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

1. Differentiate $6y=x^3+2$ w.r.t. $t$: $6\dfrac{dy}{dt}=3x^2\dfrac{dx}{dt}$. 2. Condition $\dfrac{dy}{dt}=8\dfrac{dx}{dt}$, so $6(8)=3x^2 \Rightarrow x^2=16 \Rightarrow x=\pm 4$. 3. At $x=4$: $y=\dfrac{4^3+2}{6}=\dfrac{66}{6}=11$. 4. At $x=-4$: $y=\dfrac{-64+2}{6}=\dfrac{-62}{6}=-\dfrac{31}{3}$. 5. The trap options use wrong roots or drop one point. _Source: NCERT Class 12 Mathematics Ch 6 "Application of Derivatives", p.4_