The point on the curve $y^2=16 x$ for which the y-coordinate is changing 2 times as fast as the x-coordinate is:

(1, 4)

The correct answer is Option (3) - (1, 4)

$y^2 = 16x$

$2y\frac{dy}{dt} = 16\frac{dx}{dt}$

$\frac{dy}{dt} = \frac{8}{y}\frac{dx}{dt}$

$\text{Given } \frac{dy}{dt} = 2\frac{dx}{dt}$

$\frac{8}{y}\frac{dx}{dt} = 2\frac{dx}{dt}$

$\frac{8}{y} = 2$

$y = 4$

$y^2 = 16x \Rightarrow 16 = 16x \Rightarrow x = 1$

The required point is $(1, 4)$.