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(2): Solution by Harry)

A more direct method

After constructing the axes…

Use points P(2a, 0) on the x axis and Q(2a, sqrt(3)b) on the

curve, to construct a 30^o, 60^o, 90^o triangle PQR so that

PQ = sqrt(3)b, QR = b, PR = 2b.

Then use the length 2b to construct the point S(2a, 2b) so that

OS is an asymptote.

Posted by Harry on 2014-10-09 12:08:53