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KCET2023MathematicsHyperbola

The distance between the foci of a hyperbola is 16 and its eccentricity is 2 . Its equation is

Options

  1. Ax^2 4 - y^2 9 =1
  2. B2 x^2-3 y^2=7
  3. Cy^2-x^2=32
  4. Dx^2-y^2=32

Correct answer

D. x^2-y^2=32

Step-by-step solution

Given, distance between the foci =16 and aligned e & = 2 2 a e & =16 a e & =8 aligned a= 8 2 =4 2 So, aligned & b^2=a^2 (e^2-1 ) & b^2=32(2-1)=32 aligned Equation of hyperbola is aligned & x^2 a^2 - y^2 b^2 =1 & x^2 32 - y^2 32 =1 & x^2-y^2=32 aligned

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