A particle moves along the curve $6x=y^3+2$. The points on the curve at which the $x$ coordinate is changing 8 times as fast as y coordinate are:

$(11, 4),\left(-\frac{31}{3},-4\right)$

The correct answer is Option (4) → $(11, 4),\left(-\frac{31}{3},-4\right)$

$6x=y^3+2$ [Given Curve]

$\frac{d}{dt}(6x)=\frac{d}{dt}(y^3+2)$

$6\frac{dx}{dt}=3y^2\frac{dy}{dt}$

and,

$\frac{dx}{dt}=3y^2\frac{dy}{dt}$  [Given]

$⇒6\left(8\frac{dy}{dt}\right)=3y^2\left(\frac{dy}{dt}\right)$

$⇒48\frac{dy}{dt}=3y^2\frac{dy}{dt}$

$⇒3y^2=48$

$⇒y=±4$

Substituting $±4$ in the given equation,

$6x=(±4)^3+2$

$⇒x=11$ and $x=-\frac{31}{3}$