21check the qstn
Let e be the eccentricity of hyperbola. Let f(e) be the eccentricity of its conjugate hyperbola, then is equal to
(A)3
(B)1
(C)3/2
(D)2
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11 Answers
1question needs to be changed.........if not.then beyond jee and i m not concerned!
6somethin ' s surely missing frm it
it wud probably require u to use 1/e2 + 1/e' 2 = 1
:P
but yeah , u knw this quesn is wrong!!!:P
11@kunl, it is not high level maths as u r thinking..
@vinay, is the answer 3/2??
as akhil said,
1e2 + 1e'2 =1.
then
e' =√e2e2 - 1 = f(e)
f(f(e)) = e.
1∫√2[ f(e) + f(f(e))] de
=1∫√2 [e de√e2-1 + e de]
= [√e2-1 + e22]√21
=1+1-1/2 = 3/2.
1@joydoot i did not get how exactly u got to that differential equation?
cn u explain these "two steps"
f(f(e)) = e.
1∫√2[ f(e) + f(f(e))] de
11well kunl,
if u know f(e) , u can find f(f(e)) easily,
f(f(e)) =√f(e)2f(e)2-1
put the value of f(e) = √e2e2-1 in the above equation...
and we need to find out its value 1∫√2[ f(e) + f(f(e))] de as asked by vinay in post #5
11∫√2[ f(e) + f(f(e))] de.this was given!![3] alright i was wondering from where did it appear suddenly[1]