PQ is a double ordinate of the hyperbola x2a2−y2b2=1 such that OPQ is an equilateral triangle, O being the centre of the hyperbola, then the range of the eccentricity e of the hyperbola is
A
1<e<2
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B
1<e<√2
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C
e≥2√3
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D
e<4
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Solution
The correct option is Ce≥2√3
Let double ordinate PQ be such that P=(asecθ,btanθ) and Q=(asecθ,−btanθ) and O is the centre (0,0). From △OPR tan30∘=btanθasecθ ⇒1√3=basinθ ⇒3b2a2=cosec2θ ⇒3(e2−1)=cosec2θ≥1 ⇒e≥2√3