椭圆C:x^2/3+y^2/2=1
1个回答

(1)

c^2=3-2=1,F1 (-1,0),F2 (1,0).

由线段PF2的垂直平分线交L2于M得|MP|=|MF2|,根据抛物线定义,点M的轨迹方程为y^2=4x.

(2)

设R(x1,y1),S(x2,y2),(y1,y2不为0且不相等)而Q(0,0),OR垂直RS,故向量RS=(x2-x1,y2-y1)=(y2^2/4-y1^2/4,y2-y1)=( (y1+y2)/4,1),QR=(x1,y1)=(y1^2/4,y1)=(y1/4,1),

RS*QR=(y1^2+y1y2)/16+1=0,即y2=-(16/y1+y1);

x2=y2^2/4=(64/y1^2+y1^2

+32)/4=16/x1+x1+8>=2根号16+8=16,即x2>=16.

|QS|^2=x2^2+y2^2=x2^2+4x2=(x2+2)^2-4>=18^2-4=320,|QS|>=8根号5.