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...答: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).
已知抛物线y^2=4x,直线l与抛物线相交于A,B两点,若线段AB中点为(2,2...答:抛物线的方程为y =4x,A(x 1 ,y 1 ),B(x 2 ,y 2 ),则有x 1 ≠x 2 (y1)=4x1 (y2) = 4x2 两式相减得,(y1)-(y2)=4(x 1 -x 2 ), 所以(y1-y2)/(x1-x2)=4/(y1+y2)=1 所以直线l的方程为y-2=x-2,即y=x 故答案为:y=x ...