For the ellipse `3x^(2) + 4y^(2) + 6x - 8y - 5 = 0` the eccentrically, is

Options

A. `1//3`
B. `1//2`
C. `1//4`
D. `1//5`

Answer

To find the eccentricity of the ellipse given by the equation \(3x^2 + 4y^2 + 6x - 8y - 5 = 0\), we will follow these steps:

### Step 1: Rearranging the equation
We start with the equation:
\[ 3x^2 + 4y^2 + 6x - 8y - 5 = 0 \]
We can rearrange this equation by grouping the \(x\) terms and \(y\) terms together:
\[ 3x^2 + 6x + 4y^2 - 8y = 5 \]

### Step 2: Completing the square for \(x\) terms
For the \(x\) terms \(3x^2 + 6x\), we factor out the 3:
\[ 3(x^2 + 2x) \]
Now, we complete the square inside the parentheses:
\[ x^2 + 2x = (x + 1)^2 - 1 \]
Thus, we have:
\[ 3((x + 1)^2 - 1) = 3(x + 1)^2 - 3 \]

### Step 3: Completing the square for \(y\) terms
For the \(y\) terms \(4y^2 - 8y\), we factor out the 4:
\[ 4(y^2 - 2y) \]
Now, we complete the square:
\[ y^2 - 2y = (y - 1)^2 - 1 \]
Thus, we have:
\[ 4((y - 1)^2 - 1) = 4(y - 1)^2 - 4 \]

### Step 4: Substitute back into the equation
Now substituting back into our equation:
\[ 3(x + 1)^2 - 3 + 4(y - 1)^2 - 4 = 5 \]
Simplifying this gives:
\[ 3(x + 1)^2 + 4(y - 1)^2 - 7 = 5 \]
\[ 3(x + 1)^2 + 4(y - 1)^2 = 12 \]

### Step 5: Divide by 12 to get the standard form
To convert this into the standard form of an ellipse, we divide the entire equation by 12:
\[ \frac{3(x + 1)^2}{12} + \frac{4(y - 1)^2}{12} = 1 \]
This simplifies to:
\[ \frac{(x + 1)^2}{4} + \frac{(y - 1)^2}{3} = 1 \]

### Step 6: Identify \(a^2\) and \(b^2\)
From the standard form of the ellipse \(\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1\), we identify:
- \(a^2 = 4\) (thus \(a = 2\))
- \(b^2 = 3\) (thus \(b = \sqrt{3}\))

### Step 7: Calculate the eccentricity
The eccentricity \(e\) of an ellipse is given by the formula:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Substituting the values we have:
\[ e = \sqrt{1 - \frac{3}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \]

### Final Answer
Thus, the eccentricity of the ellipse is:
\[ \boxed{\frac{1}{2}} \]