Question

A particle moves along the curve $6y=x^3+2$. The point '$P$' on the curve at which the y-coordinate is changing $8$ times as fast as the $x$ -coordinate, are $(4,11)$ and $\left(-4,-\frac{31}{3}\right)$.

• A. $x$ -coordinates at the point $P$ are $\pm 4$
• B. y-coordinates at the point $P$ are 11 and $\frac{-31}{3}$
• C. Both (a) and (b)
• D. None of the above

Answer 1

Answer: (C)

Given, $6y=x^3+2$
On differentiating w.r.t. t, we get
$6\frac{dy}{dt}=3x^2\frac{dx}{dt}$
$\Rightarrow 6\times 8\frac{dx}{dt}=3x^2\frac{dx}{dt}$
$\Rightarrow 3x^2=48$
$\Rightarrow x^2=16$
$\Rightarrow x=\pm 4$
When $x=4$, then $6y=(4{)}^3+2$
$\Rightarrow 6y=64+2$
$\Rightarrow y=\frac{66}{6}=11$
When $x=-4$, then $6y=(-4{)}^3+2$
$\Rightarrow 6y=-64+2$
$\Rightarrow y=\frac{-62}{6}=\frac{-31}{3}$
Hence, the required points on the curve are
$(4,11)$ and $\left(-4,\frac{-31}{3}\right)$