已知直线l与抛物线 y^2=4x相交于A,B两点,且OA垂直于OB,求证:直线l碧过定点,并写出这个定点的坐标

设,直线L的方程为:Y=KX+b,则有Y=K(X+b/k),即直线必过定点(-b/k,0).
y^2=4x,
令,点A坐标为(t1^2/2p,t1),点B坐标为(t2^2/2p,t2).
Koa=t1/(t1^2/2p)=2p/t1,
Kob=t2/(t2^2/2p)=2p/t2.
而,Koa*Kob=-1.有
2p/t1*2p/t2=-1,
4P^2/(t1*t2)=-1.

y^2=4x,Y=KX+b.x=(y-b)/k,
y^2=4(y-b)/k,
K*y^2-4y+4b=0,
y1+y2=t1+t2=4/k,
y1*y2=t1*t2=4b/k.
kp^2=-b,
-k/b=1/p^2,
而,4=2P,P=2,
-K/b=1/(2^2)=1/4.
即有:直线l必过定点(1/4,0).