.
∴PQ中点M的轨迹方程为y=x2++1(x≠0).
(Ⅱ)设直线l:y=kx+b,依题意k≠0,b≠0,则T(0,b).
分别过P、Q作PP'⊥x轴,QQ'⊥y轴,垂足分别为P'、Q',则
y0=x02++1(x0≠0),
∴x1=-,
将上式代入②并整理,得
则x0==kl=-,
得y1-y2=x12-x22=(x1+x2)(x1-x2)=x0(x1-x2),
由y1=x12,y2=x22,x0=,
方法二:
消去x1,得y0=x02++1(x0≠0),
y0=x12-(x0-x1).