Kepler gave the following construction for a hyperbola with

Kepler gave the following construction for a hyperbola with foci at A and B and with one vertex at C: Let pins be placed at A and B. To A let a thread with length AC be tied and to B a thread with length BC. Let each thread be lengthened by an amount equal to itself. Then grasp the two threads together with one hand (starting at C) and little by little move away from C, paying out the two threads. With the other hand, draw the path of the join of the two threads at the fingers Show that the path is a hyperbola.

Solution

One of properties of hyperbola is the magnitude of difference in distance between a point in locus of hyperbola and and the focii remains constant.Therefore |AC-BC|=constant. in the above method we are catching the two threads at point C and letting loose some thread away from point C. As long as the length of loosed length in two strings is same the drawn path using the method in question leads to hyperbola.

And the proof that |AC-BC|=constant is this

For standard hyperbola (x²/a²)-(y²/b²)=1 is

focal distance r1 is |(xe+a)|

focal distance r2 is |(xe-a)|

so the diference is r1-r2=2a=constant

Generalising it is vaild to any hyperbola