Question

# Let 0<θ<π2. If the eccentricity of the hyperbola x2cos2θ−y2sin2θ=1 is greater than 2, then the length of its latus rectum lies in the interval:

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

Solution

## The correct option is D (3,∞)Here, a2=cos2θ and b2=sin2θ Eccentricity, e=√1+(ba)2=√1+tan2θ=secθ Given, e>2∴secθ>2 ⇒cosθ<12⇒θ∈(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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