A、B是椭圆x^2/16+y^2/4=1上不同的两点,线段AB的中垂线与x轴交于P(p,0),求p的取值范围?

1个回答

  • 设A(x1,y1),B(x2,y2),设直线AB的中点为点D,

    点D的坐标为((x1+x2)/2,(y1+y2)/2),

    直线AB的斜率为,(y2-y1)/(x2-x1)

    直线DP的斜率为,k=-(x2-x1)/ (y2-y1)

    直线DP的方程为y-(y1+y2)/2=-[(x2-x1)/ (y2-y1)]*(x-(x1+x2)/2)

    0-(y1+y2)/2=-[(x2-x1)/ (y2-y1)]*(x0-(x1+x2)/2)

    (y1+y2)/2=[(x2-x1)/ (y2-y1)]*(x0-(x1+x2)/2)

    (y2^2-y1^2)/ [2(x2-x1)]= x0-(x1+x2)/2

    X0=(y2^2-y1^2)/ [2(x2-x1)]+ (x1+x2)/2

    =1/2[(y2^2-y1^2)+ (x2^2-x1^2)]/ (x2-x1)

    X1^2/16+y1^2/9=1,X2^2/16+y2^2/9=1,两式相减

    (y2^2-y1^2)/9+(x2^2-x1^2)/16=0

    (y2^2-y1^2)=-9(x2^2-x1^2)/16

    X0=1/2[(y2^2-y1^2)+ (x2^2-x1^2)]/ (x2-x1)

    =1/2[7(x2^2-x1^2)/16] / (x2-x1)

    =(7/16)*(x1+x2)/2

    (x1+x2)/2是点D的横坐标,点D在椭圆内部,-4