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
Start by rearranging the equation to isolate the constant on one side:
\[ 3x^2 + 4y^2 + 12x + 16y = 8 \]

### Step 2: Group the x and y terms
Group the \(x\) terms and the \(y\) terms:
\[ 3(x^2 + 4x) + 4(y^2 + 4y) = 8 \]

### Step 3: Complete the square for x and y
Now, we will complete the square for both \(x\) and \(y\).

For \(x\):
\[ x^2 + 4x \rightarrow (x + 2)^2 - 4 \]
For \(y\):
\[ y^2 + 4y \rightarrow (y + 2)^2 - 4 \]

### Step 4: Substitute back into the equation
Substituting these completed squares back into the equation gives:
\[ 3((x + 2)^2 - 4) + 4((y + 2)^2 - 4) = 8 \]
This simplifies to:
\[ 3(x + 2)^2 - 12 + 4(y + 2)^2 - 16 = 8 \]
\[ 3(x + 2)^2 + 4(y + 2)^2 - 28 = 8 \]
\[ 3(x + 2)^2 + 4(y + 2)^2 = 36 \]

### Step 5: Divide by 36 to get the standard form
Dividing the entire equation by 36:
\[ \frac{3(x + 2)^2}{36} + \frac{4(y + 2)^2}{36} = 1 \]
This simplifies to:
\[ \frac{(x + 2)^2}{12} + \frac{(y + 2)^2}{9} = 1 \]

### Step 6: Identify a and b
From the standard form of the ellipse \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), we identify:
- \(a^2 = 12\) (thus \(a = 2\sqrt{3}\))
- \(b^2 = 9\) (thus \(b = 3\))

### Step 7: Calculate the eccentricity
The eccentricity \(e\) of the ellipse is given by:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{9}{12}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \]

### Step 8: Find the coordinates of the foci
The coordinates of the foci for an ellipse in standard form are given by \((h \pm ae, k)\). Here, \(h = -2\), \(k = -2\), \(a = 2\sqrt{3}\), and \(e = \frac{1}{2}\):
\[ ae = 2\sqrt{3} \cdot \frac{1}{2} = \sqrt{3} \]
Thus, the coordinates of the foci are:
\[ (-2 \pm \sqrt{3}, -2) \]
This gives us the final coordinates of the foci:
\[ (-2 + \sqrt{3}, -2) \quad \text{and} \quad (-2 - \sqrt{3}, -2) \]

### Final Answer:
The coordinates of the foci are \((-2 + \sqrt{3}, -2)\) and \((-2 - \sqrt{3}, -2)\).