If the line ax + by + c = 0 is normal to the curve xy + 5 = 0 then

a > 0, b < 0 b > 0 , a < 0 a < 0, b < 0 None of these

a < 0, b < 0

$xy = -5 ⇒ x \frac{dy}{dx}+ y = 0 ⇒ \frac{dy}{dx}=-\frac{y}{x} > 0$ (as xy = -5 < 0) ⇒ The slope of the normal is negative $⇒-\frac{a}{b}< 0 ⇒ \frac{a}{b}> 0$ ⇒ a > 0, b > 0 or a < 0, b < 0. Hence (C) is the correct answer.