5.15 052 .

Obviously, two parabolas of the form $y=a x^{2}+b x+c$ can have at most two intersection points, and two parabolas of the form $x=a_{1} y^{2}+b_{1} y+c_{1}$ can have at most two intersection points. A parabola $y=a x^{2}+b x+c$ and a parabola $x=a_{1} y^{2}+b_{1} y+c_{1}$ can have at most four intersection points;
50 different parabolas $y=a x^{2}+b x+c$ can divide the plane $x O y$ into at most
$$
\begin{array}{l}
2+(2+1)+(2 \times 2+1)+(2 \times 3+1)+\cdots+ \\
(2 \times 49+1)=2051
\end{array}
$$

parts.
Similarly, 50 different parabolas $x=a_{1} y^{2}+b_{1} y+c_{1}$ can divide the plane $x O y$ into 2501 parts.

A parabola $x=a_{1} y^{2}+b_{1} y+c_{1}$ and 50 parabolas $y=a x^{2}+b x+c$ can have at most 200 intersection points, which divide the parabola $x=a_{1} y^{2}+b_{1} y+c_{1}$ into 201 segments. These 201 segments can further divide the plane $x O y$ into 201 additional parts. Therefore, 50 parabolas $x=a_{1} y^{2}+b_{1} y+c_{1}$ and 50 parabolas $y=a x^{2}+b x+c$ can further divide the plane $x O y$ into $201 \times 50=10050$ additional parts.
Thus, the total number of parts is
$$
2501+2501+10050=15052 .
$$
5. In the Cartesian coordinate system $x O y$, there are 50 different parabolas $y=a x^{2}+b x+c$ and another 50 different parabolas $x=a_{1} y^{2}+b_{1} y+c_{1}$. These 100 parabolas divide the coordinate plane into a maximum of $\qquad$ parts.