JEE Advanced2006MathematicsHyperbolaActual
If e₁ is the eccentricity of the ellipse x^2 16 + y^2 25 =1 and e₂ is the eccentricity of the hyperbola passing through the focii of the ellipse and e₁ e₂=1 , then equation of the hyperbola is
Options
- Ax^2 9 - y^2 16 =1
- Bx^2 16 - y^2 9 =-1
- Cx^2 9 - y^2 25 =1
- DNone of these
Correct answer
B. x^2 16 - y^2 9 =-1
Step-by-step solution
The eccentricity of x^2 16 + y^2 25 =1 is e₁= 1- 16 25 = 3 5 e₂= 5 3 Foci of ellipse (0, 3) Equation of hyperbola is x^2 16 - y^2 9 =-1 Hence (b) is the correct answer.