The equation of the hyperbola whose centre is `(6,2)` one focus is `(4,2)` and of eccentricity 2 is
(A) `3(x-6)^2-(y-2)^2=3`
(B) `(x-6)^2-3(y-2)^2=1`
(C) `(x-6)^2-2(y-2)^2=1`
(D) `2(x-6)^2-(y-2)^2=1`

Explanation

equation of the hyperbola with centre `(x_{0}, y_{0})` is given by `frac{(x-x_{0})^{2}}{a^{2}}-frac{(y-y_{0})^{2}}{b^{2}}=1`

Focus `=(a e+x_{0}, y_{0})`


`therefore a e=-2`


`Rightarrow a=-1`


`b^{2}=(2)^{2}-a^{2}`
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