7. Given the ellipse $\Gamma: \frac{x^{2}}{4}+y^{2}=1$, a line with a non-zero slope is drawn through the right focus $F$, intersecting the ellipse $\Gamma$ at points $A$ and $B$. If $A O^{2}=A B$, then the coordinates of point $A$ are $\qquad$ .
7. $\left(\frac{\sqrt{3}}{2}, \pm \frac{\sqrt{13}}{4}\right)$
From the problem, we know $F(\sqrt{3}, 0)$.
Let $l_{A B}: y=k(x-\sqrt{3})$, and combine it with the equation of the ellipse $\Gamma$ to get
$$
\left(\frac{1}{4}+k^{2}\right) x^{2}-2 \sqrt{3} k^{2} x+3 k^{2}-1=0 .
$$

Thus, $A B=\frac{1+k^{2}}{\frac{1}{4}+k^{2}}$.
Let point $A\left(x_{A}, y_{A}\right)$.
From $A O^{2}=\frac{3}{4} x_{A}^{2}+1$, we know
$$
k^{2}=\frac{1}{x_{A}^{2}}-\frac{1}{4} \neq 0 \text {. }
$$

Substituting into equation (1) yields $x_{A}=\frac{\sqrt{3}}{2}$.
Therefore, point $A\left(\frac{\sqrt{3}}{2}, \pm \frac{\sqrt{13}}{4}\right)$.