Example 1 Suppose $a>b>0, \alpha, \beta \in\left(0, \frac{\pi}{2}\right), a \tan \beta=b \tan \alpha$, find the maximum value of $\alpha-\beta$.


This is an easy problem combining trigonometric functions and the inequality of averages. The solution is as follows:
$\tan (\alpha-\beta)=\frac{\tan \alpha-\tan \beta}{1+\tan \alpha \tan \beta}=\frac{\tan \alpha-\frac{b}{a} \tan \alpha}{1+\tan \alpha \cdot \frac{b}{a} \tan \alpha}=\frac{a-b}{\frac{a}{\tan \alpha}+b \tan \alpha} \leqslant \frac{a-b}{2 \sqrt{\frac{a}{\tan \alpha} \cdot b \tan \alpha}}$ $=\frac{a-b}{2 \sqrt{a b}}$, equality holds when $\tan \alpha=\sqrt{\frac{a}{b}}$. Since $\alpha-\beta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$,
the maximum value of $\alpha-\beta$ is $\arctan \frac{a-b}{2 \sqrt{a b}}$.
(1) If we place this in the context of vectors, we have:

Given $a>b>0, \alpha, \beta \in\left(0, \frac{\pi}{2}\right), \boldsymbol{a}=(a, b), \boldsymbol{b}=(\tan \alpha, \tan \beta)$, and $\boldsymbol{a} / / \boldsymbol{b}$, find the maximum value of $\alpha-\beta$.
(Hint: $\boldsymbol{a} / / \boldsymbol{b} \Leftrightarrow a \tan \beta=b \tan \alpha$)
(2) If we place this in the context of complex numbers, we have:

Given $a>b>0, \theta \in\left(0, \frac{\pi}{2}\right)$, the complex number $z=a+b \tan \theta \cdot \mathrm{i}$, find the maximum value of $\theta-\arg z$.
(Hint: By the problem, $\arg z \in\left(0, \frac{\pi}{2}\right)$, then $\tan (\arg z)=\frac{b \tan \theta}{a}=\frac{b}{a} \tan \theta$, which converts to the original proposition form.)
(3) If we place this in the context of an ellipse, we have:

As shown in Figure 1-12, $P$ is a point on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ in the first quadrant, the eccentric angle of point $P$ is $\alpha$ $\left(0<\alpha<\frac{\pi}{2}\right)$, and $\angle x O P=\beta$, find the maximum value of $\alpha-\beta$.
(Hint: As shown in the figure, since the eccentric angle of point $P$ is $\alpha\left(0<\alpha<\frac{\pi}{2}\right)$, then $\angle M O x=\alpha$.
The coordinates of point $P$ are $(a \cos \alpha, b \sin \alpha)$, thus
$\tan \beta=\frac{|P Q|}{|O Q|}=\frac{b \sin \alpha}{a \cos \alpha}=\frac{b}{a} \tan \alpha$, which converts to the original proposition form.)
(4) If we place this in the context of plane geometry, we have:

As shown in Figure 1-13, $M$ is any point on $\overparen{A_{1} A_{2}}$, connect $O M$ to intersect $B_{1} B_{2}$ at $N$, draw $M P \perp O A_{1}$, and draw $N P \perp M P$ at $P$, given $O A_{1}=a, O B_{1}=b$, find the maximum value of $\angle M O P$.
(Hint: As in (3), set $\angle M O A_{1}=\alpha, \angle P O Q=\beta$, then
$$
\tan \beta=\frac{|P Q|}{|O Q|}=\frac{b \sin \alpha}{a \cos \alpha}=\frac{b}{a} \tan \alpha .
$$