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 b^{2}/(a^{2}m), it follows that if chords at 45^{o}
to the axes are
used, the diameter produced will have gradient b^{2}/a^{2}.
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 b^{2}/a and area b^{2}/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 20141007 12:16:47 