Y=X^2,Y=X+b,
设,点A坐标为(t1,t1^2),点B坐标为(t2,t2^2),点C坐标为(X,Y),有
X^2+X-b=0,
x1+x2=t1+t2=-1,
x1*x2=t1*t2=-b.
AB^2=(x2-x1)^2+(y2-y1)^2
=(t2-t1)^2+(t2^2-t1^2)^2,
而,(t2-t1)^2=(t1+t2)^2-4t1*t2=1+4b.
(t2^2-t1^2)^2=(t1+t2)^2*(t2-t1)^2
=(-1)^2*(1+4)=1+4b.
即有,AB^2=2(1+4b).
BC^2=(X-t2)^2+(y-t2^2)^2=2(1+4b)=AB^2.(1)式
AC^2=(x-t1)^2+(y-t1)^2=2(1+4b)=AB^2.(2)式
由(1)-(2)式得,
-2X(t2-t1)+(t2^2-t1^2)-2y(t2^2-t1^2)+(t2^4-t1^4)=0
-2x-1+2y-(t2^2+t1^2)=0,
-2x+2y-2-b=0,
b=2x-2y+2.
而,Y=X+b,
则有Y=X+2x-2y+2,
即,3X-3Y+2=0.
顶点C的轨迹方程是:3X-3Y+2=0.