Q105 / 180 · KCET 2023 · Mathematics

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
  • Correct answer
    D
    I or III

Solution

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.

Hence, the correct answer is option: 'D'.

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