# Fibre Choptic Solution

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

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.

This 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.