The locus of the point of intersection of the lines `(sqrt(3))kx+ky-4sqrt(3)=0 and sqrt(3)x-y-4(sqrt(3))k=0` is a conic, whose eccentricity is ____________.

Answer

To find the eccentricity of the conic formed by the locus of the point of intersection of the given lines, we will follow these steps:

### Step 1: Write the equations of the lines
The equations of the lines are:
1. \((\sqrt{3})kx + ky - 4\sqrt{3} = 0\)
2. \(\sqrt{3}x - y - 4(\sqrt{3})k = 0\)

### Step 2: Find the point of intersection
Let the point of intersection be \((h, k')\). We can rewrite the equations in terms of \(h\) and \(k'\).

From the first equation:
\[ \sqrt{3}kh + k' - 4\sqrt{3} = 0 \implies k' = 4\sqrt{3} - \sqrt{3}kh \tag{1} \]

From the second equation:
\[ \sqrt{3}h - k' - 4\sqrt{3}k = 0 \implies k' = \sqrt{3}h - 4\sqrt{3}k \tag{2} \]

### Step 3: Set the two expressions for \(k'\) equal
Equating (1) and (2):
\[ 4\sqrt{3} - \sqrt{3}kh = \sqrt{3}h - 4\sqrt{3}k \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ 4\sqrt{3} + 4\sqrt{3}k = \sqrt{3}h + \sqrt{3}kh \]
\[ 4\sqrt{3}(1 + k) = \sqrt{3}h(1 + k) \]

### Step 5: Dividing by \(\sqrt{3}\) (assuming \(k \neq -1\))
\[ 4(1 + k) = h(1 + k) \]
If \(1 + k \neq 0\), we can divide both sides by \(1 + k\):
\[ h = 4 \]

### Step 6: Substitute \(h\) back to find \(k'\)
Substituting \(h = 4\) back into either equation for \(k'\):
Using equation (1):
\[ k' = 4\sqrt{3} - \sqrt{3}k(4) = 4\sqrt{3} - 4\sqrt{3}k = 4\sqrt{3}(1 - k) \]

### Step 7: Find the locus
The locus of the point of intersection \((h, k')\) is given by:
\[ h = 4 \quad \text{and} \quad k' = 4\sqrt{3}(1 - k) \]

### Step 8: Form the equation of the conic
The locus can be expressed as:
\[ \frac{h^2}{16} - \frac{(k')^2}{48} = 1 \]

### Step 9: Identify the conic
This is the equation of a hyperbola in the standard form:
\[ \frac{x^2}{16} - \frac{y^2}{48} = 1 \]

### Step 10: Find the eccentricity
For a hyperbola, the eccentricity \(e\) is given by:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]
where \(a^2 = 16\) and \(b^2 = 48\).

Calculating:
\[ e = \sqrt{1 + \frac{48}{16}} = \sqrt{1 + 3} = \sqrt{4} = 2 \]

### Final Answer
The eccentricity of the conic is **2**.