$(BEL 1)$ A parabola $P_1$ with equation $x^2 - 2py = 0$ and parabola $P_2$ with equation $x^2 + 2py = 0, p > 0$, are given. A line $t$ is tangent to $P_2.$ Find the locus of pole $M$ of the line $t$ with respect to $P_1.$
1. **Identify the given parabolas and their equations:**
   - Parabola \( P_1 \) has the equation \( x^2 - 2py = 0 \).
   - Parabola \( P_2 \) has the equation \( x^2 + 2py = 0 \), where \( p > 0 \).

2. **Determine the equation of the tangent line to \( P_2 \):**
   - The general form of the tangent line to the parabola \( x^2 + 2py = 0 \) at the point \( T(2p\lambda, -2p\lambda^2) \) is given by:
     \[
     2\lambda x + y - 2p\lambda^2 = 0
     \]

3. **Find points on the tangent line:**
   - Two points on this tangent line are:
     - \( P(0, 2p\lambda^2) \)
     - \( Q(p\lambda, 0) \)

4. **Determine the polar lines of these points with respect to \( P_1 \):**
   - The polar line of point \( P(0, 2p\lambda^2) \) with respect to \( P_1 \) is:
     \[
     y + 2p\lambda^2 = 0
     \]
   - The polar line of point \( Q(p\lambda, 0) \) with respect to \( P_1 \) is:
     \[
     \lambda x - y = 0
     \]

5. **Find the intersection of the polar lines to determine the pole \( M \):**
   - Solve the system of equations:
     \[
     \begin{cases}
     y + 2p\lambda^2 = 0 \\
     \lambda x - y = 0
     \end{cases}
     \]
   - Substitute \( y = \lambda x \) from the second equation into the first equation:
     \[
     \lambda x + 2p\lambda^2 = 0
     \]
   - Solving for \( x \):
     \[
     x(\lambda + 2p\lambda^2) = 0 \implies x = 0 \text{ or } \lambda + 2p\lambda = 0
     \]
   - Since \( \lambda \neq 0 \), we have:
     \[
     \lambda + 2p\lambda = 0 \implies \lambda(1 + 2p) = 0 \implies \lambda = 0
     \]
   - This implies \( x = 0 \) and \( y = 0 \).

6. **Eliminate the parameter \( \lambda \) to find the locus of \( M \):**
   - The locus of the pole \( M \) is given by the equation:
     \[
     x^2 + 2py = 0
     \]

The final answer is \( \boxed{ x^2 + 2py = 0 } \).