Question

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. III or IV
• B. I or III
• C. II or III
• D. II or IV

Hint:

When dealing with a particle moving along a curve, use implicit differentiation to find relationships between the rates of change of the coordinates. In this case, the rate of change of the abscissa is given as 4 times the rate of the ordinate, and this condition helps determine the quadrant where the particle lies. Always remember to consider the signs of the coordinates when analyzing the situation geometrically.

Answer 1

The Correct Option is D

The correct answer is (D) : II or IV.

Answer 2

The correct answer is: (D) II or IV.

A particle moves along the curve:

$\frac{x^2}{16} + \frac{y^2}{4} = 1$

We are told that the rate of change of the abscissa (i.e., $\frac{dx}{dt}$) is 4 times that of its ordinate (i.e., $\frac{dy}{dt}$).
Step 1: Differentiate the equation with respect to time $t$

Differentiate the given equation implicitly with respect to $t$: $$\frac{d}{dt} \left( \frac{x^2}{16} + \frac{y^2}{4} \right) = \frac{d}{dt}(1)$$ Applying the chain rule: $$\frac{2x}{16} \frac{dx}{dt} + \frac{2y}{4} \frac{dy}{dt} = 0$$ Simplifying: $$\frac{x}{8} \frac{dx}{dt} + \frac{y}{2} \frac{dy}{dt} = 0$$ Step 2: Use the given condition that $\frac{dx}{dt} = 4 \frac{dy}{dt}$

Substituting $\frac{dx}{dt} = 4 \frac{dy}{dt}$ into the equation: $$\frac{x}{8} (4 \frac{dy}{dt}) + \frac{y}{2} \frac{dy}{dt} = 0$$ Simplifying: $$\frac{x}{2} \frac{dy}{dt} + \frac{y}{2} \frac{dy}{dt} = 0$$ Factor out $\frac{dy}{dt}$: $$\frac{dy}{dt} \left( \frac{x}{2} + \frac{y}{2} \right) = 0$$ Since $\frac{dy}{dt} \neq 0$, we have: $$\frac{x}{2} + \frac{y}{2} = 0$$ Therefore: $$x + y = 0$$ Step 3: Determine the quadrant

The equation $x + y = 0$ represents a line passing through the origin with a slope of -1. 
This line divides the coordinate plane into two parts. 
The particle lies on this line, and we are asked to determine in which quadrant the particle lies when $x + y = 0$ and the given condition holds. 
- If $x + y = 0$ and $x > 0$, then $y < 0$, which places the particle in the IV quadrant. 
- If $x + y = 0$ and $x < 0$, then $y > 0$, which places the particle in the II quadrant. 
Therefore, the particle can lie in either the II or IV quadrant. Thus, the correct answer is (D) II or IV.