Consider an arbitrary hyperbola without a marked center or foci.
Using only a straightedge and compass construct the transverse and conjugate axes.

(In reply to re: Solution by Bractals)

Ah. Hadn’t realised that this conic theme had been explored earlier.

This is the only way I can think of to construct an asymptote:

Using my result that if midpoints of parallel chords with gradient

m are used to construct a diameter, then the gradient of the

diameter is b²/(a²m), it follows that if chords at 45^o to the axes are

used, the diameter produced will have gradient b²/a².

If A is the ‘vertex’ of the hyperbola, with coordinates (a, 0), then

a right angled triangle OAB with hypotenuse OB collinear with that

diameter will have base a, height b²/a and area b²/2, so a square

can be constructed, equal in area to the rectangle OABC and with

side length b.

This length can be used to draw a line through O with gradient b/a
which is asymptotic to the hyperbola.

Posted by Harry on 2014-10-07 12:16:47