The co-ordinates of the foci of the ellipse `3x ^(2) + 4y ^(2) -12 x -8y +4=0,` are

Options

A. `(1,2), (3,4)`
B. `(1,4), (3,1)`
C. `(1,1) , (3,1)`
D. `(2,3), (5,4)`

Answer

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

### Step 1: Rearranging the Equation
First, we rearrange the equation to group the \(x\) and \(y\) terms:
\[ 3x^2 - 12x + 4y^2 - 8y + 4 = 0 \]

### Step 2: Completing the Square
Next, we complete the square for the \(x\) and \(y\) terms.

**For \(x\):**
\[ 3(x^2 - 4x) = 3((x - 2)^2 - 4) = 3(x - 2)^2 - 12 \]

**For \(y\):**
\[ 4(y^2 - 2y) = 4((y - 1)^2 - 1) = 4(y - 1)^2 - 4 \]

### Step 3: Substitute Back
Substituting these back into the equation gives:
\[ 3((x - 2)^2 - 4) + 4((y - 1)^2 - 1) + 4 = 0 \]
This simplifies to:
\[ 3(x - 2)^2 - 12 + 4(y - 1)^2 - 4 + 4 = 0 \]
\[ 3(x - 2)^2 + 4(y - 1)^2 - 12 = 0 \]
\[ 3(x - 2)^2 + 4(y - 1)^2 = 12 \]

### Step 4: Divide by 12
Dividing the entire equation by 12 gives:
\[ \frac{(x - 2)^2}{4} + \frac{(y - 1)^2}{3} = 1 \]

### Step 5: Identify Parameters
From the standard form of the ellipse \(\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1\), we can identify:
- \(h = 2\), \(k = 1\)
- \(a^2 = 4 \Rightarrow a = 2\)
- \(b^2 = 3 \Rightarrow b = \sqrt{3}\)

### Step 6: Calculate the Eccentricity
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 7: Find the Foci
The coordinates of the foci for an ellipse where \(a > b\) are given by:
\[ (h \pm ae, k) \]
Substituting the values we found:
\[ h = 2, \quad k = 1, \quad a = 2, \quad e = \frac{1}{2} \]
Calculating the foci:
\[ F_1 = (2 + 2 \cdot \frac{1}{2}, 1) = (3, 1) \]
\[ F_2 = (2 - 2 \cdot \frac{1}{2}, 1) = (1, 1) \]

### Final Answer
Thus, the coordinates of the foci are:
\[ (3, 1) \text{ and } (1, 1) \]