Explanation

(b)
The intersection of two curves
${ax}^{2} + 2hxy + {by}^{2} + 2gx + \lambda(a^{'}x^{2} + 2h^{'}xy + b'y^{2} + 2g'x) = 0$
$\Rightarrow \ x^{2}\left( a + a^{'}\lambda \right) + 2xy\left( h + h^{'}\lambda \right) + y^{2}\left( b + \lambda b^{'} \right) + 2x\left( g + \lambda g^{'} \right) = 0$
For making homogeneous equating, $g + \lambda g^{'} = 0$
$\Rightarrow \ \lambda = - \frac{g}{g^{'}}$
Since, lines are perpendicular.
$\therefore$ Coefficient of $x^{2} +$ Coefficient of $y^{2} = 0$
$\Rightarrow \ a + a^{'}\lambda + b + b^{'}\lambda = 0$
$\Rightarrow \ a + b = - \left( a^{'} + b^{'} \right)\left( - \frac{g}{g^{'}} \right)$
$\Rightarrow \ (a + b)g^{'} = \left( a^{'} + b^{'} \right)g$