Key Concepts and Formulas

• Slope of the Normal: If the slope of the tangent to a curve at a point is $m_t$, then the slope of the normal at that point is $m_n = -\frac{1}{m_t}$, provided $m_t \neq 0$.
• Angle Between Two Lines: If $\beta$ is the angle between two lines with slopes $m_1$ and $m_2$, then $\tan \beta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
• Trigonometric Identities: $\sin^2 \theta + \cos^2 \theta = 1$, $\tan \theta = \frac{\sin \theta}{\cos \theta}$, $\cot \theta = \frac{\cos \theta}{\sin \theta}$, $\sin 2\theta = 2 \sin \theta \cos \theta$, $\tan \theta \cot \theta = 1$.

Step-by-Step Solution

1. Implicit Differentiation to Find the Tangent Slope:

We are given the equation of the ellipse: $x^2 + 3y^2 = 9$. We need to find the slope of the tangent, so we differentiate implicitly with respect to $x$: $\frac{d}{dx}(x^2 + 3y^2) = \frac{d}{dx}(9)$ $2x + 6y \frac{dy}{dx} = 0$ Now, solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = -\frac{2x}{6y} = -\frac{x}{3y}$ This represents the slope of the tangent at any point $(x, y)$ on the ellipse.

2. Find the Slope of the Normal:

The slope of the normal ($m_n$) is the negative reciprocal of the slope of the tangent. Thus, $m_n = -\frac{1}{\frac{dy}{dx}} = -\frac{1}{-\frac{x}{3y}} = \frac{3y}{x}$ This formula gives the slope of the normal at any point $(x, y)$ on the ellipse.

3. Calculate Normal Slopes at Given Points:

We have two points: $P_1 = (3 \cos \theta, \sqrt{3} \sin \theta)$ and $P_2 = (-3 \sin \theta, \sqrt{3} \cos \theta)$.

• At point $P_1(3 \cos \theta, \sqrt{3} \sin \theta)$: Let $m_1$ be the slope of the normal at $P_1$. Substituting $x = 3 \cos \theta$ and $y = \sqrt{3} \sin \theta$ into $m_n = \frac{3y}{x}$, we get: $m_1 = \frac{3(\sqrt{3} \sin \theta)}{3 \cos \theta} = \frac{\sqrt{3} \sin \theta}{\cos \theta} = \sqrt{3} \tan \theta$

• At point $P_2(-3 \sin \theta, \sqrt{3} \cos \theta)$: Let $m_2$ be the slope of the normal at $P_2$. Substituting $x = -3 \sin \theta$ and $y = \sqrt{3} \cos \theta$ into $m_n = \frac{3y}{x}$, we get: $m_2 = \frac{3(\sqrt{3} \cos \theta)}{-3 \sin \theta} = -\frac{\sqrt{3} \cos \theta}{\sin \theta} = -\sqrt{3} \cot \theta$

4. Apply the Angle Between Lines Formula:

Let $\beta$ be the angle between the two normals. Using the formula $\tan \beta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$, we first find $m_1 m_2$: $m_1 m_2 = (\sqrt{3} \tan \theta)(-\sqrt{3} \cot \theta) = -3 \tan \theta \cot \theta = -3$ Now, substitute $m_1$, $m_2$, and $m_1 m_2$ into the formula for $\tan \beta$: $\tan \beta = \left| \frac{\sqrt{3} \tan \theta - (-\sqrt{3} \cot \theta)}{1 + (-3)} \right| = \left| \frac{\sqrt{3} (\tan \theta + \cot \theta)}{-2} \right|$ Since $\theta \in (0, \frac{\pi}{2})$, $\tan \theta > 0$ and $\cot \theta > 0$, so $\tan \theta + \cot \theta > 0$. Thus, $\tan \beta = \frac{\sqrt{3}}{2} (\tan \theta + \cot \theta)$

5. Simplify Using Trigonometric Identities:

We simplify the term $(\tan \theta + \cot \theta)$: $\tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta}$ Substitute this back into the expression for $\tan \beta$: $\tan \beta = \frac{\sqrt{3}}{2} \cdot \frac{1}{\sin \theta \cos \theta}$ Using the identity $\sin 2\theta = 2 \sin \theta \cos \theta$, we have $\sin \theta \cos \theta = \frac{\sin 2\theta}{2}$. $\tan \beta = \frac{\sqrt{3}}{2} \cdot \frac{1}{\frac{\sin 2\theta}{2}} = \frac{\sqrt{3}}{2} \cdot \frac{2}{\sin 2\theta} = \frac{\sqrt{3}}{\sin 2\theta}$

6. Calculate the Required Expression:

We need to find the value of $\frac{2 \cot \beta}{\sin 2\theta}$. Since $\tan \beta = \frac{\sqrt{3}}{\sin 2\theta}$, we have $\cot \beta = \frac{\sin 2\theta}{\sqrt{3}}$. Substituting this into the expression: $\frac{2 \cot \beta}{\sin 2\theta} = \frac{2 \cdot \frac{\sin 2\theta}{\sqrt{3}}}{\sin 2\theta} = \frac{2}{\sqrt{3}}$

Common Mistakes & Tips

• Be careful when finding the slope of the normal. Remember it's the negative reciprocal of the tangent's slope.
• Pay attention to the given range of $\theta$ to avoid incorrect simplification of absolute values.
• Remember trigonometric identities, especially $\sin 2\theta = 2 \sin \theta \cos \theta$, as they are frequently used in these types of problems.

Summary

We found the slopes of the normals at the given points using implicit differentiation and the negative reciprocal relationship. We then used the formula for the angle between two lines and trigonometric identities to simplify the expression and find the value of $\frac{2 \cot \beta}{\sin 2\theta}$. The final answer is $\frac{2}{\sqrt{3}}$.

Final Answer: The final answer is $\boxed{\frac{2}{\sqrt{3}}}$, which corresponds to option (A).