A particle moves along the curve $6y= x^3 + 2$. Find the points on the curve at which y-coordinate is changing 8 times as fast as the x-coordinate.

$(4,11)$ and $(−4,−\frac{31}{3})$

The correct answer is Option (1) → $(4,11)$ and $(−4,−\frac{31}{3})$

The given curve is $6y=x^3+2$   …(i)

Diff. (i) w.r.t. t, we get $6\frac{dy}{dx}=3x^2\frac{dx}{dt}$   …(ii)

But $\frac{dy}{dt}=8\frac{dx}{dt}$ (given),

$6×8\frac{dx}{dt}=3x^2\frac{dx}{dt}$ (Using (ii))

$⇒48=3x^2 ⇒ x^2 = 16⇒x=4,-4$

From (i), when $x = 4, 6y = 4^3 + 2 = 66⇒ y = 11,$

when $x = -4, 6y= (-4)^3 + 2 = −62 ⇒y=-\frac{31}{3}$

Hence, the required points on the curve are $(4,11),(−4,−\frac{31}{3})$.