设BC中点为M(x,y),而右焦点F2(2,0),A(4,0)
2=2,故x=3,y=-2,即M(3,-2) 故设BC方程为:y=k(x-3)-2代入椭圆4x2+5y2=80中
(4+5k2)x2-10k(3k+2)x+5(3k+2)2-80=0 故(x1+x2)/2=3 即5k(3k+2)/(4+5k2)=3 所以K=6/5
即BC方程为:6x-5y-28=0
②=(x1,y1-4),=(x2,y2-4).∵AB⊥AC,∴x1x2+y1y2-4(y1+y2)+16=0 (2)
设直线BC的方程为y=kx+b,代入4x2+5y2=80,得(4+5k2)x2+10bkx+5b2-80=0
∴x1+x2=(-10kb)/(4+5k2),x1x2=(5b2-80)/(4+5k2)
故y1+y2=k(x1+x2)+2b=k(-10kb)/(4+5k2)+2b=8b/(4+5k2) ,y1y2=k2x1x2+bk(x1+x2)+b2=(5k2b2-80k2)/(4+5k2)+(-10 b2k2)/(4+5k2)+b2=(4b2-80k2)/(4+5k2),
把上述各式代入(2)得 (5b2-80)/(4+5k2)+(4b2-80k2)/(4+5k2)-32b/(4+5k2)+16 =0
∴9b2-32b-16=0 ∴b=4(舍去)或b= - 4/9 故直线经过定点(0,-4/9),设D(x,y),
则(y+4/9)/x●(y-4)/x= -1,即9y2+9x2-32y-16=0,所以所求的轨迹方程是x2+(y-16/9)2=(20/9)2(y≠4)