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The director circle of a hyperbola is `x^(2) + y^(2) - 4y =0`. One end of the major axis is (2,0) then a focus isA. `(sqrt(3),2-sqrt(3))`B. `(-sqrt(3),2+sqrt(3))`C. `(sqrt(6),2-sqrt(6))`D. `(-sqrt(6),2+sqrt(6))` |
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Answer» Correct Answer - C::D Radius of director circle `sqrt(a^(2)-b^(2)) =2` Axis of hyperbola is line joining the center `C(0,2)` and `A(2,0)` (end of major axis) `:. a = CA = 2 sqrt(2)` `:. (2sqrt(2))^(2)-b^(2)=4` `:. b^(2) =4` `:. e = (sqrt(3))/(sqrt(2))` Center of the hyperbola is center of the director circle (0,2) Focus lies on this line at distance ae from center `:.` If foci are (x,y) then `(x-0)/(-(1)/(sqrt(2))) = (y-2)/((1)/(sqrt(2))) = +- 2 sqrt(3)` `(x,y) = (sqrt(6),2-sqrt(6))` or `(-sqrt(6),2+sqrt(6))` |
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