solution
