Example 4 Given $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$, find the minimum value of $\frac{x+y-1}{y+2}$.

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Parse $\frac{\text { Let } \frac{x+y-1}{y \pm 2}}{y}=t$, rearranging gives $x-y-t y=2 t +1$. The given equation is $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$. Using the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
9+4(1-t)^{2}=\left[9+4(1-t)^{2}\right]\left[\frac{x^{2}}{9}+\frac{y^{2}}{4}\right] \\
\geqq \underline{(x+y-t y)^{2}} \equiv \underline{(2 t+1)^{2}}
\end{array}$$

Solving gives $t \leqslant 1$, hence the maximum value sought is 1.