The eccentricity of ellipse $3x^2+4y^2=12$ is changed at the rate of $0.1$/sec. The time in seconds such that the ellipse becomes an auxiliary circle is
Step-by-Step Solution
Key Concept: The ellipse $x^2/4+y^2/3=1$ has initial eccentricity $e_0=1/2$. The auxiliary circle is a circle, i.e., $e=0$. Find time for $e$ to decrease from $1/2$ to $0$ at rate $0.1$/sec.
Initial $e=1/2$, target $e=0$. Rate $=0.1$/sec. Time $=(1/2)/0.1=5$ sec.
Correct Answer: 5