The equation of a line passing through the intersection of straight lines $\frac{x}{\alpha}+\frac{y}{\beta}=1$ and $\frac{x}{\beta}+\frac{y}{\alpha}=1$ is
$\begin{aligned}
& \left(\frac{x}{\alpha}+\frac{y}{\beta}-1\right)+\lambda\left(\frac{x}{\beta}+\frac{y}{\alpha}-1\right)=0 \\
& x\left(\frac{1}{\alpha}+\frac{\lambda}{\beta}\right)+y\left(\frac{1}{\beta}+\frac{\lambda}{\alpha}\right)-\lambda-1=0
\end{aligned}$
or
This meets the axes at
$A\left(\frac{\lambda+1}{\frac{1}{\alpha}+\frac{\lambda}{\beta}}, 0\right) \text { and } B\left(0, \frac{\lambda+1}{\frac{1}{\beta}+\frac{\lambda}{\alpha}}\right)$

Let $(h, k)$ be the mid point of $A B$,
$h=\frac{1}{2} \cdot \frac{\lambda+1}{\frac{1}{\alpha}+\frac{\lambda}{\beta}}, k=\frac{1}{2} \cdot \frac{\lambda+1}{\frac{1}{\beta}+\frac{\lambda}{\alpha}}$
Eliminating $\lambda$ from these two, we get
$2 h k(\alpha+\beta)=\alpha \beta(h+k)$
$\therefore$ The locus of $(h, k)$ is $2 x y(\alpha+\beta)=\alpha \beta(x+y)$