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 be

Options

A. `3x^(2) -y^(2) =3`
B. `x^(2)-3y^(2) +3 =0`
C. `x^(2)-3y^(2) -3 =0`
D. none of these

Answer

A, B

Explanation

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`