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

Options

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

Answer

A

Explanation

The length of transverse axis is
`2sin theta=2a`
`"or "A=sin theta`
Also, for the ellipse
`3x^(2)+4y^(2)=12`
`"or "(x^(2))/(4)+(y^(2))/(3)=1`
`a^(2)=4, b^(2)=3`
`therefore" "e=sqrt(1-(b^(2))/(a^(2)))=sqrt(1-(3)/(4))=(1)/(2)`
Hence, the focus of ellipse is `(2xx1//2,0)-=(1,0)`
As the hyperbola is confocal with the ellipse, the focus of the hyperbola is (1, 0). Now,
ae' = 1
`"or "sin theta xxe'=1`
`"or "e'="cosec"theta`
`therefore" "b^(2)=a^(2)(e'^(2)-1)=sin^(2)theta("cosec"^(2)theta-1)=cos^(2)theta)`
Therefore, the equation of hyperbola is
`(x^(2))/(sin^(2)theta)-(y^(2))/(cos^(2)theta)=1`
`"or "x^(2)"cosec"^(2)theta-y^(2)sec^(2)theta=1`