If PQ is a double ordinate of the hyperbola such that OPQ is an equilateral triangle, being the centre of the hyperbola, then eccentricity e of the hyperbola satisfies

Options

A. `e=(2)/sqrt(3)`
B. `e=sqrt(3)/(2)`
C. `egt(2)/sqrt(3)`
D. `1ltelt(2)/sqrt(3)`

Answer

To solve the problem, we need to find the eccentricity \( e \) of the hyperbola given that \( OPQ \) forms an equilateral triangle with the center \( O \) of the hyperbola.

### Step-by-Step Solution:

1. **Understand the Hyperbola Equation**:
The standard equation of a hyperbola is given by:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
where \( a \) and \( b \) are the semi-major and semi-minor axes respectively.

2. **Identify the Coordinates of Points P and Q**:
Let the coordinates of points \( P \) and \( Q \) be:
\[ P = (a \sec \theta, b \tan \theta) \quad \text{and} \quad Q = (a \sec \theta, -b \tan \theta) \]
Here, \( OPQ \) is a double ordinate, meaning both points have the same x-coordinate.

3. **Properties of the Equilateral Triangle**:
Since \( OPQ \) is an equilateral triangle, the angles \( \angle OPQ \) and \( \angle OQP \) are each \( 60^\circ \). By symmetry, we can consider the angle \( \angle OPQ \) to be \( 30^\circ \).

4. **Calculate the Slope**:
The slope of line \( OP \) is given by:
\[ \text{slope} = \frac{b \tan \theta}{a \sec \theta} \]
Since \( \tan 30^\circ = \frac{1}{\sqrt{3}} \), we have:
\[ \frac{b \tan \theta}{a \sec \theta} = \frac{1}{\sqrt{3}} \]

5. **Rearranging the Slope Equation**:
Rearranging the above equation gives:
\[ b \tan \theta = \frac{a \sec \theta}{\sqrt{3}} \]
Squaring both sides leads to:
\[ b^2 \tan^2 \theta = \frac{a^2 \sec^2 \theta}{3} \]

6. **Using Trigonometric Identity**:
We know that \( \sec^2 \theta = 1 + \tan^2 \theta \). Substituting this into the equation gives:
\[ 3b^2 \tan^2 \theta = a^2 (1 + \tan^2 \theta) \]

7. **Rearranging the Equation**:
Rearranging this results in:
\[ (3b^2 - a^2) \tan^2 \theta = a^2 \]
Hence,
\[ \tan^2 \theta = \frac{a^2}{3b^2 - a^2} \]

8. **Relating \( a \), \( b \), and \( e \)**:
The eccentricity \( e \) of the hyperbola is given by:
\[ e^2 = 1 + \frac{b^2}{a^2} \]
From the previous step, we can derive that:
\[ 3b^2 \geq a^2 \implies 3e^2 - 1 \geq 1 \implies 3e^2 \geq 2 \implies e^2 \geq \frac{2}{3} \]

9. **Final Calculation**:
Solving for \( e \):
\[ e^2 \geq \frac{4}{3} \implies e \geq \frac{2}{\sqrt{3}} \]

### Conclusion:
Thus, the eccentricity \( e \) of the hyperbola satisfies:
\[ e \geq \frac{2}{\sqrt{3}} \]