2. Point $P$ is on the plane of the base $A B C D$ of the cube $A_{1} B_{1} C_{1} D_{1}-A B C D$, $E$ is the midpoint of $A A_{1}$, and $\angle E P A=\angle D_{1} P D$. Then the locus of point $P$ is ( ).
(A) line
(B) circle
(C) parabola
(D) hyperbola
2.B.
$$
\begin{array}{l}
\angle E P A=\angle D_{1} P D \\
\Rightarrow \tan \angle E P A=\tan \angle D_{1} P D \\
\Rightarrow \frac{E A}{A P}=\frac{D D_{1}}{P D} \Rightarrow P D=2 P A .
\end{array}
$$
$D$ and $A$ are fixed points. From equation (1) and the definition of the Apollonian circle, the locus of point $P$ is a circle.