An insect moves along the ellipse $\frac{x^2}{16}+\frac{y^2}{4}=1$, when the rate of change of abscissa is 4 times that of ordinate, then the insect lies in the quadrant :

A. I or II

B. III or IV

C. II or III

D. II or IV

Choose the correct answer from the options given below :

D only

The correct answer is Option (1) → D only

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

$\frac{2x}{16}+\frac{2y}{4}\frac{dy}{dx}=0$

$\frac{y}{2}\frac{dy}{dx}=-\frac{x}{8}$

$\frac{dy}{dx}=-\frac{x}{y}$

we are given rate of change of abscissa is 4 times ordinate $⇒\frac{dx}{dy}=4$

so $-\frac{x}{y}=\frac{1}{4}$ so $-4x=y$

$4x+y=0$

from (1)

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

$\frac{65x^2}{16}=1⇒x=±\sqrt{\frac{16}{65}}⇒y=±4\sqrt{\frac{16}{65}}$

⇒ Point in 2nd or 4th quadrant