已知椭圆x^2/a^2+y^2/b^2=1(a>b>0)上两点A,B,直线OA‘OB的斜率之积为-(b^2/a^2)

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  • 设A(x1,y1) B(x2,y2)

    y1/x1*y2/x2=-(b^2/a^2) (1)

    x1^2/a^2+y1^2/b^2=1 (2)

    x2^2/a^2+y2^2/b^2=1 (3)

    由(2)(3) (x1x2)^2/a^4=1-y1^2/b^2-y2^2/b^2+(y1y2)^2/b^4 (4)

    即(x1x2)^2/a^4-(y1y2)^2/b^4=1-y1^2/b^2-y2^2/b^2

    即(x1x2/a^2-y1y2/b^2)(x1x2/a^2+y1y2/b^2)=1-y1^2/b^2-y2^2/b^2 (5)

    由(1)变形得到 x1x2/a^2+y1y2/b^2=0

    再结合(5) 1-y1^2/b^2-y2^2/b^2=0

    即y1^2+y2^2=b^2 (6)

    (2)+(3) (x1^2+x2^2)/a^2+(y1^2+y2^2)/b^2=2

    再结合(6) x1^2+x2^2=a^2 (7)

    由(6)得 (y1+y2)^2-b^2=2y1y2 (8)

    由(7)得 (x1+x2)^2-a^2=2x1x2 (9)

    由(8)(9)得 ((2y中)^2-b^2)/((2x中)^2-a^2)=y1y2/(x1x2)=-b^2/a^2

    化简得 2x中^2/a^2+2y中^2/b^2=1

    即轨迹方程为2x^2/a^2+2y^2/b^2=1