过椭圆x2/5+y2/4=1的右焦点作直线l与椭圆相交于A,B两点,若弦长|AB|=5/3根号5,则直线L的斜率为

椭圆右焦点坐标为（1,0）,设直线方程为 y = k(x-1),
代入椭圆方程得 x^2/5+k^2*(x-1)^2/4 =1 ,
化简得 (5k^2+4)x^2-10k^2*x+5k^2-20 = 0 ,
设 A（x1,y1）,B（x2,y2）,
则 x1+x2 = 10k^2/(5k^2+4),x1*x2 = (5k^2-20)/(5k^2+4) ,
因此 |AB|^2 = (x2-x1)^2+(y2-y1)^2
= (x2-x1)^2+k^2*(x2-x1)^2
= (1+k^2)*(x2-x1)^2
= (1+k^2)*[(x1+x2)^2-4x1*x2]
= (1+k^2)*[100k^4/(5k^2+4)^2-4(5k^2-20)/(5k^2+4)]
= 125/9 ,
解得 k = ±2 ,
即直线的斜率为 -2 或 2 .