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Analytical Geometry

Blazing goIITian

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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.

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iitaspirant10

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11 Mar 2007 19:06:13 IST

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whoa!who's askin u 2 reply?!..there...u get ur boink!...njoi

aman adakha

Hot goIITian

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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-b

e^2-i

equate both of them you will getb the ans

bye plzz rate me

vishak p

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11 Mar 2007 20:05:06 IST

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find dy/ dx and then equate the tqo values !!!!

bhargavi

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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)

Bipin Dubey

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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.

SUDEEP KUMAR

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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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