This solution provides a comprehensive, step-by-step approach to finding the eccentricity of an ellipse by utilizing its geometric properties, parametric representation, and calculus-based optimization techniques to determine the maximum area of an inscribed triangle.

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1. Understanding the Ellipse and its Parameters

The given equation of the ellipse is: $\frac{x^2}{a^2} + \frac{y^2}{4} = 1$

Our first step is to correctly identify the semi-major and semi-minor axes of this ellipse. We compare this to the standard form of an ellipse centered at the origin, which is $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$.

From the given equation, we have $A^2 = a^2$ and $B^2 = 4$. This means the semi-axis lengths are $A = |a|$ and $B = 2$. Since $a$ is typically taken as a positive length, we have $A = a$.

The problem states a crucial condition: $a > 2$.

• Since $a > 2$, it implies $A > B$ (i.e., $a > 2$).
• Why this is important: When $A > B$, the major axis of the ellipse lies along the x-axis, and its length is $2A = 2a$. The ends of the major axis (also called vertices) are $(\pm a, 0)$. The minor axis lies along the y-axis, and its length is $2B = 4$. The ends of the minor axis are $(0, \pm 2)$.

Key Concept: Eccentricity of an Ellipse For an ellipse with a horizontal major axis (i.e., $A > B$), the eccentricity $e$ is given by the formula: $e = \sqrt{1 - \frac{B^2}{A^2}}$ Substituting our values $A=a$ and $B=2$: $e = \sqrt{1 - \frac{2^2}{a^2}} = \sqrt{1 - \frac{4}{a^2}}$ Our ultimate goal is to find the value of $e$, which requires us to first determine the value of $a$.

2. Parametric Representation of Points on the Ellipse

To work with points on the ellipse, especially when dealing with geometric properties and optimization, it's often easiest to use the parametric form. Why use parametric form? It allows us to represent any point on the ellipse using a single parameter (an angle $\theta$), which simplifies differentiation later. For the ellipse $\frac{x^2}{a^2} + \frac{y^2}{B^2} = 1$, a point $(x, y)$ on the ellipse can be written as $(A \cos \theta, B \sin \theta)$. For our ellipse, $(x, y) = (a \cos \theta, 2 \sin \theta)$.

3. Setting Up the Triangle's Vertices

The problem defines specific conditions for the triangle:

• One vertex at an end of the major axis: The ends of the major axis are $(\pm a, 0)$. Due to the symmetry of the ellipse and the problem setup, choosing either $(a, 0)$ or $(-a, 0)$ will lead to the same maximum area. Let's choose the vertex $V = (a, 0)$.

• One of its sides parallel to the y-axis: Let the other two vertices of the triangle be $P$ and $Q$. For the side $PQ$ to be parallel to the y-axis, $P$ and $Q$ must have the same x-coordinate. Why this implies symmetry: Since the ellipse is symmetric with respect to the x-axis, if $P = (x_0, y_0)$ is on the ellipse, then $Q = (x_0, -y_0)$ must also be on the ellipse. This ensures $PQ$ is a vertical chord, parallel to the y-axis, and centered on the x-axis.

Let's use the parametric form for vertex $P$: $P = (a \cos \theta, 2 \sin \theta)$ Then, based on the symmetry for a side parallel to the y-axis, the third vertex $Q$ will be: $Q = (a \cos \theta, -2 \sin \theta)$

Tip: For a non-degenerate triangle (a triangle with non-zero area), we must ensure that $P$, $Q$, and $V$ are distinct. This means $\sin \theta \neq 0$ (so $P$ and $Q$ are not on the x-axis) and $a \cos \theta \neq a$ (so $P$ and $Q$ are not the same point as $V$). We'll typically look for $\theta \in (0, \pi)$ or $\theta \in (\pi, 2\pi)$ for area calculations, as $\sin \theta$ would have a consistent sign.

So, the three vertices of our triangle $VPQ$ are: $V = (a, 0)$ $P = (a \cos \theta, 2 \sin \theta)$ $Q = (a \cos \theta, -2 \sin \theta)$

4. Calculating the Area of the Triangle $VPQ$

We can calculate the area of triangle $VPQ$ using the standard formula: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$

• Base: Let's choose the side $PQ$ as the base. Since $P$ and $Q$ have the same x-coordinate, $PQ$ is a vertical segment. The length of $PQ$ is the difference in their y-coordinates: $\text{Length of base } PQ = |2 \sin \theta - (-2 \sin \theta)| = |4 \sin \theta|$ For area calculations, we usually consider $\theta \in (0, \pi)$, where $\sin \theta > 0$. This ensures a positive base length. So, $PQ = 4 \sin \theta$.

• Height: The height of the triangle is the perpendicular distance from vertex $V(a,0)$ to the line containing the base $PQ$. The line containing $PQ$ is a vertical line with equation $x = a \cos \theta$. The perpendicular distance from $V(a,0)$ to the line $x = a \cos \theta$ is: $\text{Height } h = |a - a \cos \theta|$ Since $\cos \theta \le 1$ for any real $\theta$, we have $1 - \cos \theta \ge 0$. Also, $a > 0$. So, the height $h = a(1 - \cos \theta)$.

Now, substitute these into the area formula: $K(\theta) = \frac{1}{2} \times (4 \sin \theta) \times (a(1 - \cos \theta))$ $K(\theta) = 2a \sin \theta (1 - \cos \theta)$ This expression represents the area of the triangle as a function of $\theta$.

5. Maximizing the Area of the Triangle using Calculus

To find the maximum area, we need to find the value of $\theta$ that maximizes the function $K(\theta)$. We will use differential calculus for this.

First, let's expand $K(\theta)$ to make differentiation easier: $K(\theta) = 2a (\sin \theta - \sin \theta \cos \theta)$ Using the identity $\sin(2\theta) = 2 \sin \theta \cos \theta$, we can write $\sin \theta \cos \theta = \frac{1}{2} \sin(2\theta)$. So, $K(\theta) = 2a \left( \sin \theta - \frac{1}{2} \sin(2\theta) \right)$

Now, differentiate $K(\theta)$ with respect to $\theta$: $\frac{dK}{d\theta} = 2a \left( \frac{d}{d\theta}(\sin \theta) - \frac{1}{2} \frac{d}{d\theta}(\sin(2\theta)) \right)$ $\frac{dK}{d\theta} = 2a \left( \cos \theta - \frac{1}{2} (2 \cos(2\theta)) \right)$ $\frac{dK}{d\theta} = 2a (\cos \theta - \cos(2\theta))$

To find the critical points (where the maximum or minimum area might occur), we set the derivative to zero: $\frac{dK}{d\theta} = 0 \implies 2a (\cos \theta - \cos(2\theta)) = 0$ Since $a > 0$, we must have: $\cos \theta - \cos(2\theta) = 0$ $\cos \theta = \cos(2\theta)$

Solving the trigonometric equation: The general solution for $\cos A = \cos B$ is $A = \pm B + 2n\pi$, where $n$ is an integer. So, $\theta = \pm 2\theta + 2n\pi$.

• Case 1: $\theta = 2\theta + 2n\pi$ $-\theta = 2n\pi \implies \theta = -2n\pi$ If $\theta = 0$ (for $n=0$), then $\sin \theta = 0$. This would mean $P$ and $Q$ coincide with points on the x-axis, making the base $PQ$ zero, and thus the area $K(0) = 0$. This corresponds to a minimum area (a degenerate triangle).

• Case 2: $\theta = -2\theta + 2n\pi$ $3\theta = 2n\pi \implies \theta = \frac{2n\pi}{3}$ For a non-degenerate triangle with positive area, we typically consider $\theta \in (0, \pi)$. If $n=1$, we get $\theta = \frac{2\pi}{3}$. Let's check this value:

  • $\cos(2\pi/3) = -1/2$
  • $\sin(2\pi/3) = \sqrt{3}/2$ This value of $\theta$ leads to distinct vertices and a non-zero area. This is the value that will yield the maximum area. (A second derivative test could confirm this is a maximum, but for common optimization problems like this, the non-degenerate critical point is usually the maximum).

Now, substitute $\theta = \frac{2\pi}{3}$ back into the area formula $K(\theta)$: $K_{max} = 2a \sin\left(\frac{2\pi}{3}\right) \left(1 - \cos\left(\frac{2\pi}{3}\right)\right)$ $K_{max} = 2a \left(\frac{\sqrt{3}}{2}\right) \left(1 - \left(-\frac{1}{2}\right)\right)$ $K_{max} = 2a \left(\frac{\sqrt{3}}{2}\right) \left(1 + \frac{1}{2}\right)$ $K_{max} = 2a \left(\frac{\sqrt{3}}{2}\right) \left(\frac{3}{2}\right)$ $K_{max} = \frac{3\sqrt{3}}{2} a$

6. Determining the Value of 'a'

The problem states that the maximum area of the triangle is $6\sqrt{3}$. We equate our calculated maximum area to this given value: $\frac{3\sqrt{3}}{2} a = 6\sqrt{3}$ To solve for $a$, we can divide both sides by $\sqrt{3}$:

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