2
$\begingroup$

Is it possible to construct the asymptotes of a hyperbola from the points on the hyperbola by compass and straightedge alone? And if so, how to construct them?

I have no idea how to approach the first question. It seems it should be possible as it is similar to constructing a tangent to an ellipse, but I haven't been able to adjust such a construction to work for the asymptotes of a hyperbola. An illustration or a reference would be welcome.

$\endgroup$
  • $\begingroup$ What would be the "givens" of the construction? Do you mean, for example, that we would be "given" the points on the hyperbola, so for example I could pick any eleven points on the hyperbola and attempt to base my construction of the asymptote on those points? $\endgroup$ – Lee Mosher Apr 19 '16 at 11:58
  • $\begingroup$ Good question; one is given only the hyperbola, and one can choose arbitrary points on it. So your example is perfectly allowed. $\endgroup$ – Servaes Apr 19 '16 at 12:05
2
$\begingroup$

See in https://fr.answers.yahoo.com/question/index?qid=20130810183100AArTMlg Warning, not the first solution but the second one, by Pope, which constructs the hyperbola's centre first, which looks compulsory. For fully understanding it, you need to have seen a minimum of properties of what is called a diameter in a conic curve, whose definition I recall:

When a line is moved paralelly to itsef, i.e., keeping the same direction (D), the midpoints of its points of intersection with the conic section belong to a same line (D') ; if the conic section possesses a centre (this is the case here, or for an ellipse, but not for a parabola) the centre belongs to this diameter. The line with direction (D) passing through the centre is called the conjugate diameter of (D').

$\endgroup$
  • $\begingroup$ Thank you for the good reference. I'm not assuming my hyperbola to be rectangular, but it seems the construction can be made to work in the general case as well. I'll look into this later today. $\endgroup$ – Servaes Apr 19 '16 at 12:09
  • $\begingroup$ I don't think this algorithm is specific to rectangular hyperbolas. $\endgroup$ – Jean Marie Apr 19 '16 at 16:16
  • $\begingroup$ The last step of the algorithm is explicitly specified to work only for rectangular hyperbolas; this step does not work if the hyperbola is not rectangular. $\endgroup$ – Servaes Apr 21 '16 at 13:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.