Example 3 Given $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$, find the maximum value of $\frac{x+y-1}{y+2}$.

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Let $t=\frac{x+y-1}{y+2}$, then $x+y-t y=2 t+1$. The original
equation can be transformed into
$$\frac{x^{2}}{9}+\frac{(1-t)^{2} y^{2}}{4(1-t)^{2}}=1$$

By Theorem 1, we get
$$9+4(1-t)^{2} \geqslant(x+y-t y)^{2}=(2 t+1)^{2}$$

Solving this, we get $t \leqslant 1$, hence the maximum value of $\frac{x+y-1}{y+2}$ is 1.