Question

The eccentricity of the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ is changed at the rate of $0.1 \,m / \sec$. The time at which it will touch the auxiliary circle is

• A. 2 seconds
• B. 3 seconds
• C. 6 seconds
• D. 5 seconds

Answer 1

Answer: (D)

$\frac{de}{dt}=-0.1 \,m / \sec$
$\Rightarrow e=-0.1 \,t+e_{0}$
When $e_{0}=1 / 2$
$ e_{t}=-0.1 t+\frac{1}{2}$, given $e_{t}=0$
$\Rightarrow 0.1 t=\frac{1}{2}$ or $t=5$ seconds