Let the maximum area of the triangle that can be inscribed in the ellipse `(x^(2))/(a^(2)) +(y^(2))/4 = 1, a gt 2 ` , having one of its vertices at one end of the major axis of the ellipse and one of its parallel to the y - axis , be `6sqrt(3)` . Then the eccentricity of the ellipse is :

Options

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

Answer

To solve the problem, we need to find the eccentricity of the ellipse given that the maximum area of a triangle inscribed in the ellipse is \(6\sqrt{3}\). The ellipse is defined by the equation:

\[ \frac{x^2}{a^2} + \frac{y^2}{4} = 1 \]

where \(a > 2\).

### Step 1: Understanding the maximum area of the triangle inscribed in the ellipse

The formula for the maximum area \(A\) of a triangle inscribed in an ellipse is given by:

\[ A = \frac{3\sqrt{3}}{4} \cdot a \cdot b \]

where \(a\) is the semi-major axis and \(b\) is the semi-minor axis of the ellipse. In our case, \(b = 2\) (since the coefficient of \(y^2\) is 4, which means \(b^2 = 4\)).

### Step 2: Setting up the equation for the area

Given that the maximum area of the triangle is \(6\sqrt{3}\), we can set up the equation:

\[ \frac{3\sqrt{3}}{4} \cdot a \cdot 2 = 6\sqrt{3} \]

### Step 3: Simplifying the equation

Now, we simplify the equation:

\[ \frac{3\sqrt{3}}{4} \cdot 2a = 6\sqrt{3} \]

This simplifies to:

\[ \frac{3\sqrt{3}}{2} a = 6\sqrt{3} \]

### Step 4: Solving for \(a\)

Now, we can divide both sides by \(3\sqrt{3}\):

\[ \frac{1}{2} a = 2 \]

Multiplying both sides by 2 gives:

\[ a = 4 \]

### Step 5: Finding the eccentricity of the ellipse

The eccentricity \(e\) of the ellipse is given by the formula:

\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]

Substituting \(b = 2\) and \(a = 4\):

\[ e = \sqrt{1 - \frac{2^2}{4^2}} = \sqrt{1 - \frac{4}{16}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]

### Conclusion

Thus, the eccentricity of the ellipse is:

\[ \frac{\sqrt{3}}{2} \]