Given curve is $x y=1$
So, $y=\frac{1}{x}$ ...(i)
Differentiating with respect to $x$, we get,
$x \frac{\mathrm{d} y}{\mathrm{~d} x}+y=0$ $[\because(u v) \prime=u v \prime+u \prime v]$
$\Rightarrow \quad \frac{\mathrm{d} y}{\mathrm{~d} x}=-\frac{y}{x}$
Given line is, $a x+b y+c=0$
The slope of a line is the coefficient of $x$ in
$y=m x+c$ form.
Thus, the slope of the given line is, $-\frac{a}{b}$
Thus the slope of tangent is,
$\frac{b}{a},\left[\because\right.$ slope of normal $\left.=-\frac{1}{\text { slope of line }}\right]$
Since the slope is normal to the curve, the normal of the line will be tangent to the curve. Thus,
$-\frac{y}{x}=\frac{b}{a}$
$\Rightarrow-\frac{\frac{1}{x}}{x}=\frac{b}{a}$ $\left[\because y=\frac{1}{x}\right]$
So, LHS will always be a positive value, so $\frac{a}{b}$ in the RHS must be negative. For that to happen, $a$ and $b$ must have the opposite signs.
Thus, $a>0, b < 0$ or $a < 0, b>0$.