The eccentricity of the curve `x^(2)-4x+4y^(2)=12` is

Options

A. `sqrt(3)//2`
B. `2//sqrt(3)`
C. `sqrt(3)`
D. none of these

Answer

To find the eccentricity of the curve given by the equation \(x^2 - 4x + 4y^2 = 12\), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ x^2 - 4x + 4y^2 = 12 \]
To make it easier to identify the conic section, we will complete the square for the \(x\) terms.

### Step 2: Complete the square for \(x\)
The \(x\) terms are \(x^2 - 4x\). We can complete the square:
\[ x^2 - 4x = (x - 2)^2 - 4 \]
Now, substituting this back into the equation gives:
\[ (x - 2)^2 - 4 + 4y^2 = 12 \]
This simplifies to:
\[ (x - 2)^2 + 4y^2 = 16 \]

### Step 3: Divide by 16
To put the equation in standard form, we divide the entire equation by 16:
\[ \frac{(x - 2)^2}{16} + \frac{4y^2}{16} = 1 \]
This simplifies to:
\[ \frac{(x - 2)^2}{16} + \frac{y^2}{4} = 1 \]

### Step 4: 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 can identify:
- \(a^2 = 16\) (thus \(a = 4\))
- \(b^2 = 4\) (thus \(b = 2\))

### Step 5: 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 of \(b^2\) and \(a^2\):
\[ e = \sqrt{1 - \frac{4}{16}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]

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

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