4. Given $F_{1}, F_{2}$ are the two foci of the hyperbola $x^{2}-y^{2}=1$, $M$ is a point on the right branch of the hyperbola, and $O$ is the origin. If $\frac{\left|M F_{1}\right|+\left|M F_{2}\right|}{|M O|}=\sqrt{6}$, then the coordinates of point $M$ are $\qquad$.
4. $\left(\frac{\sqrt{6}}{2}, \pm \frac{\sqrt{2}}{2}\right)$.

Let $M(x, y)(x \geqslant 1)$. It is easy to know that the foci of the hyperbola are $F_{1}(-\sqrt{2}, 0), F_{2}(\sqrt{2}, 0)$. Then $\left|M F_{1}\right|=$ $\sqrt{(x+\sqrt{2})^{2}+y^{2}},\left|M F_{2}\right|=\sqrt{(x-\sqrt{2})^{2}+y^{2}},|M O|=\sqrt{x^{2}+y^{2}}$. Therefore,
$$
\begin{aligned}
\left(\frac{\left|M F_{1}\right|+\left|M F_{2}\right|}{|M O|}\right)^{2} & =\frac{(x+\sqrt{2})^{2}+y^{2}+(x-\sqrt{2})^{2}+y^{2}+2 \sqrt{\left(x^{2}+y^{2}+2\right)^{2}-8 x^{2}}}{x^{2}+y^{2}} \\
& =\frac{4 x^{2}+2+2 \sqrt{4 x^{4}-4 x^{2}+1}}{2 x^{2}-1}=\frac{8 x^{2}}{2 x^{2}-1} .
\end{aligned}
$$

Solve the equation $\frac{8 x^{2}}{2 x^{2}-1}=6(x \geqslant 1)$, we get $x=\frac{\sqrt{6}}{2}$. Then $y^{2}=x^{2}-1=\frac{1}{2}$. Hence $M\left(\frac{\sqrt{6}}{2}, \pm \frac{\sqrt{2}}{2}\right)$.