In a Mahakumbh, a drone camera is moving along \(3y = x^3 - 3\). When \(y\)-coordinate changes \(9\) times as fast as \(x\)-coordinate, it captures good quality pictures. Then one of the precise positions of the drone at that instant is

The equation of the curve is given by \(3y = x^3 - 3\).

Differentiating both sides with respect to time \(t\), we get:

\(3 \dfrac{dy}{dt} = 3x^2 \dfrac{dx}{dt}\)

\(\dfrac{dy}{dt} = x^2 \dfrac{dx}{dt}\)

Given that the \(y\)-coordinate changes \(9\) times as fast as the \(x\)-coordinate:

\(\dfrac{dy}{dt} = 9 \dfrac{dx}{dt}\)

Equating the two expressions for \(\dfrac{dy}{dt}\):

\(x^2 \dfrac{dx}{dt} = 9 \dfrac{dx}{dt}\)

Assuming \(\dfrac{dx}{dt} \neq 0\), we get \(x^2 = 9\), which gives \(x = 3\) or \(x = -3\).

For \(x = 3\), substituting into the curve equation gives \(3y = (3)^3 - 3 = 24 \Rightarrow y = 8\). The point is \((3, 8)\).

For \(x = -3\), substituting into the curve equation gives \(3y = (-3)^3 - 3 = -30 \Rightarrow y = -10\). The point is \((-3, -10)\).

Among the given options, \((3, 8)\) is the correct position.

Answer: \((3, 8)\)