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The equation of a hyperbola with co-ordinate axes as principal axes, and the distances of one of its vertices from the foci are 3 and 1 can beA. `3x^(2) -y^(2) =3`B. `x^(2)-3y^(2) +3 =0`C. `x^(2)-3y^(2) -3 =0`D. none of these |
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Answer» Correct Answer - A::B Consider `(x^(2))/(a^(2)) - (y^(2))/(b^(2)) =1` Let one of the vertices be `(a,0)` Foci are `(+- ae,0)` According to the question we have `ae - a = 1` and `ae + a=3` Solving we get `a = 1` and `e =2` `:. e^(2) =1 + (b^(2))/(a^(2)) rArr b^(2) =3` `:.` Equation of hyperbola is `(x^(2))/(1) -(y^(2))/(3) =1` or `3x^(2) - y^(2) =3` For `(x^(2))/(a^(2)) -(y^(2))/(b^(2)) =-1` is `e^(2) =1 + (a^(2))/(b^(2)) rArr b^(2) = (1)/(3)` `:.` Hyperbola will be `x^(2) - 3y^(2) + 3=0` |
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