经过椭圆x^2/2+y^2=1的左焦点F1作倾斜角为60度的直线l,直线l与椭圆相交于A,B两点,求AB的长. 资源ID: 解:c^2=2-1=1,c=1,即F1坐标为(-1,0),直线l的斜率雀简伏顷携为:tg60=√3则直线l方程咐渗为:y=√3(x+1),由x^2/2+y^2=1y=√3(x+1)可知7x^2+12x+4=0x1+x2=-12/7,x1x2=4/7(x1-x2)^2=(x1+x2)^2-4x1x2=32/49(y1-y2)^2=3(x1-x2)^2=96/49AB=√[(x1-x2)^2+(y1-y2)^2]=8√2/7 ← 上一篇: 下一篇: →