Question

已知抛物线y^2=2x,过点M(1,0)做直线和抛物线交于A,B两点,O是坐标原点,如果只限OA,OB的斜率之和为-1,

Answer 1

设直线方程为x=ty+1

{x=ty+1;y²=2x

==>y²=2ty+2

==>y²-2ty-2=0

设A(x1,y1),B(x2,y2)

则y1+y2=2t,y1y2=-2

x1=y²1/2,x2=y²2/2

x1x2=(y1y2)²/4=1

∵OA,OB的斜率之和为-1

∴y1/x1+y2/x2=-1

∴y1x2+y2x1=-x1x2

∴y1y²2+y2y²1=-2

∴y1y2(y1+y2)=-2

∴-2*(-2t)=-2

∴t=-1/2

∴该直线方程为

x=-1/2y+1 即 2x+y-2=0