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Question

Let 0<θ<π2. If the eccentricity of the hyperbola x2cos2θy2sin2θ=1 is greater than 2, then the length of its latus rectum lies in the interval:


A
(1,32]
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B
(32,2]
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C
(2,3]
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D
(3,)
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Solution

The correct option is D (3,)
Here, a2=cos2θ and b2=sin2θ
Eccentricity, e=1+(ba)2=1+tan2θ=secθ
Given, e>2secθ>2
cosθ<12θ(π3,π2)

Let L be the length of the latus rectum.
Then L=2b2a=2×sin2θcosθ=2tanθ.sinθ, which is strictly increasing in θ(π3,π2).
So, minimum and maximum value of latus rectum occurs at θ=π3 and θ=π2 respectively.
Lmin=2×sin2π3cosπ3=3
Lmax=2×sin2π2cosπ2 
L(3,)
 

Mathematics

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