If `beta` is one of the angles between the normals to the ellipse, `x^2+3y^2=9` at the points `(3 cos theta, sqrt(3) sin theta)" and "(-3 sin theta, sqrt(3) cos theta), theta in (0,(pi)/(2))`, then `(2 cot beta)/(sin 2 theta)` is equal to:

Options

A. `(1)/(sqrt(3))`
B. `(2)/(sqrt(3))`
C. `sqrt(2)`
D. `(sqrt(3))/(4)`

Answer

To solve the problem, we need to find the value of \( \frac{2 \cot \beta}{\sin 2\theta} \) where \( \beta \) is the angle between the normals to the ellipse \( x^2 + 3y^2 = 9 \) at the given points. Let's go through the solution step by step.

### Step 1: Write the equation of the ellipse
The equation of the ellipse is given as:
\[ x^2 + 3y^2 = 9 \]

### Step 2: Differentiate the equation of the ellipse
We differentiate both sides of the equation with respect to \( x \):
\[ \frac{d}{dx}(x^2) + \frac{d}{dx}(3y^2) = \frac{d}{dx}(9) \]
This gives:
\[ 2x + 3 \cdot 2y \frac{dy}{dx} = 0 \]
Simplifying, we find:
\[ 2x + 6y \frac{dy}{dx} = 0 \]
Thus,
\[ \frac{dy}{dx} = -\frac{2x}{6y} = -\frac{x}{3y} \]

### Step 3: Find the slope of the normal
The slope of the normal is the negative reciprocal of the slope of the tangent:
\[ \text{slope of normal} = -\frac{1}{\frac{dy}{dx}} = \frac{3y}{x} \]

### Step 4: Calculate the slopes at the given points
1. For the point \( (3 \cos \theta, \sqrt{3} \sin \theta) \):
\[ m_1 = \frac{3(\sqrt{3} \sin \theta)}{3 \cos \theta} = \sqrt{3} \tan \theta \]

2. For the point \( (-3 \sin \theta, \sqrt{3} \cos \theta) \):
\[ m_2 = \frac{3(\sqrt{3} \cos \theta)}{-3 \sin \theta} = -\sqrt{3} \cot \theta \]

### Step 5: Find \( \tan \beta \)
Using the formula for the tangent of the angle between two lines:
\[ \tan \beta = \frac{m_1 - m_2}{1 + m_1 m_2} \]
Substituting \( m_1 \) and \( m_2 \):
\[ \tan \beta = \frac{\sqrt{3} \tan \theta + \sqrt{3} \cot \theta}{1 - (\sqrt{3} \tan \theta)(-\sqrt{3} \cot \theta)} \]
This simplifies to:
\[ \tan \beta = \frac{\sqrt{3} (\tan \theta + \cot \theta)}{1 + 3} \]
\[ \tan \beta = \frac{\sqrt{3} (\tan \theta + \cot \theta)}{4} \]

### Step 6: Express \( \cot \beta \)
Since \( \cot \beta = \frac{1}{\tan \beta} \):
\[ \cot \beta = \frac{4}{\sqrt{3} (\tan \theta + \cot \theta)} \]

### Step 7: Find \( \frac{2 \cot \beta}{\sin 2\theta} \)
We know that \( \sin 2\theta = 2 \sin \theta \cos \theta \). Therefore:
\[ \frac{2 \cot \beta}{\sin 2\theta} = \frac{2 \cdot \frac{4}{\sqrt{3} (\tan \theta + \cot \theta)}}{2 \sin \theta \cos \theta} \]
This simplifies to:
\[ \frac{8}{\sqrt{3} (\tan \theta + \cot \theta) \cdot 2 \sin \theta \cos \theta} \]
Using the identity \( \tan \theta + \cot \theta = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} \):
\[ \frac{2 \cot \beta}{\sin 2\theta} = \frac{8}{\sqrt{3} \cdot 2} = \frac{4}{\sqrt{3}} \]

### Final Result
Thus, the value of \( \frac{2 \cot \beta}{\sin 2\theta} \) is:
\[ \frac{8}{\sqrt{3}} \]