JEE Main202623 January 2026Evening ShiftMathematicsHyperbolaActual
Let PQ be a chord of the hyperbola x² 4 - y² b² =1 , perpendicular to the x -axis such that OPQ is an equilateral triangle, O being the centre of the hyperbola. If the eccentricity of the hyperbola is 3 , then the area of the triangle OPQ is
Options
- A2 3
- B11 5
- C9 5
- D8 3 5
Correct answer
D. 8 3 5
Step-by-step solution
Hyperbola x^2 4 - y^2 b^2 = 1 with e = 3 . e^2 = 1 + b^2 4 = 3 b^2 = 8 . Let P = (x₀, y₀) , Q = (x₀, -y₀) . For equilateral OPQ : OP = PQ x₀^2 + y₀^2 = 4y₀^2 x₀^2 = 3y₀^2 . From the hyperbola: 3y₀^2 4 - y₀^2 8 = 1 5y₀^2 8 = 1 y₀^2 = 8 5 . Area = 3 4 (2y₀)^2 = 3 y₀^2 = 8 3 5 .