KCET · Maths · Application of Derivatives
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
- A II or IV
- B III or IV
- C II or III
- D I or III
Answer & Solution
Correct Answer
(D) I or III
Step-by-step Solution
Detailed explanation
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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