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The equation to the hyperbola having its eccentricity 2 and the distance between its foci is 8 isA. `(x^(2))/(12)-(y^(2))/(4)=1`B. `(x^(2))/(4)-(y^(2))/(12)=1`C. `(x^(2))/(8)-(y^(2))/(2)=1`D. `(x^(2))/(16)-(y^(2))/(9)=1` |
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Answer» Correct Answer - B Distance between foci = 8 `:. 2ae = 8` also `e = 2, :. 2a = 4` `rArr a = 2 rArr a^(2) = 4 :. b^(2) = 4(4-1) = 12` `:.` Equation of hyperbola is `(x^(2))/(4) -(y^(2))/(12) =1`. |
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