The eccentricity of the ellipse (x2/4)+(y2/3)=1 is changed at the rate of 0.1 m / sec. The time at which it will touch the auxiliary circle is

Question

The eccentricity of the ellipse (x2/4)+(y2/3)=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

Tardigrade Admin · Accepted Answer

(de/dt)=-0.1 m / sec ⇒ e=-0.1 t+e0 When e0=1 / 2 et=-0.1 t+(1/2), given et=0 ⇒ 0.1 t=(1/2) or t=5 seconds

Q. 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

2 seconds

3 seconds

6 seconds

5 seconds

$\frac{d e}{d t} = - 0.1 m / sec$
$\Rightarrow e = - 0.1 t + e_{0}$
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

Q. The eccentricity of the ellipse 4x2​+3y2​=1 is changed at the rate of 0.1m/sec. The time at which it will touch the auxiliary circle is