6. Given the ellipse $C_{1}: \frac{x^{2}}{4}+y^{2}=1$, ellipse $C_{2}:(x-2)^{2}+4 y^{2}=1, C_{1}, C_{2}$ have a common tangent line that intersects the $x$-axis at point $A$, then the coordinates of point $A$ are $\qquad$ .
Solve:
As shown in the figure, perform the transformation $\left\{\begin{array}{l}x^{\prime}=x, \\ y^{\prime}=2 y,\end{array}\right.$ then the ellipses $C_{1}, C_{2}$ are transformed into the circles $C_{1}^{\prime}: x^{\prime 2}+y^{\prime 2}=4$, and $C_{2}^{\prime}:\left(x^{\prime}-2\right)^{2}+y^{\prime 2}=1$.
Assume the equation of the common tangent line is $y^{\prime}=k x^{\prime}+b$,
Thus, $\left\{\begin{array}{l}\frac{|b|}{\sqrt{1+k^{2}}}=2, \\ \frac{|2 k+b|}{\sqrt{1+k^{2}}}=1\end{array} \Rightarrow\left|\frac{b}{2 k+b}\right|=2 \Rightarrow 16 k^{2}+16 k b+3 b^{2}=0 \Rightarrow 16\left(\frac{k}{b}\right)^{2}+16 \cdot \frac{k}{b}+3=0\right.$ $\Rightarrow \frac{k}{b}=-\frac{1}{4}$ or $\frac{k}{b}=-\frac{3}{4}$ (discard) $\Rightarrow b=-4 k$, hence $\frac{|-4 k|}{\sqrt{1+k^{2}}}=2 \Rightarrow k= \pm \frac{\sqrt{3}}{3}$. Therefore, the equation of the common tangent line is $y^{\prime}= \pm \frac{\sqrt{3}}{3}\left(x^{\prime}-4\right) \Rightarrow 2 y= \pm \frac{\sqrt{3}}{3}(x-4) \Rightarrow y= \pm \frac{\sqrt{3}}{6}(x-4)$. Thus, the equation of the common tangent line of the ellipses $C_{1}, C_{2}$ is $y= \pm \frac{\sqrt{3}}{6}(x-4) \Rightarrow A(4,0)$.