A particle moves along the curve $8y=x^3+7.$ The points on the curve at which y-coordinate is changing 6 times as fast as the x-coordinate are :

$(4, \frac{71}{8}) \left(-4, -\frac{57}{8}\right)$

The correct answer is Option (2) → $(4, \frac{71}{8}) \left(-4, -\frac{57}{8}\right)$

$8y=x^3+7$   [Given]

$\frac{d(8y)}{dx}=\frac{d(x^3+7)}{dx}$

$8\frac{dy}{dx}=3x^2⇒\frac{dy}{dx}=\frac{3x^2}{8}$

and,

$\frac{3x^2}{8}=6$

$x^2=16⇒x±4$

At $x = 4$,

$8y=4^3+7$

$y=\frac{71}{8}$

At $x=-4$

$8y=(-4)^3+7$

$y=-\frac{57}{8}$