A tangent drawn to the hyperbola x2a2−y2b2=1 at Pπ6 forms a triangle of area 3a2 sq. units with the coordinate axes. Then the square of its eccentricity is
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a
15
b
24
c
17
d
14
answer is C.
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Detailed Solution
Pasecπ6,btanπ6≡P2a3,b3Therefore, the equation of tangent at P is x3a/2−y3b=1∴ Area of triangle =12×3a2×3b=3a2 or ba=4 or e2=1+b2a2=17