The tangents at the extremities of the latus rectum of the ellipse `3x^(2)+4y^(2)=12` form a rhombus PQRS. Area (in sq. units) of the rhombus PQRS outside the ellipse is equal to

Options

A. `8-2sqrt3pi`
B. `12-2sqrt3pi`
C. `14-pi`
D. `16-2sqrt3pi`

Answer

To solve the problem, we need to find the area of the rhombus formed by the tangents at the extremities of the latus rectum of the ellipse given by the equation \(3x^2 + 4y^2 = 12\), and then determine the area of the rhombus outside the ellipse.

### Step 1: Convert the ellipse equation to standard form
We start with the equation of the ellipse:
\[ 3x^2 + 4y^2 = 12 \]
To convert it to standard form, we divide the entire equation by 12:
\[ \frac{3x^2}{12} + \frac{4y^2}{12} = 1 \implies \frac{x^2}{4} + \frac{y^2}{3} = 1 \]
Here, we identify \(a^2 = 4\) and \(b^2 = 3\).

### Step 2: Identify the values of \(a\) and \(b\)
From the above, we find:
\[ a = \sqrt{4} = 2 \quad \text{and} \quad b = \sqrt{3} \]

### Step 3: Calculate the eccentricity \(e\)
The eccentricity \(e\) of the ellipse is given by:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{3}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \]

### Step 4: Calculate the area of the rhombus PQRS
The area \(A\) of the rhombus formed by the tangents at the extremities of the latus rectum is given by the formula:
\[ A = \frac{2a^2}{e} \]
Substituting the values of \(a^2\) and \(e\):
\[ A = \frac{2 \times 4}{\frac{1}{2}} = \frac{8}{\frac{1}{2}} = 16 \]

### Step 5: Calculate the area of the ellipse
The area \(A_e\) of the ellipse is given by:
\[ A_e = \pi a b = \pi \times 2 \times \sqrt{3} = 2\sqrt{3}\pi \]

### Step 6: Calculate the area outside the ellipse but inside the rhombus
The area outside the ellipse and enclosed by the rhombus is:
\[ A_0 = A - A_e = 16 - 2\sqrt{3}\pi \]

### Final Answer
Thus, the area of the rhombus PQRS outside the ellipse is:
\[ \boxed{16 - 2\sqrt{3}\pi} \]