If the line $y - 2 = 0$ is the directrix of the parabola $x^2 - ky + 32 = 0, k ≠ 0$ and the parabola intersects the circle $x^2 + y^2 = 8$ at two real distinct points, then the absolute value of $k$ is
Step-by-Step Solution
Key Concept: The directrix of a parabola $x^2 = 4a(y - h) + k$ is located at a distance $a$ from the vertex on the opposite side of the focus.
The parabola $x^2 = ky - 32$ can be rewritten as $x^2 = k(y - \frac{32}{k})$. Comparing with standard form, the vertex is at $(0, \frac{32}{k})$ and $4a = k$. The directrix is $y - \frac{32}{k} - \frac{k}{4} = 2$. For the parabola to intersect the circle at two distinct points with the given directrix condition and using the constraint from the intersection, we find $k = -16$ or $k = 8$. Therefore, the absolute value of $k$ is $16$.
Correct Answer: 16