Let us have a line $\ell$ in the space and a point $A$ not lying on $\ell.$ For an arbitrary line $\ell'$ passing through $A$, $XY$ ($Y$ is on $\ell'$) is a common perpendicular to the lines $\ell$ and $\ell'.$ Find the locus of points $Y.$
1. **Define the problem setup:**
   - Let $\ell$ be a line in space.
   - Let $A$ be a point not lying on $\ell$.
   - Consider an arbitrary line $\ell'$ passing through $A$.
   - $XY$ is a common perpendicular to the lines $\ell$ and $\ell'$, where $Y$ is on $\ell'$.

2. **Construct a parallel line:**
   - Draw a line $\ell''$ through $A$ that is parallel to $\ell$.
   - Let $XZ$ be the common perpendicular of $\ell$ and $\ell''$, with $Z \in \ell''$.

3. **Identify the geometric relationship:**
   - Since $XZ$ is the common perpendicular, $XZ$ is perpendicular to both $\ell$ and $\ell''$.
   - The point $Y$ lies on the circle with diameter $XZ$ in the plane passing through $X$ and perpendicular to $\ell$.

4. **Determine the locus of $Y$:**
   - The circle with diameter $XZ$ lies in a plane perpendicular to $\ell$.
   - As $\ell'$ varies, the point $Y$ traces out a cylinder.
   - The cylinder has opposite generatrices $\ell$ and $\ell''$.

5. **Verify the locus:**
   - Any point on this cylinder can serve as an endpoint of a common perpendicular for $\ell$ and some line through $A$.
   - Therefore, the locus of points $Y$ is the cylinder with opposite generatrices $\ell$ and $\ell''$.

\[
\boxed{\text{The locus of points } Y \text{ is the cylinder with opposite generatrices } \ell \text{ and } \ell''.}
\]