分析：1）设圆心O(x,y),它到定点N(0,2)和到定直线y=-2距离相等,由抛物线定义得其轨迹为抛物线,且P/4=2,焦点在y轴上,于是轨迹方程为8y=x^2.2）设A(x1,y1),B(x2,y2),Q(xo,yo),N(0,2),显然AB斜率存在且过N(0,2),设其直线方程为y=kx+2,联立8y=x^2消去y得:x^2-8kx-16=0,判别式恒大于零.于是x1+x2=8k,x1x2=-16,又曲线4y=x^2上任意一点斜率为y'=x/4,则易得切线AQ,BQ方程分别为y=(1/4)x1(x-x1)+y1,y=(1/4)x2(x-x2)+y2,其中8y1=x1^2,8y2=x2^2,联立方程易解得交点Q坐标,xo=(x1+x2)/2=4k,yo=(x1x2)/8=-2,又向量AB=[x2-x1,(x2^2-x1^2)/8],向量NQ=(4k,-4),于是,向量NQ*向量AB=4k(x2-x1)-(x2^2-x1^2)/2=(x2-x1)[4k-(x1+x2)/2]=(x2-x1)[4k-(8k)/2]=0,定值,得证!也即AB垂直NQ