A hyperbola having the transverse axis of length 2 sin `theta` is confocal with the ellipse `3x^(2)+4y^(2)=12.` Then its equation is

Options

A. `x^(2)cosec^(2)theta-y^(2)sec^(2)theta=1`
B. `x^(2)sec^(2)theta-y^(2)cosec^(2)theta=1`
C. `x^(2)sin^(2)theta-y^(2)cos^(2)theta=1`
D. `x^(2)cos^(2)theta-y^(2)sin^(2)theta=1`

Explanation

The equation of the ellipse is `(x^(2))/(2^(2))+(y^(2))/((sqrt(3))^(2))=1`.
Let `e` be its eccentricity. Then,
`e=sqrt(1-(3)/(4))=(1)/(2)`
So, coordinates of its foci are `(1,0)` and `(-1,0)`.
Let the equation of the hyperbola be `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1`.
It is given that `a=sin theta` and the hyperbola is confocal to the ellipse.
`:.e sin theta=1impliese=cosec theta`
`:.b^(2)=a^(2)(e^(2)-1)impliesb=costheta`
Hence, the equation of the hyperbola is
`x^(2)cosec^(2)theta-y^(2)sec^(2)theta=1`.