6. Given that point $P$ is a moving point on the line $x+2 y=4$, and through point $P$ two tangents are drawn to the ellipse $x^{2}+4 y^{2}=4$, with the points of tangency being $A, B$. When point $P$ moves, the line $A B$ passes through a fixed point with coordinates $\qquad$ .
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), P\left(x_{0}, y_{0}\right)$, then the tangents to the ellipse $x^{2}+4 y^{2}=4$ at points $A$ and $B$ are $x_{1} x+4 y_{1} y=4, x_{1} x+4 y_{1} y=4$, respectively. Thus, the line $A B: x_{0} x+4 y_{0} y=4$.
Notice that $x_{0}+2 y_{0}=4 \Rightarrow x_{0}=4-2 y_{0}$, substituting into the equation of line $A B$ gives $\left(2-y_{0}\right) x+2 y_{0} y=2$ $\Rightarrow 2(x-1)-y_{0}(x-2 y)=0$.
Since the above equation holds for all $y_{0} \in \mathbf{R}$, the line $A B$ passes through the fixed point $\left(1, \frac{1}{2}\right)$.