The equation of the hyperbola whose foci are (-2, 0) and (2,0) and eccentricity is 2 is given by

Options

A. `3x^(2)-y^(2)=3`
B. `-x^(2)+3y^(2)=3`
C. `-3x^(2)+y^(2)=3`
D. `x^(2)-3y^(2)=3`

Explanation

Let the equatin of the hyperbola be `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1`,
where `b^(2)=a^(2)(e^(2)-1)`.
It is given that `ae=2` and `e=2` . Therefore , `a=1`.
Consequently, `b^(2)=1(4-1)=3`.
Hence, the equation of the hyperbola is
`(x^(2))/(1)-(y^(2))/(3)=1`, or `3x^(2)-y^(2)=3`.