Let LL’ be the latus rectum through the focus S of a hyperbola and A’ be the farther vertex of the conic.If Δ A’ LL’ is equilateral then its eccentricity e =
A
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B
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C
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D
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Solution
The correct option is D Given, A’LL’ is an equilatral triangle ∴∠SA‘L=30∘ tan30∘=SLA′S⇒1√3=b2aa(a+ae)⇒1√3=a2(e2−1)(1+e)⇒e=√3+1√3