let the eccentricity of the hyperbola `x^2/a^2-y^2/b^2=1` be reciprocal to that of the ellipse `x^2+4y^2=4.` if the hyperbola passes through a focus of the ellipse then: (a) the equation of the hyperbola is `x^2/3-y^2/2=1` (b) a focus of the hyperbola is `(2,0)` (c) the eccentricity of the hyperbola is `sqrt(5/3)` (d) the equation of the hyperbola is `x^2-3y^2=3`

Options

A. the equation of the hyperbola is `(x^(2))/(3)-(y^(2))/(2)=1`
B. a focus of the hyperbola is (2,0)
C. the eccentricity of the hyperbola is `(2)/(sqrt3)`
D. the equation of the hyperbola is `x^(2)-3y^(2)=3`

Answer

B, D

Explanation

For the ellipse
`(x^(2))/(2^(2))+(y^(2))/(1^(2))=1.`
we have `1^(2)=2^(2)(1-e^(2))`
`"or "e=(sqrt3)/(2)`
Therefore, the eccentricity of the hyperbola is `2sqrt3`. So, for hyperbola
`b^(2)=a^(2)((4)/(3)-1)` ltBrgt `"or "3b^(2)=a^(2)`
One of the foci of the ellipse is `(sqrt3,9).`
Therefore, `(3)/(a^(2))=1`
`"or "a^(2)=3 and b^(2)=1`
Therefore, the equation of the hyperbola is `(x^(2))/(3)-(y^(2))/(1)=1` ltBrgt the focus of the hyperbola is
`(ae, 0)-=(sqrt3xx(2)/(sqrt3),0)-=(2,0)`