The centre of the ellipse `4x^(2) + 9y^(2) + 16x - 18y - 11 = 0` is

Options

A. (-2, -1)
B. (-2, 1)
C. (2, -1)
D. none of these

Answer

To find the center of the ellipse given by the equation \( 4x^2 + 9y^2 + 16x - 18y - 11 = 0 \), we will follow these steps:

### Step 1: Rearrange the equation
Start by rearranging the equation to group the \(x\) and \(y\) terms together:
\[ 4x^2 + 16x + 9y^2 - 18y - 11 = 0 \]

### Step 2: Complete the square for \(x\)
To complete the square for the \(x\) terms, factor out the coefficient of \(x^2\) (which is 4):
\[ 4(x^2 + 4x) + 9y^2 - 18y - 11 = 0 \]
Now, complete the square inside the parentheses:
\[ x^2 + 4x \rightarrow (x + 2)^2 - 4 \]
Thus, we can rewrite it as:
\[ 4((x + 2)^2 - 4) + 9y^2 - 18y - 11 = 0 \]
Distributing the 4 gives:
\[ 4(x + 2)^2 - 16 + 9y^2 - 18y - 11 = 0 \]

### Step 3: Complete the square for \(y\)
Now, group the \(y\) terms:
\[ 9(y^2 - 2y) = 9((y - 1)^2 - 1) = 9(y - 1)^2 - 9 \]
Substituting this back in, we have:
\[ 4(x + 2)^2 - 16 + 9(y - 1)^2 - 9 - 11 = 0 \]
Combine the constants:
\[ 4(x + 2)^2 + 9(y - 1)^2 - 36 = 0 \]
Rearranging gives:
\[ 4(x + 2)^2 + 9(y - 1)^2 = 36 \]

### Step 4: Write in standard form
Divide the entire equation by 36 to express it in standard form:
\[ \frac{(x + 2)^2}{9} + \frac{(y - 1)^2}{4} = 1 \]

### Step 5: Identify the center
From the standard form of the ellipse \(\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1\), we can identify the center \((h, k)\):
- Here, \(h = -2\) and \(k = 1\).

Thus, the center of the ellipse is:
\[ \text{Center} = (-2, 1) \]