（2）x'^2+y'^2=8 (x',y'）=（x+2y,x)
所以 2x^2+4xy+4y^2=8 x^2+2xy+2y^2=4
(3)C (x1,y1) D(x2,y2) (x1,y1-2)=t(x2,y2-2) x1=tx2,y1-2=t(y2-2)
設(shè) CD:y=kx+2 代入 x^2+2xy+2y^2=4 中得 x^2+2x(kx+2)+2(k^2x^2+4kx+4)=4
（1+2k+4k^2）x^2+（4+8k)x+4=0
判別式=（4+8k)^2-16(1+2k+4k^2)=32k>0,k>0
x1+x2=-(4+8k)/(1+2k+4k^2)=(1+t)x2,x1*x2=4/(1+2k+4k^2)=t x2^2
4/(1+2k+4k^2)=[t/(1+t)^2][16(1+2k)^2/(1+2k+4k^2)^2
(1+t)^2/t=4(1+2k)^2/(1+2k+4k^2)=4(1+2k/(1+2k+4k^2)=4+8/(1/k+4k+2)>4
且0,t≠1 1/3