A particle moves along the curve $y^2=16x .$ A point on this curve at which the ordinate increases at twice the rate of abscissa is :

(1, 4)

The correct answer is Option (4) → (1, 4)

the curve equation,

$y^2=16x$

$⇒2y\frac{dy}{dx}=16\frac{dx}{dt}$

$⇒y\left(2\frac{dx}{dt}\right)=16\frac{dx}{dt}$

$⇒2y=8$

$⇒y=4$

and,

$⇒x=\frac{y^2}{16}=\frac{16}{16}=1$

∴ The required point, (1, 4).