AB is double ordinate of the hyperbola `x^2/a^2-y^2/b^2=1` such that `DeltaAOB`(where 'O' is the origin) is an equilateral triangle, then the eccentricity e of hyperbola satisfies:

Options

A. `e gt (2)/(sqrt(3))`
B. `e = (2)/(sqrt(3))`
C. `e lt (2)/(sqrt(3))`
D. `1 lt e lt (2)/(sqrt(3))`

Answer

To solve the problem, we need to find the eccentricity \( e \) of the hyperbola given that \( AB \) is a double ordinate and \( \Delta AOB \) forms an equilateral triangle. Let's break down the solution step by step.

### Step 1: Understand the Hyperbola and Double Ordinate
The equation of the hyperbola is given by:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
A double ordinate of a hyperbola is a line segment perpendicular to the transverse axis that intersects the hyperbola at two points.

### Step 2: Identify Points A and B
Let the points \( A \) and \( B \) be represented as:
\[ A(a \sec \theta, b \tan \theta) \quad \text{and} \quad B(a \sec \theta, -b \tan \theta) \]
This is because the double ordinate is vertical, and the coordinates are derived from the parametric form of the hyperbola.

### Step 3: Establish the Equilateral Triangle Condition
Since \( \Delta AOB \) is an equilateral triangle, the angle \( AOB \) is \( 60^\circ \). The angle \( AOM \) (where \( M \) is the midpoint of \( AB \)) is \( 30^\circ \).

### Step 4: Calculate the Slope of Line OA
The slope of line \( OA \) can be calculated as:
\[ \text{slope of } OA = \frac{b \tan \theta - 0}{a \sec \theta - 0} = \frac{b \tan \theta}{a \sec \theta} \]
Using the angle \( AOM = 30^\circ \), we have:
\[ \tan 30^\circ = \frac{1}{\sqrt{3}} \]
Thus, we can set up the equation:
\[ \frac{b \tan \theta}{a \sec \theta} = \frac{1}{\sqrt{3}} \]

### Step 5: Rearranging the Equation
From the above equation, we can rearrange to find:
\[ b \tan \theta = \frac{a \sec \theta}{\sqrt{3}} \]
Using the identities \( \sec \theta = \frac{1}{\cos \theta} \) and \( \tan \theta = \frac{\sin \theta}{\cos \theta} \), we can express this as:
\[ b \frac{\sin \theta}{\cos \theta} = \frac{a}{\sqrt{3} \cos \theta} \]
This simplifies to:
\[ b \sin \theta = \frac{a}{\sqrt{3}} \]

### Step 6: Finding the Eccentricity
The eccentricity \( e \) of the hyperbola is given by:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]
From the previous step, we have:
\[ \frac{b}{a} = \frac{1}{\sqrt{3} \sin \theta} \]
Squaring both sides gives:
\[ \frac{b^2}{a^2} = \frac{1}{3 \sin^2 \theta} \]
Substituting this into the eccentricity formula:
\[ e^2 = 1 + \frac{1}{3 \sin^2 \theta} \]

### Step 7: Analyzing the Result
Since \( \sin^2 \theta \) is always between 0 and 1, we can conclude:
\[ e^2 > 1 + \frac{1}{3} = \frac{4}{3} \]
Thus, we have:
\[ e > \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}} \]

### Final Answer
The eccentricity \( e \) of the hyperbola satisfies:
\[ e > \frac{2}{\sqrt{3}} \]