Example 4. Given the parabola $y=x^{2}-2 x+2$ (I) and another parabola $y=-x^{2}+a x+b(\mathbf{I})$ intersect, and at one intersection point $\mathrm{P}$, the tangents of the two curves are perpendicular to each other.
1. Find the relationship that $\mathrm{a}$ and $\mathrm{b}$ should satisfy;
2. Prove that the parabola (I) passes through a fixed point $Q$ (this point's position is independent of $\mathrm{a}, \mathrm{b}$), and find the coordinates of $\mathrm{Q}$.
1. Let the coordinates of the intersection point $\mathbf{P}$ of the parabolas (I) and (I) be $\left(\mathrm{x}_{0}, \mathrm{y}_{0}\right)$. Since $\mathbf{P}$ lies on both parabolas, we have
$$
\begin{array}{l}
y_{0}=x_{0}^{2}-2 \\
x_{0}+2=-x_{0}^{2}+ \\
a x_{0}+b, \\
\therefore \quad 2 x_{0}^{2}- \\
(a+2) \quad x_{0}+2 \\
-b=0 .(12)
\end{array}
$$

Furthermore, the slopes of the tangents to the parabolas (I) and (I) at point $\mathbf{P}$ are
$$
k_{1}=2 x_{0}-2, \quad k_{2}=-2 x_{0}+a .
$$

Since the two tangents are perpendicular to each other, we have
$$
k_{1} \cdot k_{d}=\left(2 x_{0}-2\right)\left(-2 x_{0}+a\right)=-1,
$$

which simplifies to
$$
4 x_{0}^{2}-2(a+2) x_{0}+2 a-1=0 \text {. }
$$

By eliminating $x_{0}$ from equations (12) and (13), we obtain
$$
\mathrm{a}+\mathrm{b}=\frac{5}{2} \text {. }
$$
Equation (14) is the relationship between $\mathrm{a}$ and $\mathrm{b}$.
2. Substituting $\mathrm{b}=\frac{5}{2}-\mathrm{a}$ into the equation of the parabola (I), we get
$$
y=-x^{2}+a(x-1)+\frac{5}{2} \text {. }
$$

From equation (15), it is clear that when $\mathrm{x}=1$, regardless of the value of $\mathrm{a}$, we always have $y=\frac{3}{2}$, meaning the parabola always passes through the fixed point $Q\left(1, \frac{3}{2}\right)$, which is independent of $a$ and $b$ (Figure 3).