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Q.

A tangent drawn to the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ at $P (\frac{π}{6})$ forms a triangle of area $3 a^{2} s q .$ 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

$P (a \sec \frac{π}{6}, b \tan \frac{π}{6}) = P (\frac{2 a}{\sqrt{3}}, \frac{b}{\sqrt{3}})$
Therefore, the equation of tangent at P is $\frac{x}{\sqrt{3} a / 2} - \frac{y}{\sqrt{3} b} = 1$
$∴$ Area of triangle $= \frac{1}{2} \times \frac{\sqrt{3} a}{2} \times \sqrt{3} b = 3 a^{2}$ Or $\frac{b}{a} = 4$ Or $e^{2} = 1 + \frac{b^{2}}{a^{2}} = 17$

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