Thank you for reporting, we will resolve it shortly
Q.
If the line $ax + by + c =0 $ is a normal to the curve $xy = 1$, then
Conic Sections
Solution:
Since $m$ of the normal is always $+ve$ in this case
$ a > 0, b < 0$ or $a < 0, b > 0$
$\quad\left( {\text{slope}}\,\, m= -\frac{a}{b}>0\right) $
[$y=\frac{1}{x} \therefore \frac{dy}{dx}= -\frac{1}{x^{2}} \therefore -\frac{1}{dy/dx } = x^{2}>0$
for all $x \therefore $ slope of normal is $ > 0$]