设直线l的方程为x=m(y+2)+5,m为实数,该直线交抛物线c;y^2=4x与P,Q两个不同的点

(1)立方程组〈
gc =my+(2m+5)y2=4a

,消去c得y-4my-4 (2
m+5) =0
没P(む1, 9i) , Q (c2, 2) ,则y1+9z=4m, 919z=-8m-20
因;内Aカ銭段PQ的中点,所以+92=2m=-2,解得m=-
21,
所以直桟l的方程カz+y-3=0.
(2)证明:因カ
c1+c2=m(y1 +y2)+2(2m+5)=4m'+4m+10,
0122=-ui.弖-(u12)2==(2m+5')2，
4416
所以Bp* Bq=(z1-1)(c2-1)+(y1-2)(uz-2),
即Bp* Bq=[2122-(z1+22)+1]+[y1y2-2(u1+92)+4]
所以
Bp' Bo =[(2m+5)-(4m2+4m+10)+ 11+l-8m-20-2(4m)+

因此BP」BQ,即以銭段PQ杓直径的圆恒辻点B(1,
6元3分狆

F(1,0)
y=k*(x-1)=kx-k,x=y/k+1，m=1/k
AB:x=my+1
y^2=4x=4(my+1)
y^2-4my-4=0
yA+yB=4m,(yA+yB)/2=2m
xA+xB=2+4m^2,(xA+xB)/2=1+2m^2
xA= yA=
xB= yB=
k(MN)=-1/m
MN:y-2m=(-1/m)*(x-1-2m^2)
x=1+4m^2-my
y^2=4x=4*(1+4m^2-my)
y^2+4my-4-4m^2=0
(yM+yN)/2=-2m
(xM+xN)/2=1+6m^2
C(1+6m^2,-2m)
CA^2=CB^2=r^2
m=±1
x=±y+1
L:y=±(x-1)