22. As shown in the figure, points $A, B$ are on the graph of the inverse proportion function $y=\frac{k_{1}}{x}$, and points $C, D$ are on the graph of the inverse proportion function $y=\frac{k_{2}}{x}$. $A C \perp x$-axis at point $E$, $B D \perp x$-axis at point $F$, $A C=2$, $B D=3$, $E F=\frac{10}{3}$, then $k_{2}-k_{1}=$ ( ).
(A) 4
(B) $\frac{14}{3}$
(C) $\frac{16}{3}$
(D) 6
Answer: (A)
Solution: Let $A\left(a, \frac{k_{1}}{a}\right), B\left(b, \frac{k_{1}}{b}\right)$, then $C\left(a, \frac{k_{2}}{a}\right), D\left(b, \frac{k_{2}}{b}\right)$.
Since

Therefore,
$$
\begin{array}{c}
A C=2, \quad B D=3, \\
\frac{k_{1}}{a}-\frac{k_{2}}{a}=2, \frac{k_{2}}{b}-\frac{k_{1}}{b}=3 .
\end{array}
$$

Thus,
$$
k_{1}-k_{2}=2 a, k_{2}-k_{1}=3 b \text {, }
$$

We get
$$
-2 a=3 b
$$

So,
$$
\begin{aligned}
b & =-\frac{2}{3} a \\
E F & =\frac{10}{3},
\end{aligned}
$$

That is,
$$
b-a=\frac{10}{3} \text {, }
$$

Therefore,
$$
-\frac{2}{3} a-a=\frac{10}{3}
$$

Solving for $a$,
$$
a=-2 \text {, }
$$

Thus,
$$
k_{2}-k_{1}=-2 a=-2 \times(-2)=4 .
$$