The eccentricity of the ellipse `x^(2)+4y^(2)+2x+16y+13=0` is

Options

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

Answer

To find the eccentricity of the ellipse given by the equation \( x^2 + 4y^2 + 2x + 16y + 13 = 0 \), we will follow these steps:

### Step 1: Rearranging the Equation
Start by rearranging the equation:
\[ x^2 + 2x + 4y^2 + 16y + 13 = 0 \]

### Step 2: Completing the Square for \(x\)
To complete the square for the \(x\) terms:
\[ x^2 + 2x \rightarrow (x + 1)^2 - 1 \]
So, we rewrite:
\[ (x + 1)^2 - 1 + 4y^2 + 16y + 13 = 0 \]

### Step 3: Completing the Square for \(y\)
Now, complete the square for the \(y\) terms:
\[ 4y^2 + 16y \rightarrow 4(y^2 + 4y) \rightarrow 4((y + 2)^2 - 4) = 4(y + 2)^2 - 16 \]
Substituting this back into the equation gives:
\[ (x + 1)^2 - 1 + 4(y + 2)^2 - 16 + 13 = 0 \]

### Step 4: Simplifying the Equation
Combine the constants:
\[ (x + 1)^2 + 4(y + 2)^2 - 4 = 0 \]
This simplifies to:
\[ (x + 1)^2 + 4(y + 2)^2 = 4 \]

### Step 5: Dividing by 4
Divide the entire equation by 4 to put it in standard form:
\[ \frac{(x + 1)^2}{4} + \frac{(y + 2)^2}{1} = 1 \]

### Step 6: 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 \(a^2 = 4\) and \(b^2 = 1\). Thus:
\[ a = 2 \quad \text{and} \quad b = 1 \]

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

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

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