Explanation

(b)
The equation of line passing through the point of intersection of
$\frac{x}{\alpha} + \frac{y}{\beta} = 1\ and\frac{x}{\beta} + \frac{y}{\alpha} = 1\ is$
$\left( \frac{x}{\alpha} + \frac{\lambda}{\beta} - 1 \right) + \lambda\left( \frac{x}{\beta} + \frac{\lambda}{\alpha} - 1 \right) = 0$
$\Rightarrow \ x\left( \frac{1}{\alpha} + \frac{\lambda}{\beta} \right) + \lambda\left( \frac{1}{\beta} + \frac{\lambda}{\alpha} \right) - \lambda - 1 = 0$
This meets the coordinate axes at $A\left( \frac{\lambda + 1}{\frac{1}{\alpha} + \frac{\lambda}{\beta}},\ 0 \right)\ \ and\ B\left( 0,\frac{\lambda + 1}{\frac{1}{\beta} + \frac{\lambda}{\alpha}} \right)$
Let $(h,\ k)$ be the mid point of $AB$. Then, $h = \frac{1}{2}\left( \frac{\lambda + 1}{\frac{1}{\beta} + \frac{\lambda}{\alpha}} \right)\ and\ k = \frac{1}{2}\left( \frac{\lambda + 1}{\frac{1}{\beta} + \frac{\lambda}{\alpha}} \right)$
On eliminating $\lambda$ from these two equations, we get
$2hk(\alpha + \beta) = \alpha\beta(h + k)$
Hence, the locus of $(h,\ k)$ is $2xy(\alpha + \beta) = \alpha\beta(x + y)$