答案2√2

9在平面直角坐标系中,求下列方程所对应的图形经过伸缩变换{■(x"'" =1/2 x"," @y"'" =1/3 y)┤后的图形.

(1)5x+2y=0;

(2)x2+y2=2.

解(1)由伸缩变换{■(x"'" =1/2 x"," @y"'" =1/3 y"," )┤得{■(x=2x"'," @y=3y"'," )┤

将其代入5x+2y=0,得到经过伸缩变换后的图形的方程是5x'+3y'=0.

所以经过伸缩变换{■(x"'" =1/2 x"," @y"'" =1/3 y)┤后,直线5x+2y=0变成直线5x'+3y'=0.

(2)将{■(x=2x"'," @y=3y"'" )┤代入x2+y2=2,得到经过伸缩变换后的图形的方程是 (x"'" ^2)/(1/4)+(y"'" ^2)/(1/9)=2,即 (x"'" ^2)/(1/2)+(y"'" ^2)/(2/9)=1.

所以经过伸缩变换{■(x"'" =1/2 x"," @y"'" =1/3 y)┤后,圆x2+y2=2变成椭圆 (x"'" ^2)/(1/2)+(y"'" ^2)/(2/9)=1.

10在△ABC中,底边BC的长为12,其他两边AB和AC上的中线CE和BD的和为30,建立适当的平面直角坐标系,求△ABC的重心G的轨迹方程.

解以BC边所在的直线为x轴,BC边上的中点为原点建立如图所示的平面直角坐标系,

则B(6,0),C(-6,0).由|BD|+|CE|=30,

得|GB|+|GC|=2/3(|BD|+|CE|)=20>12,

所以重心G的轨迹是椭圆.

设G(x,y),则其轨迹方程为 x^2/a^2 +y^2/b^2 =1(a>b>0).

易知a=10,c=6,由此可得b=8.

因为点G不可能位于x轴上,所以x≠±10.

故重心G的轨迹方程为 x^2/100+y^2/64=1(x≠±10).