4. Given the parabola $y=a x^{2}+b x+c$ passing through points $A(-1,0)$ and $B(2,0)$, for any non-zero real number $a$, the parabola never passes through point $P\left(x_{0}+1,2 x_{0}^{2}-2\right)$. Then the coordinates of point $P$ are $\qquad$
4. $(-1,6)$.

Notice that, the parabola $y=a(x+1)(x-2)$.
Since point $P$ is not on the parabola, we know
$$
2 x_{0}^{2}-2 \neq a\left(x_{0}+2\right)\left(x_{0}-1\right) \text {. }
$$

Since the above equation holds for any non-zero real number $a$, we have
$$
\left(x_{0}+2\right)\left(x_{0}-1\right)=0 \text {, and } 2-2 x_{0}^{2} \neq 0 \text {. }
$$

Therefore, $x_{0}=-2$.
Hence, point $P(-1,6)$.