construct the focii and asymptotes with compass and ruler, given a non rectangular hyperbola Problem: Given a non-rectangular hyperbola only, construct its asymptotes and focii with compass and straightedge.
Reference (1) works if the hyperbola is rectangular.
Reference (2-comment) states that reference (1) construction should work for non-rectangular hyperbolas also.
Reference (3) is similar to reference (1) and also constructs the foci.
As I follow the steps in reference (3) the "asymptote" that is constructed for a non-rectangular hyperbola is not the asymptote. It approaches the asymptote as the random point on the hyperbola gets farther from the center.  It approaches the conjugate axis as the random point on the hyperbola approaches the vertex.
Here is my geogebra construction.  https://www.geogebra.org/geometry/uycqtrrk.  I hope the link works to show you what I have done.  It seems to open better in Chrome than Safari.  And I had to sign in to GeoGebra to open the test file I sent myself.

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*Constructing the asymptotes of a hyperbola by compass and straightedge.


*Constructing the asymptotes of a hyperbola by compass and straightedge. (comment)


*http://whistleralley.com/conics/conic_construction/hyperbola_parts/
 A: We need to solve this problem in chunks.  First chunk is to find the center.
Construct any chord of the hyperbola.  Then construct another, parallel chord.  Bisect both chords and draw the line connecting the generated midpoint.  This line is called the conjugate diameter with respect to the set of parallel chords you just used, and it is guaranteed to pass through the center.  In a similar way construct another conjugate diameter using a pair of chords parallel to each other but not parallel to the first two.  The two conjugate diameters intersect at the center, which I now label $O$.
Next comes the transverse axis.  Center the compasses at $O$ and make a circle with radius $r$, big enough to hit the hyperbola at four points which, of course, are the vertices of a rectangle.  Construct the perpendicular bisector of either side that connects two points within the same branch of the hyperbola.  This cuts through $O$ and is the transverse axis.
Now we are ready to attack the asymptotes.  For the rectangle you just constructed, define $l$ as the length of transverse axis from the center to either point where it hits the rectangle.  Now construct a larger rectangle centered on $O$, with radius $r'$ and half-length $l'$ corresponding respectively to $r$ and $l$ from the first rectangle.
Construct a right triangle with a vertex at $O$, a leg from $O$  along the transverse axis measuring $\sqrt{(l')^2-l^2}$, and hypoteneuse $\sqrt{(r')^2-r^2}$.  The hypoteneuse will be directed along an asymptote, and the other asymptote is obtained via a mirror reflection through the transverse axis.
Finally we go for the foci.  Go back to the transverse axis and drop a perpendicular to it from either of the vertices where it hits the hyperbola.  This intersects an asymptote (pick either one) at point $P$.  Construct the circle through $P$ centered on $O$, which intersects the transverse axis at the foci.
Done!
