The eccentricity of the conjugate hyperbola of the hyperbola `x^2-3y^2=1` is 2 (b) `2sqrt(3)` (c) 4 (d) `4/5`

Options

A. 2
B. `2//sqrt3`
C. 4
D. `4//5`

Answer

A

Explanation

The given hyperbola is
`(x^(2))/(1)-(y^(2))/(1//3)=1`
Its eccentricity e is given by
`(1)/(3)=1(e^(2)-1)`
Hence, eccentricity e' of the hyperbola is given by
`1=(1)/(3)(e'^(2)-1)`
`"or "e'^(2)=4`
`"or "e'=2`