Example 2 Given the equation of curve $c$ is
$$
(t+1) x^{2}+y^{2}-2\left(a^{2}+2 a t\right) x+3 a t+b=0 \text {. }
$$

For any real number $t$, curve $c$ always passes through the fixed point $P(1,0)$. Find the constants $a$ and $b$.
Strategy: Substitute $P(1,0)$ into the curve equation, isolate the parameter $t$, and then set the coefficients of the parameter to zero to determine $a$ and $b$.

Solution: Substituting $P(1,0)$ into the curve equation, we get
$$
t+1-2\left(a^{2}+2 a t\right)+3 a t+b=0 \text {. }
$$

Rearranging, we have
$$
\begin{array}{l}
(1-a) t+1-2 a^{2}+b=0 . \\
\therefore\left\{\begin{array} { l } 
{ 1 - a = 0 , } \\
{ 1 - 2 a ^ { 2 } + b = 0 . }
\end{array} \text { Solving, we get } \left\{\begin{array}{l}
a=1, \\
b=1 .
\end{array}\right.\right.
\end{array}
$$