(b) 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
$\left( {\frac{x}{\alpha } + \frac{y}{\beta } – 1} \right) + \lambda \left( {\frac{x}{\beta } + \frac{y}{\alpha } – 1} \right) = 0$

or $x\,\left( {\frac{1}{\alpha } + \frac{\lambda }{\beta }} \right) + y\left( {\frac{1}{\beta } + \frac{\lambda }{\alpha }} \right) – \lambda – 1 = 0$

This meets the axes at

$A{\rm{ }}\left( {\frac{{\lambda + 1}}{{\frac{1}{\alpha } + \frac{\lambda }{\beta }}},0} \right)$ and $B{\rm{ }}\left( {0,\frac{{\lambda + 1}}{{\frac{1}{\beta } + \frac{\lambda }{\alpha }}}} \right)$.

Let $(h, k)$ be the mid point of $AB$,

then $h = \frac{1}{2}.\frac{{\lambda + 1}}{{\frac{1}{\alpha } + \frac{\lambda }{\beta }}},k = \frac{1}{2}.\frac{{\lambda + 1}}{{\frac{1}{\beta } + \frac{\lambda }{\alpha }}}$

Eliminating $\lambda $from these two, we get

$2hk(\alpha + \beta ) = \alpha \beta (h + k)$.

 The locus of $(h,k)$is $2xy(\alpha + \beta ) = \alpha \beta (x + y)$.