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Analytical Geometry

Blazing goIITian

 Joined: 16 Dec 2006 Post: 424
11 Mar 2007 18:54:24 IST
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A tangent drawn to the hyperbola x2/a2 - y2/b2=1 at 1 of the end points of its latus rectum is parallel to one of its asymptotes.Find its eccentricity.

Blazing goIITian

Joined: 16 Dec 2006
Posts: 424
11 Mar 2007 19:06:13 IST
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Hot goIITian

Joined: 22 Feb 2007
Posts: 122
11 Mar 2007 20:03:00 IST
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hi friend first find dy/dx at x tends to infinity
you will get dy/dx=b/a
now find the value of the derivative at (ae,+or-be^2-i
equate both of them you will getb the ans
bye plzz rate me

Blazing goIITian

Joined: 9 Feb 2007
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11 Mar 2007 20:05:06 IST
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find dy/ dx and then equate the tqo values !!!!

New kid on the Block

Joined: 16 Jan 2007
Posts: 24
12 Mar 2007 09:08:21 IST
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equations of asymptotes to hyperbola x2/a2- y2/b2=1 is x2/a2-y2/b2=0
implies   y=+/-(b/a)x
ends of latusrectum  are (ae,b2/a)  & (ae,-b2/a) .........(1)
equation   of tangent is xx1/a2-yy1/b2=1
substituting  (1) in this, we get  y=xe-a &y=-xe-a  since this is parallel to asymptotes, equating slopes we get e=+/-b/a

hence  eccentricity=+/-(b/a)

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Joined: 23 Jan 2007
Posts: 7958
13 Mar 2007 09:52:07 IST
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Equation os asymptotes : y = (b/a)x and y = -(b/a)x

Equation of tangents : y = xe-a and y = -xe-a

Comparing their slopes we get e = b/a.

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Joined: 23 Jan 2007
Posts: 373
14 Mar 2007 18:43:03 IST
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bhargavi is correct, but i agree with bipin, we should take the positive value of e.

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