A line $ l$ is tangent to a circle $ S$ at $ A$. For any points $ B,C$ on $ l$ on opposite sides of $ A$, let the other tangents from $ B$ and $ C$ to $ S$ intersect at a point $ P$. If $ B,C$ vary on $ l$ so that the product $ AB \cdot AC$ is constant, find the locus of $ P$.
1. **Restate the problem in a more familiar notation:**
   Let \( (I, r) \) be a fixed circle tangent to a fixed line \( \mathcal{L} \) at \( X \). Points \( B \) and \( C \) are on \( \mathcal{L} \) such that \( X \in \overline{BC} \) and \( BX \cdot CX = k^2 = \text{const} \). Tangents from \( B \) and \( C \) to \( (I) \) different from \( \mathcal{L} \) meet at \( A \). We need to find the locus of \( A \).

2. **Express the given condition in terms of distances:**
   Since \( BX \cdot CX = k^2 \), let \( BX = p - b \) and \( CX = p - c \). Therefore, we have:
   \[
   (p - b)(p - c) = k^2
   \]

3. **Use the identity for the radius \( r \) of the circle:**
   From the identity for the radius of the incircle of a triangle, we have:
   \[
   r = \sqrt{\frac{(p - a)(p - b)(p - c)}{p}}
   \]
   Substituting \( (p - b)(p - c) = k^2 \), we get:
   \[
   r^2 = \frac{(p - a)k^2}{p}
   \]
   Simplifying, we obtain:
   \[
   r^2 = k^2 \left(1 - \frac{a}{p}\right)
   \]
   Therefore:
   \[
   \frac{p}{a} = \frac{k^2}{k^2 - r^2}
   \]

4. **Find the altitude \( h_a \) from \( A \) to \( BC \):**
   Using the formula for the area of a triangle, we have:
   \[
   a \cdot h_a = 2rp
   \]
   Solving for \( h_a \), we get:
   \[
   h_a = \frac{2rp}{a} = \frac{2r \cdot k^2}{k^2 - r^2}
   \]
   Since \( k \) and \( r \) are constants, \( h_a \) is also a constant.

5. **Determine the locus of \( A \):**
   Since the altitude \( h_a \) from \( A \) to \( BC \) is constant, the locus of \( A \) is a line parallel to \( BC \) at a constant distance \( h_a \).

The final answer is the locus of \( \boxed{ A } \) is a line parallel to \( BC \) at a constant distance \( h_a \).