證明：設(shè)直線L的斜率為k,由點(diǎn)斜式可寫出L的直線方程y=k(x+1)
直線L與橢圓方程聯(lián)立消y得(4k²+1)x²+8k²x+(4k²-4)=0
設(shè)A(x1,y1)、B(x2,y2),對上式由韋達(dá)定理有
x1+x2= -8k²/(4k²+1)
x1x2=(4k²-4)/(4k²+1)
向量AQ=(-1-x1,-y1),向量QB=(x2+1,y2),向量AE=(-4-x1,……),向量EB=(x2+4,……)
由題意有：AQ=λQB,AE=μEB,即
(-1-x1,-y1)= λ(x2+1,y2),(-4-x1,……)=μ(x2+4,……),
從而-1-x1= λ(x2+1),-4-x1=μ(x2+4),
所以λ= -(x1+1)/(x2+1),μ= -(x1+4)/(x2+4),
所以λ+μ=[ -(x1+1)/(x2+1)]+[ -(x1+4)/(x2+4)]
= -[(x1+1)(x2+4)+(x2+1)(x1+4)]/[(x2+1)(x2+4)]
= -[2x1x2+5(x1+x2)+8]/[(x2+1)(x2+4)]
將韋達(dá)定理代入分子得
= -[2(4k²-4)/(4k²+1)-5*8k²/(4k²+1)+8]/[(x2+1)(x2+4)]
=0
故λ+μ為定值得證,這個(gè)定值為0.