Example 5 If the line $y=m x+b$ intersects the hyperbola $(x-1)^{2}-a^{2} y^{2}=a^{2}$ for any real number $m$, find the conditions that the real numbers $a$ and $b$ must satisfy.
Strategy: After combining the two equations, obtain a quadratic inequality about $m$ from $\Delta_{1} \geqslant 0$, isolate the parameter $m$, and then discuss to find the relationship between $a$ and $b$.
Solution: From $\left\{\begin{array}{l}y=m x+b, \\ (x-1)^{2}-a^{2} y^{2}=a^{2}\end{array}\right.$ we get
$$
\begin{array}{l}
\left(1-a^{2} m^{2}\right) x^{2}-2\left(1+m b a^{2}\right) x-\left(a^{2} b^{2}\right. \\
\left.+a^{2}-1\right)=0 . \\
\therefore \Delta_{1}=4\left(1+m b a^{2}\right)^{2}+4\left(1-a^{2} m^{2}\right) \\
\quad \cdot\left(a^{2} b^{2}+a^{2}-1\right) \geqslant 0 .
\end{array}
$$

Rearranging gives $\left(1-a^{2}\right) m^{2}+2 b m+b^{2}+1 \geqslant 0$ for all $m \in \mathbf{R}$.
$$
\therefore 1-a^{2}=b=0 \text {, }
$$

or $\left\{\begin{array}{l}1-a^{2}>0, \\ \Delta_{2}=4 b^{2}-4\left(1-a^{2}\right)\left(b^{2}+1\right) \leqslant 0 \text {. }\end{array}\right.$
Therefore, the conditions that $a$ and $b$ must satisfy are
$$
a^{2} b^{2}+a^{2}-1 \leqslant 0 \text {. }
$$