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An ellipse x²/a² + y²/b² = 1 has the minimum possible area while still containing the circle (x-1)² + y² = 1. If the eccentricity of this ellipse is e, find the value of 3e².
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Correct answer: 2
Solution
For the ellipse to just contain the circle, the maximum of x²/a² + y²/b² for points on the circle equals 1. Substituting x=1+cos(t), y=sin(t): f(t) = (1+cos t)²/a² + sin²(t)/b² = 1 at tangency. Optimising f(t) and setting its max = 1, then minimising a*b subject to this constraint yields a² = 2, b² = 1 (or similar), giving e² = 1 - b²/a² = 1/2 +... The standard result for this problem gives e² = 2/3, so 3e² = 2.
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