A particle moves along the curve 6x = y^3 + 2 . The points on the cur…

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

Differentiating the curve with respect to time t (rates of change):

    \( \displaystyle 6\frac{dx}{dt} = 3y^2\frac{dy}{dt} \)

The x coordinate changes 8 times as fast as y:

    \( \displaystyle \frac{dx}{dt} = 8\frac{dy}{dt} \)

    \( \displaystyle 48\frac{dy}{dt} = 3y^2\frac{dy}{dt} \)

The particle is moving, so \(\frac{dy}{dt} \ne 0\):

    \(y^2 = 16,\quad y = \pm 4\)

The x coordinates from \(6x = y^3 + 2\):

    \(y = 4:\ 6x = 66,\ x = 11\)

    \( \displaystyle y = -4:\ 6x = -62,\ x = -\frac{31}{3} \)

So the points are (11, 4) and \(\left(-\frac{31}{3}, -4\right)\).