坐标原点O,抛物线y^2=2x与过焦点的直线教育A,B两点,则向量OA*向量OB=?

设,点A坐标为(X1,Y1),点B坐标为(X2,Y2).
|OA|^2=X1^2+Y1^2=X1^2+2X1,
|OB|^2=X2^2+2X2.
|AB|^2=(P+X1+X2)^2.(焦半径公式,可得).
Y^2=2X,2P=2,P=1,则焦点F的坐标为(1/2,0),
令,直线AB的方程为:Y=K(X-1/2),
K^2*(X-1/2)^2=2X,
K^2*X^2-(K^2+2)X+K^2/4=0,
X1+X2=(K^2+2)/K^2,
X1*X2=1/4.
令,向量OA与向量OB的夹角为X,
向量OA*向量OB=|OA|*|OB|*COSX
=|OA|*|OB|*(OA^2+OB^2-AB^2)/(2*|OA|*|OB|)
=1/2*(OA^2+OB^2-AB^2)
=1/2*[X1^2+2X1+X2^2+2X2-(1+X1+X2)^2]
=1/2*(-3/2)
=-3/4.

设,点A坐标为(X1,Y1),点B坐标为(X2,Y2). x1*x2=p^2/4 y1*y2=-p^2上为结论 可得-0·75