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 and 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 PQRS 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 calculate the area outside the ellipse.

### Step 1: Rewrite the equation of the ellipse in standard form
The given equation of the ellipse is:
\[ 3x^2 + 4y^2 = 12 \]
Dividing the entire equation by 12, we get:
\[ \frac{x^2}{4} + \frac{y^2}{3} = 1 \]
This shows that \(a^2 = 4\) and \(b^2 = 3\), where \(a = 2\) and \(b = \sqrt{3}\).

### Step 2: Find the eccentricity of the ellipse
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 3: Determine the coordinates of the extremities of the latus rectum
The coordinates of the extremities of the latus rectum are given by:
\[ \left( ae, \frac{b^2}{a} \right) \text{ and } \left( -ae, \frac{b^2}{a} \right) \]
Calculating these coordinates:
- \(ae = 2 \cdot \frac{1}{2} = 1\)
- \(\frac{b^2}{a} = \frac{3}{2}\)

Thus, the extremities of the latus rectum are:
\[ \left( 1, \frac{3}{2} \right) \text{ and } \left( -1, \frac{3}{2} \right) \]

### Step 4: Find the equations of the tangents at these points
The equation of the tangent to the ellipse at a point \((x_1, y_1)\) is given by:
\[ \frac{x x_1}{a^2} + \frac{y y_1}{b^2} = 1 \]
For the point \((1, \frac{3}{2})\):
\[ \frac{x \cdot 1}{4} + \frac{y \cdot \frac{3}{2}}{3} = 1 \implies \frac{x}{4} + \frac{y}{2} = 1 \]
Multiplying through by 4 gives:
\[ x + 2y = 4 \quad \text{(Tangent 1)} \]

For the point \((-1, \frac{3}{2})\):
\[ \frac{x \cdot (-1)}{4} + \frac{y \cdot \frac{3}{2}}{3} = 1 \implies -\frac{x}{4} + \frac{y}{2} = 1 \]
Multiplying through by 4 gives:
\[ -x + 2y = 4 \quad \text{(Tangent 2)} \]

### Step 5: Find the points of intersection of the tangents
Now we solve the equations of the tangents:
1. \(x + 2y = 4\)
2. \(-x + 2y = 4\)

Adding these two equations:
\[ (x + 2y) + (-x + 2y) = 4 + 4 \implies 4y = 8 \implies y = 2 \]
Substituting \(y = 2\) into \(x + 2y = 4\):
\[ x + 4 = 4 \implies x = 0 \]
Thus, the intersection point is \((0, 2)\).

### Step 6: Calculate the area of the rhombus
The vertices of the rhombus PQRS are:
- \(P(4, 0)\)
- \(Q(0, 2)\)
- \(R(-4, 0)\)
- \(S(0, -2)\)

The area of the rhombus can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times d_1 \times d_2 \]
where \(d_1\) and \(d_2\) are the lengths of the diagonals. The lengths of the diagonals are:
- \(d_1 = 4 - (-4) = 8\)
- \(d_2 = 2 - (-2) = 4\)

Thus, the area of the rhombus is:
\[ \text{Area} = \frac{1}{2} \times 8 \times 4 = 16 \text{ square units} \]

### Step 7: Calculate the area of the ellipse
The area of the ellipse is given by:
\[ \text{Area}_{\text{ellipse}} = \pi \cdot a \cdot b = \pi \cdot 2 \cdot \sqrt{3} = 2\sqrt{3}\pi \text{ square units} \]

### Step 8: Calculate the area outside the ellipse
The area outside the ellipse but inside the rhombus is:
\[ \text{Area}_{\text{outside}} = \text{Area}_{\text{rhombus}} - \text{Area}_{\text{ellipse}} = 16 - 2\sqrt{3}\pi \]

### Final Answer
The area of the rhombus PQRS outside the ellipse is:
\[
\boxed{16 - 2\sqrt{3}\pi} \text{ square units}