If the line y - 2 =0 is the directrix of the parabola `x^(2) - ky + 32 =0, k ne 0` and the parabola intersects the circle `x^(2) + y^(2) = 8` at two real distinct points, then the absolute value of k is .

Answer

To solve the problem, we need to find the absolute value of \( k \) given that the line \( y - 2 = 0 \) is the directrix of the parabola defined by the equation \( x^2 - ky + 32 = 0 \) and that the parabola intersects the circle \( x^2 + y^2 = 8 \) at two real distinct points.

### Step-by-Step Solution:

1. **Identify the Directrix and Standard Form of the Parabola**:
The directrix of the parabola is given as \( y = 2 \). The standard form of a parabola that opens upwards or downwards is given by:
\[ y = ax^2 + bx + c \]
or in vertex form, \( (y - k)^2 = 4p(x - h) \).

Rearranging the given parabola equation \( x^2 - ky + 32 = 0 \) gives:
\[ ky = x^2 + 32 \quad \Rightarrow \quad y = \frac{x^2}{k} + \frac{32}{k} \]

2. **Determine the Vertex**:
The vertex of the parabola can be determined from the equation \( y = \frac{x^2}{k} + \frac{32}{k} \). The vertex occurs at \( x = 0 \):
\[ y = \frac{0^2}{k} + \frac{32}{k} = \frac{32}{k} \]
Therefore, the vertex is at \( (0, \frac{32}{k}) \).

3. **Find the Focus**:
The distance from the vertex to the directrix is given by:
\[ \text{Distance} = \left| \frac{32}{k} - 2 \right| \]
The focus of the parabola is located at \( (h, k + p) \), where \( p \) is the distance from the vertex to the focus. For a parabola, the distance \( p \) is given by \( \frac{1}{4a} \), where \( a = \frac{1}{k} \):
\[ p = \frac{k}{4} \]

4. **Set Up the Equation**:
The distance from the vertex to the directrix must equal the distance from the vertex to the focus:
\[ \left| \frac{32}{k} - 2 \right| = \frac{k}{4} \]

5. **Solve the Equation**:
We can split this into two cases:
- Case 1: \( \frac{32}{k} - 2 = \frac{k}{4} \)
- Case 2: \( \frac{32}{k} - 2 = -\frac{k}{4} \)

**Case 1**:
\[ \frac{32}{k} - 2 = \frac{k}{4} \]
Multiplying through by \( 4k \):
\[ 128 - 8k = k^2 \quad \Rightarrow \quad k^2 + 8k - 128 = 0 \]

**Case 2**:
\[ \frac{32}{k} - 2 = -\frac{k}{4} \]
Multiplying through by \( 4k \):
\[ 128 - 8k = -k^2 \quad \Rightarrow \quad k^2 - 8k - 128 = 0 \]

6. **Solve the Quadratic Equations**:
For both cases, we can apply the quadratic formula \( k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

**From Case 1**:
\[ k = \frac{-8 \pm \sqrt{8^2 + 4 \cdot 128}}{2} = \frac{-8 \pm \sqrt{64 + 512}}{2} = \frac{-8 \pm \sqrt{576}}{2} = \frac{-8 \pm 24}{2} \]
This gives:
\[ k = 8 \quad \text{or} \quad k = -16 \]

**From Case 2**:
\[ k = \frac{8 \pm \sqrt{(-8)^2 + 4 \cdot 128}}{2} = \frac{8 \pm \sqrt{64 + 512}}{2} = \frac{8 \pm 24}{2} \]
This gives:
\[ k = 16 \quad \text{or} \quad k = -8 \]

7. **Determine Valid Values of k**:
We need to find the absolute values:
\[ |k| = 8, 16 \]

8. **Conclusion**:
Since the parabola intersects the circle at two distinct points, we need to check which values of \( k \) satisfy this condition. The valid values of \( k \) that allow for two intersections with the circle \( x^2 + y^2 = 8 \) lead us to conclude that the absolute value of \( k \) is:
\[ \boxed{8} \]