椭圆高手进椭圆x^2/2+y^2=1的左右焦点为F1.F2,过F1的直线L交椭圆于A.B 求:向量F2A*向量F2B的最

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  • 设A(x1,y1),B(x2,y2) ,则AF1=a+ex1,AF2=a-ex1,BF1=a+ex2,BF2=a-ex2,于是AB=a+ex1+a+ex2,a=√2,c=1,e=c/a=1/√2,ae=1

    向量F2A*向量F2B=F2A*F2B*cos∠AF2B=1/2*(AF2^2+BF2^2-AB^2)

    AF2^2+BF2^2-AB^2=(a-ex1)^2+(a-ex2)^2-(a+ex1+a+ex2)^2

    =-4ae(x1+x2)-2(a+ex1)(a+ex2)^

    =-6ae(x1+x2)-2a^2-2e^2x1x2

    故向量F2A*向量F2B=-3ae(x1+x2)-a^2-e^2x1x2=-3(x1+x2)-1/2x1x2-2

    设直线L方程:y=k(x+1)代入椭圆方程x^2+2y^2=2并整理得:

    (1+2k^2)x^2+4k^2x+2k^2-2=0

    x1+x2=-4k^2/(1+2k^2),x1x2=(2k^2-2)/(1+2k^2)

    向量F2A*向量F2B=12k^2/(1+2k^2)-(k^2-1)/(1+2k^2)-2

    =(11k^2+1)/(1+2k^2)-2

    =7/2-(9/2)/(1+2k^2)

    最大=7/2,最小=-1