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 :
Answer & explanation
Correct answer: option 1
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