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December 6th, 2009, 08:18 AM   #1
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rectangular hyperbola

Prove that the chord joining the points P(cp , c/p) and Q (cq , c/q) on the rectangular hyperbola xy=c^2 has the equation pqy+x=c(p+q).

I can prove this part.

Given that the points P, Q and R lie on the hyperbola xy=c^2, prove that if PQ and PR are inclined equally to the coordinate axes, then QR passes through O.
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December 10th, 2009, 03:54 AM   #2
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Re: rectangular hyperbola

can someone get me started . THanks .
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December 10th, 2009, 02:26 PM   #3
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Re: rectangular hyperbola

The line PQ can be written

and the line PR as

so the slopes are -1/pq and -1/pr.

If these are equal, then q=r, which makes Q the same point as R. So assume instead that one is the negative of the other. What is the consequence for the line QR

?
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December 10th, 2009, 02:45 PM   #4
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Given distinct points P(cp, c/p), Q (cq, c/q), and R(cr, c/r) on the hyperbola, the slope of PQ is (c/q - c/p)/(cq - cp), i.e., -1/(pq), and similarly the slope of PR is -1/(pr). (These values are probably already known from the work done for the first part of the problem.)

The above slopes are unequal, but may differ only in sign, in which case q = -r.
You already know QR has equation qry + x = c(q + r), so . . . (it's easy to finish from there).
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December 10th, 2009, 06:46 PM   #5
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Re:

Quote:
Originally Posted by skipjack
Given distinct points P(cp, c/p), Q (cq, c/q), and R(cr, c/r) on the hyperbola, the slope of PQ is (c/q - c/p)/(cq - cp), i.e., -1/(pq), and similarly the slope of PR is -1/(pr). (These values are probably already known from the work done for the first part of the problem.)

The above slopes are unequal, but may differ only in sign, in which case q = -r.
You already know QR has equation qry + x = c(q + r), so . . . (it's easy to finish from there).
thank you guys .
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