5. Given real numbers $x, y$ satisfy
$$
x^{2}+3 y^{2}-12 y+12=0 \text {. }
$$

then the value of $y^{x}$ is $\qquad$
Hint: Treat the known equation as a quadratic equation in $y$ (the main variable)
$$
3 y^{2}-12 y+\left(12+x^{2}\right)=0 \text {. }
$$

From $\Delta=-12 x^{2} \geqslant 0$, and since $x^{2} \geqslant 0$, then $x=0$.
Thus, $y=2$. Therefore, $y^{x}=2^{0}=1$.