Question

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

Answer 1

The given ellipse is
` (x^(2))/(4)+(y^(2))/(3)=1 rArr a=2, b=sqrt(3)`
`therefore 3=4(1-e^(2)) rArr e=(1)/(2)`
So, `a e =2xx (1)/(2) =1`
Hence, the coccentricity `e_(1)` of the hyperbola is given by
`e_(1)="cosec"theta " "[ because ae =e sin theta]`
`rArr b^(2)=sin^(2) theta ("cosec"^(2)theta-1)=cos^(2) theta `
Hence, 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.`