A particle moves along the curve frac{x^2}{16 | vedclass

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Question

(A) Given the curve equation: $\frac{x^2}{16} + \frac{y^2}{4} = 1$. Differentiating both sides with respect to $t$: $\frac{2x}{16} \frac{dx}{dt} + \fr

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$II$ or $IV$

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(A) Given the curve equation: $\frac{x^2}{16} + \frac{y^2}{4} = 1$. Differentiating both sides with respect to $t$: $\frac{2x}{16} \frac{dx}{dt} + \frac{2y}{4} \frac{dy}{dt} = 0$. $\frac{x}{8} \frac{dx}{dt} + \frac{y}{2} \frac{dy}{dt} = 0$. Given that the rate of change of abscissa $(\frac{dx}{dt})$ is $4$ times the rate of change of ordinate $(\frac{dy}{dt})$,i.e.,$\frac{dx}{dt} = 4 \frac{dy}{dt}$. Substituting this into the differentiated equation: $\frac{x}{8} (4 \frac{dy}{dt}) + \frac{y}{2} \frac{dy}{dt} = 0$. $(\frac{x}{2} + \frac{y}{2}) \frac{dy}{dt} = 0$. Since $\frac{dy}{dt} 
eq 0$,we have $\frac{x}{2} + \frac{y}{2} = 0$,which implies $x = -y$. Substituting $x = -y$ into the original curve equation: $\frac{(-y)^2}{16} + \frac{y^2}{4} = 1$. $\frac{y^2}{16} + \frac{4y^2}{16} = 1 \Rightarrow \frac{5y^2}{16} = 1 \Rightarrow y^2 = \frac{16}{5}$. Thus,$y = \pm \frac{4}{\sqrt{5}}$. Since $x = -y$,if $y = \frac{4}{\sqrt{5}}$,then $x = -\frac{4}{\sqrt{5}}$ (Quadrant $II$). If $y = -\frac{4}{\sqrt{5}}$,then $x = \frac{4}{\sqrt{5}}$ (Quadrant $IV$). Therefore,the particle lies in the $II$ or $IV$ quadrant.

$A$ 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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