Question

数学，椭圆，求解第二问…… 挺急的……

Answer 1

(1) 

椭圆的上顶点A与两焦点的距离相等，即三点构成以A为顶点的等腰三角形，其顶角为60°，则其为等边三角形。F1(-c, 0), F2(c, 0), A(0, b)
|F1A|² = b² + c² = |F1F2|² = 4c², b² = 3c²

e² = c²/a² = c²/(b² + c²) = c²/(4c²) = 1/4

e = 1/2

(2)

直线AB与横轴和纵轴分别交于F2(c, 0), A(0, √3c), 其方程为x/c + y/(√3c) = 1

y = -√3x + b = -√3x + √3c

椭圆x²/a² + y²/b² = x²/(4c²) + y²/(3c²) = 1

联立并整理：x(5x - 8c) = 0, x = 8c/5, B(8c/5, -3√3c/5)

|F1A| = √(b² + c²) = √(3c² + c²) = 2c

|AB| = 16c/5

三角形AF1B的面积S = (1/2)|AF1|*|AB|*sin∠BAF1 = (1/2)*2c*(16c/5)*(√3/2) = 8√3c²/5 = 40√3

c = 5

a = 2c = 10

b = √3c = 5√3