JEE Main202125 Feb 2021Evening ShiftMathematicsHyperbolaActual
A hyperbola passes through the foci of the ellipse x 2 25 + y 2 16 = 1 and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is:
Options
- Ax 2 9 - y 2 16 = 1
- Bx 2 - y 2 = 9
- Cx 2 9 - y 2 25 = 1
- Dx 2 9 - y 2 4 = 1
Correct answer
A. x 2 9 - y 2 16 = 1
Step-by-step solution
For ellipse e 1 = 1 - b 2 a 2 = 3 5 for hyperbola e 2 = 5 3 Let hyperbola be x 2 a 2 - y 2 b 2 = 1 ∵ it passes through 3 , 0 ⇒ 9 a 2 = 1 ⇒ a 2 = 9 ⇒ b 2 = a 2 e 2 - 1 = 9 25 9 - 1 = 16 ∴ Hyperbola is x 2 9 - y 2 16 = 1 option 1