Question

已知抛物线y^2=-x与直线y=k（x+1）相交于AB两点,点o是坐标原点

Answer 1

(1)设A（x1,y1),B(x2,y2)

联立方程y^2=-x,y=k（x+1）得：k^2x^2+(2k^2+1)x+k^2=0,

∴x1+x2=-(2k^2+1)/(k^2),x1x2=1

∴向量OA·向量OB=x1x2+y1y2=1+k^2(x1x2+x1+x2+1)=1+k^2-2k^2-1+k^2=0

(2)S△OAB=（OA*OB）/2=根号（x1^2+y1^2)*根号(x2^2+y2^2)/2=根号（x1^2-x1)*根号(x2^2+-x2)/2=根号(x1x2)*根号[(x1-1)(x2-1)]/2=根号(x1x2-x1-x2+1)/2=根号(4+1/k^2)/2=根号10

解得k=1/6或k=-1/6