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 cos 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 step by step, we will follow the outlined approach based on the information provided in the video transcript.

### Step 1: Understand the Equation of the Ellipse
The given ellipse is represented by the equation:
\[ x^2 + 3y^2 = 9 \]

### Step 2: Differentiate the Ellipse Equation
To find the slope of the tangent line at any point on the ellipse, we differentiate the equation implicitly:
\[ 2x + 3 \cdot 2y \cdot \frac{dy}{dx} = 0 \]
This simplifies to:
\[ \frac{dy}{dx} = -\frac{2x}{6y} = -\frac{x}{3y} \]

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

### Step 4: Calculate the Slope of Normals at Given Points
The points given are:
1. \( (3 \cos \theta, \sqrt{3} \sin \theta) \)
2. \( (-3 \sin \theta, \sqrt{3} \cos \theta) \)

**For the first point \( (3 \cos \theta, \sqrt{3} \sin \theta) \):**
\[ M_1 = \frac{3(\sqrt{3} \sin \theta)}{3 \cos \theta} = \frac{\sqrt{3} \sin \theta}{\cos \theta} = \sqrt{3} \tan \theta \]

**For the second 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 the Angle Between the Normals
The angle \( \beta \) between the two normals can be found using the formula:
\[ \tan \beta = \left| \frac{M_1 - M_2}{1 + M_1 M_2} \right| \]
Substituting \( M_1 \) and \( M_2 \):
\[ \tan \beta = \left| \frac{\sqrt{3} \tan \theta + \sqrt{3} \cot \theta}{1 - \sqrt{3} \tan \theta \cdot (-\sqrt{3} \cot \theta)} \right| \]
This simplifies to:
\[ \tan \beta = \left| \frac{\sqrt{3} (\tan \theta + \cot \theta)}{1 + 3} \right| = \left| \frac{\sqrt{3} (\tan \theta + \cot \theta)}{4} \right| \]

### Step 6: Simplify \( \tan \beta \)
Using the identity \( \tan \theta + \cot \theta = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} \):
\[ \tan \beta = \left| \frac{\sqrt{3}}{4 \sin \theta \cos \theta} \right| = \frac{\sqrt{3}}{4} \cdot \frac{1}{\frac{1}{2} \sin 2\theta} = \frac{\sqrt{3}}{2 \sin 2\theta} \]

### Step 7: Find \( 2 \cot \beta \)
Since \( \cot \beta = \frac{1}{\tan \beta} \):
\[ \cot \beta = \frac{2 \sin 2\theta}{\sqrt{3}} \]
Thus,
\[ 2 \cot \beta = \frac{4 \sin 2\theta}{\sqrt{3}} \]

### Step 8: Calculate \( \frac{2 \cot \beta}{\sin 2\theta} \)
Now, we compute:
\[ \frac{2 \cot \beta}{\sin 2\theta} = \frac{\frac{4 \sin 2\theta}{\sqrt{3}}}{\sin 2\theta} = \frac{4}{\sqrt{3}} \]

### Final Answer
Thus, the final result is:
\[ \frac{2 \cot \beta}{\sin 2\theta} = \frac{4}{\sqrt{3}} \]