Let the eccentricity of the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` be the reciprocal to that of the ellipse `x^(2)+4y^(2)=4`. If the hyperbola passes through a focus of the ellipse, then the equation of the hyperbola, is

Options

A. `(x^(2))/(3)-(y^(2))/(2)=1`
B. `x^(2)-3y^(2)=3`
C. `(x^(2))/(2)-(y^(2))/(3)=1`
D. `3x^(2)-y^(2)=3`

Explanation

The equation of the ellipse is `(x^(2))/(4)+(y^(2))/(1)=1`. Let `e` be its eccentricity. Then,
`e=sqrt(1-(1)/(4))=(sqrt(3))/(2)`
The foci of the ellipse are `S(sqrt(3),0)` and `S'(-sqrt(3),0)`.
Eccentricity of the hyperbola `=(1)/(e)=(2)/(sqrt(3))`
`:.b^(2)=a^(2)((4)/(3)-1)=a^(2)/(3)`[`:'b^(2)=a^(2)(e^(2)-1)`]
The hyperbola passes through `S(sqrt(3),0)`
`:.(3)/(a^(2))=1impliesa^(2)=3`
Now, `b^(2)=(a^(2))/(3)impliesb^(2)=1`
Hence, the equation of the hyperbola is `(x^(2))/(3)-(y^(2))/(1)=1` or, `x^(2)-3y^(2)=3`.