题目内容：

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?

优质解答

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