The co-ordinates of foci of an ellipse `3x^(2)+4y^(2)+12x+16y-8=0` is :

Options

A. `(pm sqrt(3)-2,-2)`
B. `(-2, -2) `
C. `(2-sqrt(3), pm2)`
D. `(2+sqrt(3), -2)`

Explanation

To find the coordinates of the foci of the ellipse given by the equation \(3x^2 + 4y^2 + 12x + 16y - 8 = 0\), we will follow these steps:

### Step 1: Rearrange the equation
We start with the equation:
\[ 3x^2 + 4y^2 + 12x + 16y - 8 = 0 \]
Rearranging gives:
\[ 3x^2 + 12x + 4y^2 + 16y = 8 \]

### Step 2: Complete the square for \(x\) and \(y\)
For the \(x\) terms:
\[ 3(x^2 + 4x) = 3[(x + 2)^2 - 4] = 3(x + 2)^2 - 12 \]
For the \(y\) terms:
\[ 4(y^2 + 4y) = 4[(y + 2)^2 - 4] = 4(y + 2)^2 - 16 \]
Substituting these back into the equation gives:
\[ 3(x + 2)^2 - 12 + 4(y + 2)^2 - 16 = 8 \]
Simplifying this:
\[ 3(x + 2)^2 + 4(y + 2)^2 - 28 = 8 \]
\[ 3(x + 2)^2 + 4(y + 2)^2 = 36 \]

### Step 3: Divide by 36 to get the standard form
Dividing the entire equation by 36 gives:
\[ \frac{(x + 2)^2}{12} + \frac{(y + 2)^2}{9} = 1 \]
This is now in the standard form of an ellipse:
\[ \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 \]
where \(h = -2\), \(k = -2\), \(a^2 = 12\), and \(b^2 = 9\).

### Step 4: Identify \(a\) and \(b\)
From the above, we have:
\[ a = \sqrt{12} = 2\sqrt{3}, \quad b = \sqrt{9} = 3 \]

### Step 5: Calculate the eccentricity \(e\)
The eccentricity \(e\) is given by:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{9}{12}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \]

### Step 6: Find the coordinates of the foci
The foci of the ellipse are located at:
\[ (h \pm ae, k) = (-2 \pm (2\sqrt{3} \cdot \frac{1}{2}), -2) = (-2 \pm \sqrt{3}, -2) \]
Thus, the coordinates of the foci are:
\[ (-2 + \sqrt{3}, -2) \quad \text{and} \quad (-2 - \sqrt{3}, -2) \]

### Final Answer
The coordinates of the foci of the ellipse are:
\[ (-2 + \sqrt{3}, -2) \quad \text{and} \quad (-2 - \sqrt{3}, -2) \]