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,求该直线方程.

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,求该直线方程.

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