QPaperGen

We start with the equation of the curve:
\[6y = x^3 + 2\]

Differentiating both sides with respect to time \(t\) gives:
\[6\frac{dy}{dt} = 3x^2 \frac{dx}{dt}\]

Dividing both sides by 3:
\[2\frac{dy}{dt} = x^2 \frac{dx}{dt}\]

According to the problem, the \(y\)-coordinate is changing 8 times as fast as the \(x\)-coordinate, i.e.,
\[\frac{dy}{dt} = 8\frac{dx}{dt}\]

Substituting this into the equation:
\[2(8\frac{dx}{dt}) = x^2 \frac{dx}{dt}\]
\[\Rightarrow 16\frac{dx}{dt} = x^2 \frac{dx}{dt}\]

Assuming \(\frac{dx}{dt} \ne 0\), we cancel it out. So :
\[x^2 = 16 \Rightarrow x = \pm 4\]

Now, we substitute these \(x\) values back into the original equation to find \(y\):
\[\]

When \(x = 4\):
\[y = \frac{4^3 + 2}{6} = \frac{64 + 2}{6} = \frac{66}{6} = 11\]

When \(x = -4\):
\[y = \frac{(-4)^3 + 2}{6} = \frac{-64 + 2}{6} = \frac{-62}{6} = -\frac{31}{3}\]

Final Answer:
The required points on the curve are \((4,\ 11)\) and \(\left(-4,\ -\frac{31}{3}\right)\)