AF1+AF2 = BF1+BF2 = 2a = 10 (椭圆定义)
所以△ABF2的周长为C = 20.
内切圆的周长为π
所以内切圆半径为0.5
所以△ABF2的面积S = 0.5 * r * C = 5 (分割三角形得面积)
又△ABF2的面积S = 0.5 * |F1F2| * |y1-y2| = 0.5 * 6 * |y1-y2| = 3|y1-y2|
所以|y1-y2| = 5/3
选D
AF1+AF2 = BF1+BF2 = 2a = 10 (椭圆定义)
所以△ABF2的周长为C = 20.
内切圆的周长为π
所以内切圆半径为0.5
所以△ABF2的面积S = 0.5 * r * C = 5 (分割三角形得面积)
又△ABF2的面积S = 0.5 * |F1F2| * |y1-y2| = 0.5 * 6 * |y1-y2| = 3|y1-y2|
所以|y1-y2| = 5/3
选D