Let a triangle of maximum area be inscribed in the ellipse `x^2/a^2+y^2/4=1`, such that the area is `6sqrt(3)`.find the eccentricity.

Answer

To solve the problem of finding the eccentricity of the ellipse given that a triangle of maximum area is inscribed in it, we will follow these steps:

### Step 1: Understanding the ellipse
The equation of the ellipse is given as:
\[ \frac{x^2}{a^2} + \frac{y^2}{4} = 1 \]
Here, \( a \) is the semi-major axis and \( b = 2 \) is the semi-minor axis.

### Step 2: Area of the largest inscribed triangle
The maximum area of a triangle inscribed in an ellipse can be calculated using the formula:
\[ \text{Area} = \frac{3\sqrt{3}}{4} \cdot a \cdot b \]
For our ellipse, substituting \( b = 2 \):
\[ \text{Area} = \frac{3\sqrt{3}}{4} \cdot a \cdot 2 = \frac{3\sqrt{3}}{2} \cdot a \]

### Step 3: Setting the area equal to the given area
We are given that the area of the triangle is \( 6\sqrt{3} \). Therefore, we set up the equation:
\[ \frac{3\sqrt{3}}{2} \cdot a = 6\sqrt{3} \]

### Step 4: Solving for \( a \)
To solve for \( a \), we can divide both sides by \( \sqrt{3} \):
\[ \frac{3}{2} \cdot a = 6 \]
Now, multiply both sides by \( \frac{2}{3} \):
\[ a = 6 \cdot \frac{2}{3} = 4 \]

### Step 5: Finding the eccentricity
The eccentricity \( e \) of an ellipse is given by the formula:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Substituting the values \( b = 2 \) and \( a = 4 \):
\[ e = \sqrt{1 - \frac{2^2}{4^2}} = \sqrt{1 - \frac{4}{16}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]

### Final Answer
The eccentricity of the ellipse is:
\[ \boxed{\frac{\sqrt{3}}{2}} \]