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A particle moves along the curve $\frac{x^2}{16}+\frac{y^2}{4}=1$. When the rate of change of abscissa is 4 times that of its ordinate, then the quadrant in which the particle lies is
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The correct answer is:
I or III
Given, particles moves along the curve
$\frac{x^2}{16}+\frac{y^2}{4}=$
$\Rightarrow \frac{2 x}{16}+\frac{2 y}{4} \frac{d y}{d x}=0$
$\therefore \quad \frac{d y}{d x}=\frac{-2 x}{16} \times \frac{4}{2 y}=\frac{-x}{4 y}$
According to the question, $x=4 y$
From Eq. (i), we get
$\frac{d y}{d x}=-1$
When slope is negative, then the particle lies in Ist and IIIrd quadrant.
$\frac{x^2}{16}+\frac{y^2}{4}=$
$\Rightarrow \frac{2 x}{16}+\frac{2 y}{4} \frac{d y}{d x}=0$
$\therefore \quad \frac{d y}{d x}=\frac{-2 x}{16} \times \frac{4}{2 y}=\frac{-x}{4 y}$
According to the question, $x=4 y$
From Eq. (i), we get
$\frac{d y}{d x}=-1$
When slope is negative, then the particle lies in Ist and IIIrd quadrant.
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