Explanation

(b)
The given equation of curve is $xy = 1$
$\Rightarrow \ \ \ \ \ x\frac{dy}{dx} = - y\ \ \ \ \Rightarrow \ \ \ \frac{dy}{dx} = - \frac{y}{x}$
Let $\left( t,\frac{1}{t} \right)$ be any point on the curve at which normal to the curve can be drawn
$\therefore\ \ \ \ \ \left( \frac{dy}{dx} \right)_{\left( t,\frac{1}{t} \right)} = - \frac{1}{t^{2}}$
So, slope of normal$= t^{2}$
Therefore, the given line $ax + bx + c = 0$ will be normal to the curve, if
$t^{2} = \frac{- b}{a}$
Since, $t^{2} > 0$
$\therefore$ Either $b > 0,\ \ \ \ a < 0$
or $a > 0,\ \ \ b < 0$