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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`A. 15B. 24C. 17D. 14 |
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Answer» Correct Answer - C `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` |
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