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)

5 seconds

Answer 2

$\frac{de}{dt}$= – 0.1 m/sec.

⇒ e = –0.1 t + e₀

where e₀ = 1/2

⇒ e_t = – 0.1 t + $\frac{1}{2}$, given e_t = 0

⇒ 0.1t = $\frac{1}{2}$ or t = 5 seconds.