Step 1: Find the slope of the curve $xy = 1$.
Differentiate implicitly: $y + x\dfrac{dy}{dx} = 0 \Rightarrow \dfrac{dy}{dx} = -\dfrac{y}{x}$.
Step 2: Find the slope of the normal.
The normal is perpendicular to the tangent, so its slope is $\dfrac{x}{y}$. Since $xy = 1$, $y = 1/x$, so slope of normal $= \dfrac{x}{1/x} = x^2$.
Step 3: Note the sign of the normal slope.
$x^2 \geq 0$ for all real $x$. So the slope of the normal to $xy = 1$ is always non-negative. More precisely, $x^2 > 0$ for $x
e 0$, so the slope is strictly positive.
Step 4: Relate to the given line $ax + by + c = 0$.
The slope of this line is $-a/b$. For this to be a normal, $-a/b = x^2 > 0$, which means $-a/b > 0$.
Step 5: Determine the signs of $a$ and $b$.
$-a/b > 0$ means $a$ and $b$ have opposite signs. So either $a > 0, b < 0$ or $a < 0, b > 0$.
Step 6: Choose the correct option.
The answer is $a > 0$, $b < 0$ (option 2).
\[ \boxed{a > 0,\ b < 0} \]