按椭圆定义,F₁A + F₂A = 2a,F₁F₂ = 2c△AF1F2的周长为6 = 2a + 2ca + c = a + √(a² - b²) = 3a = (b² + 9)/6a² - b² = (b² - 9)²/36e² = (a²...总感觉有点问题，第三步到第四步怎么回事？看题目应当是b为已知，否则条件不够。c² = a² - b²a + c = a + √(a² - b²) = 3√(a² - b²) = 3 - a平方: -b² = 9 - 6aa = (b² + 9)/6那第四到第五步呢？你也够懒的a² - b² = (b²+ 9)²/36 - b² = [(b⁴ + 18b² + 81) - 36b²]/36 = (b⁴ - 18b² + 81)/36 = (b² - 9)²/36e² = c²/a² = (a² - b²)/a² = [(b² - 9)²/36]/[(b² + 9)²/36] = (b² - 9)²/(b² + 9)²