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JEE Main Mathematics Hyperbola 2026 JEE Main 2026 (23 January Shift 2)

JEE Main Mathematics Question (2026) — Solution

Question

Let PQ be a chord of the hyperbola x^ 2 4 - y^ 2 b^ 2 =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

  1. A. 2 3
  2. B. 11 5
  3. C. 9 5
  4. D. 8 3 5

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 = (x_0, y_0), Q = (x_0, -y_0). For equilateral OPQ: OP = PQ x_0^2 + y_0^2 = 4y_0^2 x_0^2 = 3y_0^2. From the hyperbola: 3y_0^2 4 - y_0^2 8 = 1 5y_0^2 8 = 1 y_0^2 = 8 5 . Area = 3 4 (2y_0)^2 = 3 y_0^2 = 8 3 5 .

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