直线x+y-1=0与椭圆x^2/a^2+y^2/b^2=1(a>b>0)交于两点A、B,AB中点M,直线OM斜率k=1/2,则b/a=
问题描述：
直线x+y-1=0与椭圆x^2/a^2+y^2/b^2=1(a>b>0)交于两点A、B,AB中点M,直线OM斜率k=1/2,则b/a=
答
联立x+y-1=0x^2/a^2+y^2/b^2=1解得：（a^2+b^2)x^2-2a^2x+a^2-a^2b^2=0x1+x2=2a^2/(a^2+b^2) X1X2=(a^2-a^2b^2)/(a2+b^2)y1+y2=-(x1+x2)+2所以M点坐标为（X0,y0）=（a^2/(a^2+b^2) ,b^2/(a^2+b^2) ）y0/x0=k=1/2就可...