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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
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
Updated On: Sep 17, 2024
- III or IV
- I or III
- II or III
- II or IV
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The Correct Option is D
Solution and Explanation
The correct answer is (D) : II or IV.
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