A tangent drawn to hyperbola `x^2/a^2-y^2/b^2 = 1` at `P(pi/6)` froms a triangle of area `3a^2` square units, with the coordinate axes, then the square of its eccentricity is (A) `15` (B) `24` (C) `17` (D) `14`

Options

A. 15
B. 24
C. 17
D. 14

Answer

C

Explanation

`P(a sec.(pi)/(6),b tan.(pi)/(6))-=P((2a)/(sqrt3),(b)/(sqrt3))`
Therefore, the equation of tangent at P is
`(x)/(sqrt3a//2)-(y)/(sqrt3b)=1`
`therefore" Area of triangle"=(1)/(2)xx(sqrt3a)/(2)xxsqrt3b=3a^(2)"(Given)"`
`"or "(b)/(a)=4`
`"or "e^(2)=1+(b^(2))/(a^(2))=17`