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A particle moves along the curve $y=x^2+2 x$. Then, the point on the curve such that $x$ and $y$ coordinates of the particle change with the same rate is
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Verified Answer
The correct answer is:
$\left(-\frac{1}{2},-\frac{3}{4}\right)$
Given equation of curve is
$$
y=x^2+2 x
$$
On differentiating both sides w.r.t. $t$, we get
$$
\frac{d y}{d t}=(2 x+2) \frac{d x}{d t}
$$
$$
\begin{aligned}
\Rightarrow & & 2 x & =-1 \\
\Rightarrow & & x & =-1 / 2, y=-3 / 4
\end{aligned}
$$
$\therefore$ Point on the curve is $\left(-\frac{1}{2},-\frac{3}{4}\right)$.
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