|MS|=13,|MT|=15
||NS|-|MS||=||NT|-|MT||=2a(a>7)
设|NS|=s,|NT|=t,则|s-13|=|t-15|=2a>14
解不等式得：s>27或s29或t29时,s-13=t-15 => t-s=2,N点轨迹是双曲线x²-y²/48=1的左支在圆(x-7)²+y²=29²之外的部分,下面来看分界点：设点N坐标为(x,y)
则(x-7)²+y²>29²,与x²-y²/48=1联立消去y得：7x²-2x-120>0
解得：x30/7(舍去)
综上所述,N点轨迹方程为x²-y²/48=1 (x