Question

第三问详细过程

Answer 1

由前面知，点B1(-2，0)；B2(2，0)；椭圆为(x²/20)+(y²/4)=1
设过点B1的直线为：y=k(x+2)
联立直线与椭圆方程得到：x²+5y²-20=0，y=k(x+2)
==> x²+5k²(x+2)²-20=0
==> (5k²+1)x²+20k²x+20(k²-1)=0
所以，x1+x2=-20k²/(5k²+1)；x1x2=20(k²-1)/(5k²+1)
MB2的斜率为k1=y1/(x1-2)；NB2的斜率为k2=y2/(x2-2)
已知MB2⊥NB2，则K1K2=-1
==> y1/(x1-2)*y2/(x2-2)=-1
==> y1y2+(x1-2)(x2-2)=0
==> k(x1+2)*k(x2+2)+x1x2-2(x1+x2)+4=0
==> k²[x1x2+2(x1+x2)+4]+x1x2-2(x1+x2)+4=0
==> (k²+1)x1x2+2(k²-1)(x1+x2)+4(k²+1)=0
==> (k²+1)*20(k²-1)/(5k²+1)+2(k²-1)*(-20k²)/(5k²+1)+4(k²+1)=0
==> 20(k^4-1)-40k²*(k²-1)+4(k²+1)(5k²+1)=0
==> 20k^4-20-40k^4+40k²+20k^4+24k²+4=0
==> 64k²=16
==> k²=1/4
==> k=±1/2