A particle moves along the curve $6y=x^3+2$, the point(s) on the curve at which y coordinate is/are changing 8 times as fast as x coordinate are :

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

The correct answer is Option (1) → $(4, 11)$ and $(-4, \frac{-31}{3})$

$6y=x^3+2$

so $6\frac{dy}{dx}=3x^2⇒\frac{dy}{dx}=\frac{x^2}{2}$

given y coordinate changes 8 times as fast as x

$⇒\frac{dy}{dx}=8⇒\frac{x^2}{2}=8⇒x=±4$

so $6y=±64+2$

$y=\frac{±64+2}{6}$

so $(x,y)=(4, 11)$ or $(-4, \frac{-31}{3})$