已知椭圆x²/25+y²/16=1,△ABC是椭圆的内接三角形,点A是椭圆的上定点,此椭圆的右焦点F是△ABC的重心,求BC边所在的直线方程.
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![已知椭圆x²/25+y²/16=1,△ABC是椭圆的内接三角形,点A是椭圆的上定点,此椭圆的右焦点F是△ABC的重心,求BC边所在的直线方程.](/uploads/image/z/10201771-19-1.jpg?t=%E5%B7%B2%E7%9F%A5%E6%A4%AD%E5%9C%86x%26%23178%3B%2F25%2By%26%23178%3B%2F16%3D1%2C%E2%96%B3ABC%E6%98%AF%E6%A4%AD%E5%9C%86%E7%9A%84%E5%86%85%E6%8E%A5%E4%B8%89%E8%A7%92%E5%BD%A2%2C%E7%82%B9A%E6%98%AF%E6%A4%AD%E5%9C%86%E7%9A%84%E4%B8%8A%E5%AE%9A%E7%82%B9%2C%E6%AD%A4%E6%A4%AD%E5%9C%86%E7%9A%84%E5%8F%B3%E7%84%A6%E7%82%B9F%E6%98%AF%E2%96%B3ABC%E7%9A%84%E9%87%8D%E5%BF%83%2C%E6%B1%82BC%E8%BE%B9%E6%89%80%E5%9C%A8%E7%9A%84%E7%9B%B4%E7%BA%BF%E6%96%B9%E7%A8%8B.)
已知椭圆x²/25+y²/16=1,△ABC是椭圆的内接三角形,点A是椭圆的上定点,此椭圆的右焦点F是△ABC的重心,求BC边所在的直线方程.
已知椭圆x²/25+y²/16=1,△ABC是椭圆的内接三角形,点A是椭圆的上定点,此椭圆的右焦点F是△ABC的重心,求BC边所在的直线方程.
已知椭圆x²/25+y²/16=1,△ABC是椭圆的内接三角形,点A是椭圆的上定点,此椭圆的右焦点F是△ABC的重心,求BC边所在的直线方程.
a² = 25, b² = 16, A(0, 4)
c² = 25 - 16 = 9, F(3, 0)
令BC的中点为D(p, q), 根据重心的性质: |AF| = 2|FD|, 且
(0 - 3)/(3 - p) = 2, p = 9/2
(4 - 0)/(0 - q) = 2, q = -2
令B(m, n), C(u, v)
m²/25 + n²/16 = 1 (1)
u²/25 + v²/16 = 1 (2)
D为BC的中点:
9/2 = (m + u)/2, m + u = 9 (3)
-2 = (n + v)/2, n + v = -4 (4)
(1) - (2):
(m² - u²)/25 = -(n² - v²)/16, (m + u)(m - u)/25 = -(n + v)(n - v)/16
BC的斜率k = (n - v)/(m - u)
= (-16/25)(m + u)/(n + v)
= (-16/25)*9/(-4)
= 36/25
y + 2 = (36/25)(x - 9/2)
y = 36x/25 - 212/25
见图,题中的数字有问题