The length of the major axis of the ellipse `3x^2+2y^2-6x8y-1=0` is

Options

A. `sqrt3`
B. `sqrt6`
C. `2sqrt3`
D. `2sqrt6`

Explanation

To find the length of the major axis of the ellipse given by the equation \(3x^2 + 2y^2 - 6x + 8y - 1 = 0\), we will follow these steps:

### Step 1: Rearrange the equation
We start with the equation:
\[ 3x^2 + 2y^2 - 6x + 8y - 1 = 0 \]
Rearranging gives:
\[ 3x^2 - 6x + 2y^2 + 8y = 1 \]

### Step 2: Complete the square for \(x\) and \(y\)
**For the \(x\) terms:**
\[ 3(x^2 - 2x) \]
To complete the square, we take half of the coefficient of \(x\) (which is \(-2\)), square it, and add/subtract it:
\[ 3\left((x - 1)^2 - 1\right) = 3(x - 1)^2 - 3 \]

**For the \(y\) terms:**
\[ 2(y^2 + 4y) \]
Similarly, for \(y\):
\[ 2\left((y + 2)^2 - 4\right) = 2(y + 2)^2 - 8 \]

### Step 3: Substitute back into the equation
Substituting the completed squares back into the equation gives:
\[ 3((x - 1)^2 - 1) + 2((y + 2)^2 - 4) = 1 \]
Expanding this:
\[ 3(x - 1)^2 - 3 + 2(y + 2)^2 - 8 = 1 \]
Combining like terms:
\[ 3(x - 1)^2 + 2(y + 2)^2 - 11 = 1 \]
Thus,
\[ 3(x - 1)^2 + 2(y + 2)^2 = 12 \]

### Step 4: Divide by 12 to standardize the equation
Dividing the entire equation by 12 gives:
\[ \frac{3(x - 1)^2}{12} + \frac{2(y + 2)^2}{12} = 1 \]
This simplifies to:
\[ \frac{(x - 1)^2}{4} + \frac{(y + 2)^2}{6} = 1 \]

### Step 5: Identify the lengths of the axes
The standard form of the ellipse is:
\[ \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 \]
where \(a^2 = 4\) and \(b^2 = 6\). Thus, we have:
- \(a = 2\)
- \(b = \sqrt{6}\)

### Step 6: Determine the length of the major axis
The length of the major axis is given by \(2b\) (since \(b > a\)):
\[ \text{Length of major axis} = 2\sqrt{6} \]

### Final Answer
The length of the major axis of the ellipse is:
\[ \boxed{2\sqrt{6}} \]