Step-by-step Solution:

1. Double Ordinate \( PQ \): A double ordinate is a chord perpendicular to the transverse axis. Let \( P \) and \( Q \) be points on the hyperbola with coordinates \( (x, y) \) and \( (x, -y) \) respectively. 2. Equilateral Triangle Condition: For \( OPQ \) to be equilateral, the distances \( OP \), \( OQ \), and \( PQ \) must be equal. 3. Calculate Distances: - Distance \( OP \): \[ OP = \sqrt{x^2 + y^2} \] - Distance \( PQ \): \[ PQ = 2y \] Since \( OP = PQ \): \[ \sqrt{x^2 + y^2} = 2y \] Square both sides: \[ x^2 + y^2 = 4y^2 \Rightarrow x^2 = 3y^2 \] 4. Substitute into Hyperbola Equation: Since \( P \) lies on the hyperbola: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \] Substitute \( x^2 = 3y^2 \): \[ \frac{3y^2}{a^2} - \frac{y^2}{b^2} = 1 \] Factor out \( y^2 \): \[ y^2 \left( \frac{3}{a^2} - \frac{1}{b^2} \right) = 1 \] Solve for \( y^2 \): \[ y^2 = \frac{1}{\frac{3}{a^2} - \frac{1}{b^2}} = \frac{a^2 b^2}{3b^2 - a^2} \] 5. Ensure \( y^2 \) is Positive: For \( y^2 \) to be positive, the denominator must be positive: \[ 3b^2 - a^2 > 0 \Rightarrow 3b^2 > a^2 \Rightarrow b^2 > \frac{a^2}{3} \] Therefore, the condition for \( OPQ \) to be an equilateral triangle is: \[ \boxed{b^2 > \frac{a^2}{3}} \]