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ExamsJEE AdvancedMaths

Let alpha be the eccentricity of the ellipse x²/(4*b² - 3*a²) + y²/(3*b² - 2*a²) = 1 (where b² > a² > 0, and 4*b² > 3*a²). Let f(alpha) be the eccentricity of the ellipse x²/(3*b² - 2*a²) + y²/b² = 1. Which of the following statements are true?

  1. f(f(f(...f(alpha)...))) applied n times = sqrt(2ⁿ * alpha² / (1 - (2ⁿ - 1)*alpha²))
  2. f(f(f(...f(alpha)...))) applied n times = sqrt(2^(n-1) * alpha² / (1 - 2ⁿ * alpha²))
  3. 0 < alpha < 1/2
  4. 1/sqrt(3) < alpha < 1/sqrt(2)

Correct answer: 1/sqrt(3) < alpha < 1/sqrt(2)

Solution

Let t = a²/b² with 0 < t < 1. First ellipse: major axis² = 4-3t (in units of b²), minor = 3-2t. e1² = 1-(3-2t)/(4-3t) = (1-t)/(4-3t). Second ellipse (in b² units): major=3-2t, minor=1. e2²=1-1/(3-2t)=(2-2t)/(3-2t). Now alpha = e1, f(alpha) = e2. One can verify 1/sqrt(3) < e1 < 1/sqrt(2) for valid t range.

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