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.)
 Solution | Comment 1 of 7
If Points A, B, C, D lie on a hyperbola, with AB parallel to CD, then
the line joining the mid-points of AB and CD passes through the
centre of the hyperbola.*

Choose any two points A and B on the curve, and find M, the
mid-point of AB. Construct a line parallel to AB that cuts the curve
at C and D, and find N, the midpoint of CD. Draw MN.

Now repeat the whole process starting with new points A’, B’, and
finishing with a line M’N’.

The centre, O, of the hyperbola is the intersection of MN and M’N’.

Draw a circle with centre at O which intersects the hyperbola at
P, Q, R, S. The two lines through O parallel to the sides of the
rectangle PQRS will be the two axes of the hyperbola.

_____________

* Outline Proof

The line y = mx + c crosses the hyperbola x2/a2 – y2/b2 = 1 at

points which satisfy:   (b2 – m2a2)x2 – 2a2cmx – a2(b2 + c2) = 0

The average of the two roots gives the x coordinate of the

mid-point of the chord:    X  =  a2cm/(b2 – m2a2).

By similar reasoning,   Y = b2c/( b2 – m2a2)  so that the equation

of the locus of this mid-point as c varies is:  Y = b2X/(a2m),

which passes through (0, 0), the centre of the hyperbola.

 Posted by Harry on 2014-10-05 11:42:37

 Search: Search body:
Forums (0)