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 will rearrange it to group the \(x\) and \(y\) terms together:
\[ 3x^2 + 6x + 4y^2 - 8y = 5 \]

### Step 2: Completing the square for \(x\) and \(y\)
Next, we complete the square for the \(x\) terms and the \(y\) terms.

For \(x\):
\[ 3(x^2 + 2x) = 3((x+1)^2 - 1) = 3(x+1)^2 - 3 \]

For \(y\):
\[ 4(y^2 - 2y) = 4((y-1)^2 - 1) = 4(y-1)^2 - 4 \]

Substituting these back into the equation gives:
\[ 3((x+1)^2 - 1) + 4((y-1)^2 - 1) = 5 \]
This simplifies to:
\[ 3(x+1)^2 - 3 + 4(y-1)^2 - 4 = 5 \]
\[ 3(x+1)^2 + 4(y-1)^2 - 7 = 5 \]
\[ 3(x+1)^2 + 4(y-1)^2 = 12 \]

### Step 3: Dividing by 12 to get the standard form
Now, 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 4: Identifying \(a\) and \(b\)
From the standard form of the ellipse \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), we identify:
- \(a^2 = 4\) → \(a = 2\)
- \(b^2 = 3\) → \(b = \sqrt{3}\)

### Step 5: Calculating the eccentricity
The eccentricity \(e\) of an ellipse is given by the formula:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Substituting the values of \(a\) and \(b\):
\[ e = \sqrt{1 - \frac{3}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \]

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