`3x^2+4y^2-6x+8y+k=0` represents an ellipse with eccentricity 1/2,

Options

A. for all value of k
B. for k `gt` 7
C. `k lt 7`
D. k=7

Answer

To determine the value of \( k \) for which the equation \( 3x^2 + 4y^2 - 6x + 8y + k = 0 \) represents an ellipse with eccentricity \( \frac{1}{2} \), we will follow these steps:

### Step 1: Rearranging the Equation
Start by rearranging the given equation into a standard form of an ellipse. We will group the \( x \) and \( y \) terms together.

\[ 3x^2 - 6x + 4y^2 + 8y + k = 0 \]

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

1. For \( x \):
\[ 3(x^2 - 2x) = 3\left((x - 1)^2 - 1\right) = 3(x - 1)^2 - 3 \]

2. For \( y \):
\[ 4(y^2 + 2y) = 4\left((y + 1)^2 - 1\right) = 4(y + 1)^2 - 4 \]

Now substitute these back into the equation:

\[ 3(x - 1)^2 - 3 + 4(y + 1)^2 - 4 + k = 0 \]

### Step 3: Simplifying the Equation
Combine the constant terms:

\[ 3(x - 1)^2 + 4(y + 1)^2 + (k - 7) = 0 \]

Rearranging gives:

\[ 3(x - 1)^2 + 4(y + 1)^2 = 7 - k \]

### Step 4: Standard Form of the Ellipse
To express this in standard form, we divide by \( 7 - k \):

\[ \frac{(x - 1)^2}{\frac{7 - k}{3}} + \frac{(y + 1)^2}{\frac{7 - k}{4}} = 1 \]

### Step 5: Identifying \( a^2 \) and \( b^2 \)
From the standard form, we identify:

\[ a^2 = \frac{7 - k}{3}, \quad b^2 = \frac{7 - k}{4} \]

### Step 6: Eccentricity Condition
The eccentricity \( e \) of an ellipse is given by:

\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]

Substituting \( a^2 \) and \( b^2 \):

\[ e = \sqrt{1 - \frac{\frac{7 - k}{4}}{\frac{7 - k}{3}}} = \sqrt{1 - \frac{3}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \]

### Step 7: Conditions for \( k \)
For the ellipse to be valid, both \( a^2 \) and \( b^2 \) must be positive:

1. \( \frac{7 - k}{3} > 0 \) implies \( 7 - k > 0 \) or \( k < 7 \)
2. \( \frac{7 - k}{4} > 0 \) implies \( 7 - k > 0 \) or \( k < 7 \)

Both conditions give us \( k < 7 \).

### Conclusion
Thus, the value of \( k \) for which the conic section represents an ellipse with eccentricity \( \frac{1}{2} \) is:

\[ \text{For } k < 7 \]