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

$\frac{1}{2}$ $-\frac{1}{2}$ 1 -1

$-\frac{1}{2}$

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