Question

$PQ$ is a double ordinate of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ such that $\triangle OPQ$ is an equilateral triangle, with $O$ being the center of the hyperbola. Then the eccentricity $e$ of the hyperbola satisfies

• A. 1<e <2/ √3
• B. e= 2/ √3
• C. e= 2 √3
• D. e>2/ √3

Answer 1

The Correct Option is D

We are given the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) with the center at \( O(0,0) \). The points \( P \) and \( Q \) are on a double ordinate of the hyperbola, and the triangle \( \triangle OPQ \) is equilateral.

Step 1: Geometry of the Hyperbola and Double Ordinate
A double ordinate is a line segment perpendicular to the transverse axis (the x-axis) that passes through two points on the hyperbola. Let the points \( P \) and \( Q \) have coordinates \( P(x_1, y_1) \) and \( Q(x_1, -y_1) \), as they lie on the double ordinate. For the points to lie on the hyperbola, the coordinates must satisfy the hyperbola equation: \[ \frac{x_1^2}{a^2} - \frac{y_1^2}{b^2} = 1 \] which gives the relationship between \( x_1 \) and \( y_1 \). 

Step 2: Equilateral Triangle Condition
Since \( \triangle OPQ \) is an equilateral triangle, the distances \( OP \), \( OQ \), and \( PQ \) must all be equal. The distance between \( O \) and \( P \) is: \[ OP = \sqrt{x_1^2 + y_1^2} \] Similarly, the distance between \( O \) and \( Q \) is: \[ OQ = \sqrt{x_1^2 + y_1^2} \] The distance between \( P \) and \( Q \) is: \[ PQ = 2y_1 \] For the triangle to be equilateral, we must have: \[ OP = PQ \] \[ \sqrt{x_1^2 + y_1^2} = 2y_1 \] Squaring both sides: \[ x_1^2 + y_1^2 = 4y_1^2 \] \[ x_1^2 = 3y_1^2 \] 

Step 3: Using the Equation of the Hyperbola
From the equation of the hyperbola, we have: \[ \frac{x_1^2}{a^2} - \frac{y_1^2}{b^2} = 1 \] Substitute \( x_1^2 = 3y_1^2 \) into this equation: \[ \frac{3y_1^2}{a^2} - \frac{y_1^2}{b^2} = 1 \] Factor out \( y_1^2 \): \[ y_1^2 \left( \frac{3}{a^2} - \frac{1}{b^2} \right) = 1 \] Solving for \( y_1^2 \): \[ y_1^2 = \frac{1}{\frac{3}{a^2} - \frac{1}{b^2}} \] 

Step 4: Finding the Eccentricity
The eccentricity \( e \) of a hyperbola is given by: \[ e = \sqrt{1 + \frac{b^2}{a^2}} \] Substituting values and simplifying gives: \[ e > \frac{2}{\sqrt{3}} \]

Answer:

\[ \boxed{e > \frac{2}{\sqrt{3}}} \]