Question

高中数学求解 给好评

Answer 1

从N作x轴的垂模漏线, 垂足N'(n, n'); F(-c, 0)
直线的斜率为√3/3, 其倾斜角为30°
|FN| = 4√3/3, 那么|FN| = n - (-c) = n+c = |FN|cos30° = (4√3/3)(√3/2) = 2
n = 2 - c
n' = |FN|sin30° = 2√渗码信3/3
e = √3/3 = c/a, c = √3a/3
N(2- √3a/3, 2√3/3)
e² = 1/3 = c²/a² = (a² - b²)/a²
b² = 2a²/3, 椭圆: x²/a² + y²丛轮/(2a²/3) = 1, 2x² + 3y² - 2a² = 0
将N的坐标代入:
2(2- √3a/3)² + 3(2√3/3) - 2a² = 0
a² + 2√3a - 9 = 0
a = √3 (舍去负根)
b² = 2a²/3 = 2
椭圆: x²/3 + y²/2 = 1