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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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$(1, 3)$

$(\frac{1}{2}, \frac{3}{4})$

$(- \frac{1}{2}, - \frac{3}{4})$

$(- 1, - 1)$

answer is C.

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Detailed Solution

$\frac{d y}{d t} = (2 x + 2) \frac{d x}{d t}$

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