分析：1）設(shè)圓心O(x,y),它到定點(diǎn)N(0,2)和到定直線y=-2距離相等,由拋物線定義得其軌跡為拋物線,且P/4=2,焦點(diǎn)在y軸上,于是軌跡方程為8y=x^2.2）設(shè)A(x1,y1),B(x2,y2),Q(xo,yo),N(0,2),顯然AB斜率存在且過(guò)N(0,2),設(shè)其直線方程為y=kx+2,聯(lián)立8y=x^2消去y得:x^2-8kx-16=0,判別式恒大于零.于是x1+x2=8k,x1x2=-16,又曲線4y=x^2上任意一點(diǎn)斜率為y'=x/4,則易得切線AQ,BQ方程分別為y=(1/4)x1(x-x1)+y1,y=(1/4)x2(x-x2)+y2,其中8y1=x1^2,8y2=x2^2,聯(lián)立方程易解得交點(diǎn)Q坐標(biāo),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,得證!也即AB垂直NQ