OM斜率为2,∴直线l的斜率为2
则l的方程为y=2x+m,带入椭圆得
x²+4(2x+m)²=8,化简得17x²+16mx+4m²-8=0........①
记点A坐标为(xA,yA),点B为(xB,yB)
则MA的斜率为(1-yA)/(2-xA),MB的斜率为(1-yB)/(2-xB)
MA⊥MB,则(1-yA)(1-yB)+(2-xA)(2-xB)=0
带入yA=2xA+m,yB=2xB+m,得
(1-2xA-m)(1-2xB-m)+(2-xA)(2-xB)=0
化简得:(1-m)²-2(xA+xB)(2-m)+5xAxB+4=0
而xA,xB为方程①的两个解,所以由韦达定理得
xA+xB=-16m/17,xAxB=(4m²-8)/17,带入上式得
(1-m)²+32m(2-m)/17+5(4m²-8)/17+4=0
=> 17-34m+17m²+64m-32m²+20m²-40+68=0
=> 5m²+30m+45=0 => m²+6m+9=0
=> (m+3)²=0 ,∴解得m=-3