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|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.
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Submitted by K Sengupta
Rating: 4.0000 (1 votes)
| Comment 1 of 7
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
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
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