A tangent drawn to hyperbola x2a2−y2b2=1 at P(π6) forms a triangle of area a2 square units, with coordinate axes, then the square of its eccentricity is
A
15
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B
24
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C
17
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D
14
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Solution
The correct option is C17 (A) Tangent to the given hyperbola at p(6) is 2x√3a−a√3yb=1⇒2xb−ya=√3ab It cuts x-axis at (√3a2,0) and yaxis at (0,√3b) ∴ area of triangle=34ab 3a2=34abba=4 e2=17