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![[Post New]](/templates/default/images/icon_minipost_new.gif) 29 Nov 2007 09:51:52 IST
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The foci of hyperbola xy = (a2 + b2) / 4 are given by??
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is it (+/- rt(a^2+b^2)/2),+/- rt((a^2+b^2)/2))
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29 Nov 2007 14:15:49 IST
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there's no root there rhd..but ans's close..wts ur logic??
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29 Nov 2007 20:50:16 IST
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what's the answer in buk
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- Edward Everett Hale
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29 Nov 2007 21:11:37 IST
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ans's x = y = +/- (a^2 + b^2) / 2a
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29 Nov 2007 21:29:34 IST
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a n b r not sm special parameters dis r just constants right!
then just by looking at the question,
the points shud come symmetrical in a n b (juzz coz a n b has no special relations to x or y or to any axis)
dis is a rectangular hyperbola. i applied two methods:
1st) rotated it by 45 deg anticl. n got the eqn
x^2-y^2 = (a^2+b^2)/2 n found its focus
den again rotated it by 45deg clockwise n found the original focus.
2nd) if u draw it, u'll find vertices r points closest to origin on the hyperbola
applied distance formula, found its minima n got the vertices
added ae to it n got the focus
note that ae is dist of focus frm vertices along y=x
by no way i'm getting a soln w'out roots
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I cannot do everything,
But still I can do something;
And because I cannot do everything
I will not refuse to do the something that I can do.
- Edward Everett Hale
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29 Nov 2007 22:16:35 IST
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ur ans's right after all..its just a little bit of rearrangement..put 2=e^2=(a^2+b^2)/ a^2..u get the ans..
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