设,点A坐标为(t1,t1^2),点B坐标为(t2,t2^2),
OA^2=t1^2+(t1^2)^2=t1^2+t1^4.
OB^2=t2^2+t2^4.
∵OA⊥OB,有t1^2/t1*t2^2/t2=-1,
t1*t2=-1.
直线AB的方程为:(Y-t1^2)/(t2^2-t1^2)=(x-t1)/(t2-t1),
即,Y=(t1+t2)x+1.
(t1+t2)=(y-1)/x,
令,点C的坐标为(X,Y),OA中点为点E,坐标为(Xa,ya),OB点为F,坐标为(Xf,yf).
Xa=t1/2,ya=t1^2/2,
xf=t2/2,yf=t2^2/2.
(OA^2)/2=(xa-x)^2+(ya-y)^2,即有,
X^2+Y^2-(t1x+t1^2y)=(t1^2+t1^4)/4,.(1)
同理,
X^2+y^2-(t2x+t2^2y)=(t2^2+t2^4)/4,.(2)
(1)+(2)式,得
2(X^2+Y^2)-(t1+t2)x-(t1^2+t2^4)y=[(t1^2+t2^2)+(t1^4+t2^4)]/4,
而,t1+t2=(y-1)/x,t1*t2=-1.
2(x^2+y^2)-(y-1)-[(t1+t2)^2-2t*t2)]*y=[(t1+t2)^2-2t1*t2+(t1^2+t2^2)^2-2(t1*t2)^2]/4
2(x^2+y^2)-(y-1)-[(y-1)^2/x^2+2]y={(y-1)^2/x^2+2+[(t1+t2)^2-2t1*t2]^2-2}/4.
即,点C的轨迹方程为:
2(x^2+y^2)-(y-1)-[(y-1)^2/x^2+2]y={(y-1)^2/X^2+2+[(Y-1)^2/X^2+2]^2-2}/4.
"繁"啊.