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![[Post New]](/templates/default/images/icon_minipost_new.gif) 6 Apr 2007 14:39:40 IST
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the angle betweeen asymtodes = 2 tan-1(b/a) how does this come----- the asymts. are x/a + y/b = 0 and x/a - y/b =0 my method :: tan  = | (m2 - m1) / 1 + m1 m2 | => m1 = -b/a and m2 = b/a putting this in the above equation we are getting different result !!! please help ! somebody and anybody !
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6 Apr 2007 14:43:44 IST
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please helppp !
only two days are left........ please have mercy !!
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6 Apr 2007 14:47:50 IST
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somebody help pls...............................................................................
me a newcomer...................................................................................
please have a little compassion
thank u !
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6 Apr 2007 14:56:07 IST
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arey yaaroooooooooooo............ atleast tell me...................................
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6 Apr 2007 14:59:24 IST
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m1= b/a and m2= -b/a So tanQ=(m1-m2)/(1+m1m2) tanQ=((b/a)+(b/a))/(1-(b/a)2) tanQ=2(b/a)/(1-(b/a)2) or Q=tan-12(b/a)/(1-(b/a)2) ie Q=2tan-1(b/a) Because 2tan-1x=tan-1(2x/(1-x2))
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ADARSH
NITK Surathkal
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6 Apr 2007 15:23:21 IST
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oop !! forgot that property !!
thank u so much kab !!!
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6 Apr 2007 15:39:07 IST
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friend.. i had one more doubt..............
that formula is v alid only when x belongs to (-1,1)
here is it necassary that b/a is in that range ??
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6 Apr 2007 16:00:53 IST
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m1 = -b/a and m2 = b/a
tan  = | (m2 - m1) / 1 + m1 m2 |
= tan-1[{(b/a) - (-b/a)}/{1 + (b/a)(-b/a)}]
= tan-1{(2b/a)/(1 - b2/a2)}
= 2tan-1(b/a)
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Bipin Kumar Dubey
Chemical Dept.
IIT Kharagpur
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6 Apr 2007 19:14:02 IST
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friend.. i had one more doubt..............
that formula is v alid only when x belongs to (-1,1)
here is it necassary that b/a is in that range ??
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6 Apr 2007 21:46:44 IST
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help !
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7 Apr 2007 01:28:59 IST
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help naa pls !
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7 Apr 2007 15:26:49 IST
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helppppppppppp !
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7 Apr 2007 15:29:55 IST
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tan-1x+tan-1y=tan-1((x+y)/(1-xy)) when x>=0,y>=0 and xy<1 When xy>1 tan-1x+tan-1y=pi+tan-1((x+y)/(1-xy)) Here tanQ=|2(b/a)/(1-(b/a)2)| So whether (b/a)2 is less than 1 or greater than 1 there will be no difference because we consider the positive value. So tanQ will always be positive. So that condition is not required. You can also prove this analytically by drawing the hyperbola and its asymptotes.
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ADARSH
NITK Surathkal
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