JEE Advanced2007MathematicsEllipseActual
A hyperbola, having the transverse axis of length 2 , is confocal with the ellipse 3 x^2+4 y^2=12 . Then, its equation is
Options
- Ax^2 cosec ^2 -y^2 ^2 =1
- Bx^2 ^2 -y^2 cosec ^2 =1
- Cx^2 ^2 -y^2 ^2 =1
- Dx^2 ^2 -y^2 ^2 =1
Correct answer
A. x^2 cosec ^2 -y^2 ^2 =1
Step-by-step solution
The given ellipse is aligned & x^2 4 + y^2 3 =1 =2, = 3 & 3=4 (1-e^2 ) e= 1 2 & aligned a e=1 Hence the eccentricity e₁ , of the hyperbola is given by array rlrl & & 1 & =e₁ & & e₁ & = cosec & b^2 & = ^2 ( cosec ^2 -1 )= ^2 array Hence, the hyperbola is x^2 ^2 - y^2 ^2 =1 or x^2 cosec ^2 -y^2 ^2 =1