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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

  1. Ax^2 cosec ^2 -y^2 ^2 =1
  2. Bx^2 ^2 -y^2 cosec ^2 =1
  3. Cx^2 ^2 -y^2 ^2 =1
  4. 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

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