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If e₁ is the eccentricity of the ellipse x^2 16 + y^2 25 =1 and e₂ is the eccentricity of a hyperbola passing through the foci of the given ellipse and e₁ e₂=1 , then the equation of such a hyperbola among the following is

Options

  1. Ax^2 9 - y^2 16 =1
  2. By^2 9 - x^2 16 =1
  3. Cx^2 9 - y^2 25 =1
  4. Dx^2 25 - y^2 9 =1

Correct answer

B. y^2 9 - x^2 16 =1

Step-by-step solution

Equation of ellipse x^2 16 + y^2 25 =1 Foci =(0, 3) e₂= 1- 16 25 = 3 5 given e₂ e₂=1 (where e₂ is eccentricity of hyperbola) Let equation of hyperbola - x^2 a^2 + y^2 b^2 =1 its passes through (0, 3) array ll & b^2=9 & e₂= 1+ a^2 b^2 & e₂^2=1+ a^2 b^2 & 1 e₂^2 =1+ a^2 9 25 9 =1+ a^2 9 & a^2=16 array Hence, the equation of hyperbola is y^2 9 - x^2 16 =1

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