6. Draw a normal line (a normal line is a line passing through the point of tangency and perpendicular to the tangent line) at point $A$ on the parabola $y=x^{2}$, intersecting the parabola at another point $B$, and $O$ is the origin. When the area of $\triangle O A B$ is minimized, the y-coordinate of point $A$ is $\qquad$ (Liu Kaifeng)
6. $\frac{-3+\sqrt{33}}{24}$.

Let $A\left(x_{0}, x_{0}^{2}\right), B\left(x_{1}, x_{1}^{2}\right)$, the slope of the tangent line to the parabola $y=x^{2}$ at point $A$ is $2 x_{0}$, so the slope of the line is $-\frac{1}{2 x_{0}}$, i.e.,
$$
-\frac{1}{2 x_{0}}=\frac{x_{1}^{2}-x_{0}^{2}}{x_{1}-x_{0}}=x_{1}+x_{0}
$$

Thus, $x_{1}=-\frac{1}{2 x_{0}}-x_{0}$. It is easy to find that the equation of line $A B$ is $y-x_{0}^{2}=-\frac{1}{2 x_{0}}\left(x-x_{0}\right)$, and the intersection point with the $y$-axis is $\left(0, x_{0}^{2}+\frac{1}{2}\right)$. Without loss of generality, assume $x_{0}>0$, then
$$
S_{\triangle O A B}=\frac{1}{2}\left(x_{0}-x_{1}\right) \cdot\left(x_{0}^{2}+\frac{1}{2}\right)=\frac{1}{2}\left(2 x_{0}+\frac{1}{2 x_{0}}\right)\left(x_{0}^{2}+\frac{1}{2}\right) .
$$

Consider the function $f(x)=\left(2 x+\frac{1}{2 x}\right)\left(x^{2}+\frac{1}{2}\right)=2 x^{3}+\frac{3}{2} x+\frac{1}{4 x}$, so, $f^{\prime}(x)=6 x^{2}+\frac{3}{2}-\frac{1}{4 x^{2}}$. Let $f^{\prime}(x)=0$, and let $t=x^{2}$, then $6 t+\frac{3}{2}-\frac{1}{4 t}=0 \Rightarrow 24 t^{2}+6 t-1=0$, solving for $t=\frac{-3+\sqrt{33}}{24}$.

Therefore, when the area of $\triangle O A B$ is minimized, the $y$-coordinate of point $A$ is $x_{0}^{2}=\frac{-3+\sqrt{33}}{24}$.