13%6 * Given: The parabola $y=-x^{2}+b x+c$ is tangent to the parabola $y=x^{2}$. Try to find: The geometric position of the vertex of the parabola.
Since the parabola $y=-x^{2}+b x+c$ is tangent to $y=x^{2}$, the equation $-x^{2}+b x+c=$ $x^{2}, 2 x^{2}-b x-c=0$ has a unique real root, so its discriminant $\Delta=b^{2}+8 c=0, c=-\frac{1}{8} b^{2}$. Therefore, the equation of the parabola is $y=-x^{2}+b x-\frac{1}{8} b^{2}$. Its vertex coordinates are $\left(\frac{1}{2} b, \frac{1}{8} b^{2}\right)$. Thus, the vertex lies on the parabola with the equation $y=\frac{1}{2} x^{2}$.
Conversely, if the point $\left(x_{0}, y_{0}\right)$ lies on the parabola $y=\frac{1}{2} x^{2}$, i.e., $y_{0}=\frac{1}{2} x_{0}^{2}$, then the parabola
$$
y=-x^{2}+2 x_{0} x-\frac{1}{2} x_{0}^{2}=-\left(x-x_{0}\right)^{2}+\frac{1}{2} x_{0}^{2},
$$
has the vertex $\left(x_{0}, \frac{1}{2} x_{0}^{2}\right)$, and is tangent to the parabola $y=x^{2}$ at the point $\left(\frac{1}{2} x_{0}, \frac{1}{4} x_{0}^{2}\right)$, so it is a parabola of the form $y=\frac{1}{2} x^{2}$.