All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Hyperbola Axes Hinder (Posted on 2014-10-03)
Consider an arbitrary hyperbola without a marked center or foci.
Using only a straightedge and compass construct the transverse and conjugate axes.

 No Solution Yet Submitted by K Sengupta Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re(2): Solution | Comment 3 of 7 |
(In reply to re: Solution by Bractals)

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 45o 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

 Search: Search body:
Forums (0)