The slope of the normal to the curve y^2=16x at the point (1, 4) is :

The correct answer is Option (2) → $-\frac{1}{2}$

$y^2=16x$

differentiating wrt x

$2y\frac{dy}{dx}=16⇒\frac{dy}{dx}=\frac{8}{y}$ slope of normal

→ $-\frac{dx}{dy}=-\frac{y}{8}$

$\left.-\frac{dx}{dy}\right]_{(1,4)}=-\frac{4}{8}=-\frac{1}{2}$