The equation of the normal passing through a point $(x_0,y_0)$ of the ellipse and a generic point $(x,y)$ is
This must be combined with the ellipse equation:
where $a$ and $b$ are the semi-axes.
We can solve $(1)$ for $y_0$ and plug the result into $(2)$, thus obtaining a resolvent quartic equation in $x_0$:
We know by hypothesis that this equation admits $4$ real solutions for $x_0$. The sum $S_x$ of the solutions is given by $-A_3/A_4$, where $A_3$ is the coefficient of $x_0^3$ in the polynomial on the left hand side and $A_4$ is the coefficient of $x_0^4$. By expanding $(3)$ we can find this to be: $S_x=2a^2x/(a^2-b^2)$.
We can repeat the same process, solving $(1)$ for $x_0$ and plugging the result into $(2)$, to obtain a resolvent quartic equation in $y_0$ with four real solutions. The sum $S_y$ of the solutions turns out to be:
Finally, the coordinates of the centroid $G$ of the four points are given by $(S_x/4,S_y/4)$, that is:
Notice that $P=(x,y)$ in the above formula is the point where the four normals intersect.