Show that the equation `3x^(2)+4y^(2)-12x-8y+4=0` represents an ellipse. Find the focus also

Answer

To show that the equation \(3x^2 + 4y^2 - 12x - 8y + 4 = 0\) represents an ellipse and to find its focus, we will follow these steps:

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

### Step 2: Grouping terms
Group the \(x\) terms and the \(y\) terms:
\[ 3(x^2 - 4x) + 4(y^2 - 2y) + 4 = 0 \]

### Step 3: Completing the square for \(x\)
To complete the square for \(x^2 - 4x\):
\[ x^2 - 4x = (x - 2)^2 - 4 \]
Substituting back gives:
\[ 3((x - 2)^2 - 4) + 4(y^2 - 2y) + 4 = 0 \]
This simplifies to:
\[ 3(x - 2)^2 - 12 + 4(y^2 - 2y) + 4 = 0 \]
\[ 3(x - 2)^2 + 4(y^2 - 2y) - 8 = 0 \]

### Step 4: Completing the square for \(y\)
Now, complete the square for \(y^2 - 2y\):
\[ y^2 - 2y = (y - 1)^2 - 1 \]
Substituting back gives:
\[ 3(x - 2)^2 + 4((y - 1)^2 - 1) - 8 = 0 \]
This simplifies to:
\[ 3(x - 2)^2 + 4(y - 1)^2 - 4 - 8 = 0 \]
\[ 3(x - 2)^2 + 4(y - 1)^2 - 12 = 0 \]
\[ 3(x - 2)^2 + 4(y - 1)^2 = 12 \]

### Step 5: Dividing by 12
Now, divide the entire equation by 12:
\[ \frac{3(x - 2)^2}{12} + \frac{4(y - 1)^2}{12} = 1 \]
This simplifies to:
\[ \frac{(x - 2)^2}{4} + \frac{(y - 1)^2}{3} = 1 \]

### Step 6: Identifying the ellipse
Now we have the equation in the standard form of an ellipse:
\[ \frac{(x - 2)^2}{4} + \frac{(y - 1)^2}{3} = 1 \]
Here, \(a^2 = 4\) and \(b^2 = 3\). Since \(a^2 > b^2\), this confirms that it is indeed an ellipse.

### Step 7: Finding the foci
The distance of the foci from the center is given by:
\[ c = \sqrt{a^2 - b^2} = \sqrt{4 - 3} = \sqrt{1} = 1 \]
The center of the ellipse is at \((2, 1)\). Therefore, the foci are located at:
\[ (2 \pm c, 1) = (2 \pm 1, 1) = (3, 1) \text{ and } (1, 1) \]

### Final Result
Thus, the foci of the ellipse are \((3, 1)\) and \((1, 1)\).