The eccentricity of the conic `3x^(2)+4y^(2)-6x-8y+4=0` is

Options

A. `1/2`
B. `1//sqrt(2)`
C. `sqrt2`
D. none

Answer

To find the eccentricity of the conic given by the equation \(3x^2 + 4y^2 - 6x - 8y + 4 = 0\), we will first rewrite the equation in standard form.

### Step 1: Rearranging the equation
We start with the equation:
\[ 3x^2 + 4y^2 - 6x - 8y + 4 = 0 \]
Rearranging gives:
\[ 3x^2 - 6x + 4y^2 - 8y + 4 = 0 \]

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

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

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

### Step 3: Substitute back into the equation
Substituting these completed squares back into the equation gives:
\[ 3((x - 1)^2 - 1) + 4((y - 1)^2 - 1) + 4 = 0 \]
Simplifying this:
\[ 3(x - 1)^2 - 3 + 4(y - 1)^2 - 4 + 4 = 0 \]
\[ 3(x - 1)^2 + 4(y - 1)^2 - 3 = 0 \]
\[ 3(x - 1)^2 + 4(y - 1)^2 = 3 \]

### Step 4: Divide by 3 to get standard form
Dividing the entire equation by 3:
\[ \frac{(x - 1)^2}{1} + \frac{(y - 1)^2}{\frac{3}{4}} = 1 \]

### Step 5: Identify the conic type
This is the standard form of an ellipse:
\[ \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 \]
where \(h = 1\), \(k = 1\), \(a^2 = 1\), and \(b^2 = \frac{3}{4}\).

### Step 6: Calculate eccentricity
The eccentricity \(e\) of an ellipse is given by:
\[ e = \sqrt{1 - \frac{a^2}{b^2}} \]
Here, \(a^2 = 1\) and \(b^2 = \frac{3}{4}\). Thus:
\[ e = \sqrt{1 - \frac{1}{\frac{3}{4}}} = \sqrt{1 - \frac{4}{3}} = \sqrt{-\frac{1}{3}} \]
Since \(e\) cannot be imaginary, we need to check our calculations.

### Step 7: Correct calculation of eccentricity
We should actually use:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} \text{ (since } b > a \text{ for this ellipse)} \]
So:
\[ e = \sqrt{1 - \frac{\frac{3}{4}}{1}} = \sqrt{1 - \frac{3}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \]

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