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A hyperbola having the transverse axis o...

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

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`

Text Solution

Verified by Experts

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.`
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