1
2c=√15,c=√15,
F1(-√15,0),F2(√15,0)
橢圓經(jīng)過點M(4.1),根據(jù)定義,
2a=|MF1|+|MF2|
=√[(4+√15)²+1]+√[(4-√15)²+1]
=√(32+8√15)+√(32-8√15)
=√(2√5+2√3)²+√(2√5-2√3)²
=4√5
∴a=2√5,b²=a²-c²=5
∴橢圓方程為x²/20+y²/5=1
{x²/20+y²/5=1
｛y=x+m
==>
x²+4(x+m)²-20=0
==>
5x²+8mx+4m²-20=0
Δ=64m²-20(4m²-20)>0
即16m²-400<0解得-52
設(shè)A(x1,y1),B(x2,y2)
那么x1+x2=-8m/5,x1x2=(4m²-20)/5
kMA+kMB
=(y1-1)/(x1-4)+(y2-1)/(x2-4)
=[(x1+m-1)(x2-4)+(x2+m-1)(x1-4)]/[(x1-4)(x2-4)]
=[2x1x2+(m-5)(x1+x2)-8(m-1)]/[(x1-4)(x2-4)]
=[2(4m²-20)/5+(m-5)(-8m/5)-8(m-1)]/[(x1-4)(x2-4)]
=(8m²/5-8-8m²/5+8m-8m+8)/[(x1-4)(x2-4)]
=0
即直線MA、MB的斜率互為相反數(shù)