Fibre Choptic Solution

The following is a solution to Puzzle #13 – Fibre Choptic. Take a look if you haven’t already.

snip4

All that’s needed is to consider a single fibre a transverse distance x from the point at which the fibres are cut.

Suppose the point is a distance a above the base of the fibres. Then the length l to which the fibre at x is cut is given by

x^2+a^2=l^2(x)

from Pythagoras’ theorem.

After the snipped fibre is straightened out, the height of the fibre above its base, y(x), is equal to its length l(x). Hence

x^2+a^2=y^2

y^2-x^2=a^2

This is the Cartesian equation for a hyperbola.

snip5This gives an interesting geometric way of interpreting a hyperbola. Every point on a hyperbola can be rotated about its ‘shadow’ on the x axis, and be made to pass through a common point at the hyperbola’s vertex.

Conversely, you can construct a hyperbola by rotating the vertex point about a series of points on the x axis.

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