The eccentricity of the ellipse
`x^(2)+4y^(2)+8y-2x+1=0`, is

Options

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

Explanation

To find the eccentricity of the ellipse given by the equation \( x^2 + 4y^2 + 8y - 2x + 1 = 0 \), we will follow these steps:

### Step 1: Rearranging the Equation
We start with the equation:
\[ x^2 + 4y^2 + 8y - 2x + 1 = 0 \]
We can rearrange this to group the \(x\) and \(y\) terms:
\[ x^2 - 2x + 4y^2 + 8y + 1 = 0 \]

### Step 2: Completing the Square for \(x\)
Next, we complete the square for the \(x\) terms:
\[ x^2 - 2x = (x - 1)^2 - 1 \]
Substituting this back into the equation gives:
\[ (x - 1)^2 - 1 + 4y^2 + 8y + 1 = 0 \]
This simplifies to:
\[ (x - 1)^2 + 4y^2 + 8y = 0 \]

### Step 3: Completing the Square for \(y\)
Now, we complete the square for the \(y\) terms:
\[ 4y^2 + 8y = 4(y^2 + 2y) = 4((y + 1)^2 - 1) = 4(y + 1)^2 - 4 \]
Substituting this back gives:
\[ (x - 1)^2 + 4(y + 1)^2 - 4 = 0 \]
This simplifies to:
\[ (x - 1)^2 + 4(y + 1)^2 = 4 \]

### Step 4: Dividing by 4
To convert this into standard form, we divide the entire equation by 4:
\[ \frac{(x - 1)^2}{4} + \frac{(y + 1)^2}{1} = 1 \]
This is now in the standard form of an ellipse:
\[ \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 \]
where \(h = 1\), \(k = -1\), \(a^2 = 4\), and \(b^2 = 1\).

### Step 5: Identifying \(a\) and \(b\)
From the standard form, we identify:
\[ a = \sqrt{4} = 2 \quad \text{and} \quad b = \sqrt{1} = 1 \]

### Step 6: 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 \(a^2\) and \(b^2\):
\[ e = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]

### Final Answer
Thus, the eccentricity of the ellipse is:
\[ \frac{\sqrt{3}}{2} \]