$\begin{array}{1 1} \big(-4,\frac{-31}{3}\big)\; and (4,11) \\ \big(-4,\frac{31}{3}\big)\; and\;(4,11) \\\big(-4,\frac{-31}{3}\big)\; and \;(-4,11) \\ \big(-4,\frac{-31}{3}\big)\; and \;(4,-11) \end{array} $

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- If $y=f(x)$,then $\large\frac{dy}{dx}$ measures the rate of change of $y$ w.r.t $x$.
- $\big(\large\frac{dy}{dx}\big)_{x=x_0}$ represents the rate of change of $y$ w.r.t $x$ at $x=x_0$

Step 1:

Let the required point be $P(x,y)$.It is given that rate of change of $y$-coordinate =8(Rate of change of $x$-coordinate)

(i.e)$\large\frac{dy}{dt}$$=8.\large\frac{dx}{dt}$

It is given $6y=x^3+2$

On differentiating w.r.t $t$ on both sides,

$6\large\frac{dy}{dt}$$=3x^2\large\frac{dx}{dt}$$+2$

$\Rightarrow 6.\big(8\large\frac{dx}{dt}\big)=$$3x^2\large\frac{dx}{dt}$

$\Rightarrow 3x^2=48$

$\Rightarrow x^2=16$

$\Rightarrow x=\pm 4$

Step 2:

When $x=4$,then $6y=4^3+2$

$\Rightarrow 66$

Therefore $y=11$

When $x=-4$,then $6y=(-4)^3+2$

$\Rightarrow -62$

$y=\large\frac{-31}{3}$

Hence the required points are $\big(-4,\large\frac{-31}{3}\big)$ and $(4,11)$.

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