Centre of the ellipse `3x^(2)+4y^(2)-12x-8y+4=0`

Options

A. (1,1)
B. (2,1)
C. (3,1)
D. None of these

Explanation

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

### Step 1: Rearrange the equation
First, we rearrange the equation to group the \(x\) and \(y\) terms together:
\[ 3x^2 - 12x + 4y^2 - 8y + 4 = 0 \]
We can move the constant to the other side:
\[ 3x^2 - 12x + 4y^2 - 8y = -4 \]

### Step 2: Complete the square for \(x\) terms
Next, we complete the square for the \(x\) terms:
\[ 3(x^2 - 4x) + 4y^2 - 8y = -4 \]
To complete the square for \(x^2 - 4x\), we take half of \(-4\) (which is \(-2\)), square it (getting \(4\)), and add/subtract it inside the parentheses:
\[ 3((x^2 - 4x + 4) - 4) + 4y^2 - 8y = -4 \]
This simplifies to:
\[ 3((x - 2)^2 - 4) + 4y^2 - 8y = -4 \]
Expanding gives:
\[ 3(x - 2)^2 - 12 + 4y^2 - 8y = -4 \]

### Step 3: Complete the square for \(y\) terms
Now, we complete the square for the \(y\) terms:
\[ 4(y^2 - 2y) = 4((y^2 - 2y + 1) - 1) = 4((y - 1)^2 - 1) \]
Substituting this back into the equation gives:
\[ 3(x - 2)^2 - 12 + 4((y - 1)^2 - 1) = -4 \]
This simplifies to:
\[ 3(x - 2)^2 - 12 + 4(y - 1)^2 - 4 = -4 \]
Combining the constants:
\[ 3(x - 2)^2 + 4(y - 1)^2 - 16 = -4 \]
Thus:
\[ 3(x - 2)^2 + 4(y - 1)^2 = 12 \]

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

### Step 5: Identify the center
The standard form of the ellipse is:
\[ \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 \]
From this, we can identify the center \((h, k)\):
\[ h = 2, \quad k = 1 \]
Thus, the center of the ellipse is:
\[ \boxed{(2, 1)} \]