2019-2020学年人教A版选修4-4 第一讲坐标系1.1平面直角坐标系 课时作业
2019-2020学年人教A版选修4-4  第一讲坐标系1.1平面直角坐标系   课时作业第2页

解析将伸缩变换{■(x"'" =1/7 x"," @y"'" =4y)┤代入x'2+y'2=1,

得 x^2/49+16y2=1.

答案C

5已知有相距1 400 m的甲、乙两个观测站,在甲站听到爆炸声的时间比在乙站听到爆炸声的时间早4 s.若当时声音的传播速度为340 m/s,建立适当的平面直角坐标系,则爆炸点所在的曲线为(  )

A.双曲线的一支 B.直线

C.椭圆 D.抛物线

答案A

6在平面直角坐标系中,将点P(-2,2)变换为P'(-6,1)的伸缩变换公式为(  )

A.{■(x"'" =1/3 x"," @y"'" =2y)┤B.{■(x"'" =1/2 x"," @y"'" =3y)┤

C.{■(x"'" =3x"," @y"'" =1/2 y)┤D.{■(x"'" =3x"," @y"'" =2y)┤

解析由伸缩变换公式{■(x"'" =λx"(" λ>0")," @y"'" =μy"(" μ>0")," )┤

得{■("-" 6=λ×"(-" 2")," @1=μ×2"." )┤所以{■(λ=3"," @μ=1/2 "," )┤故伸缩变换公式为{■(x"'" =3x"," @y"'" =1/2 y"." )┤

答案C

7在平面直角坐标系中,方程3x-2y+1=0所对应的直线经过伸缩变换{■(x"'" =1/3 x"," @y"'" =2y)┤后的直线方程为 .

解析由伸缩变换{■(x"'" =1/3 x"," @y"'" =2y)┤得{■(x=3x"'," @y=1/2 y"'," )┤将其代入方程3x-2y+1=0,得9x'-y'+1=0.

答案9x'-y'+1=0

8已知函数f(x)=√("(" x"-" 1")" ^2+1)+√("(" x+1")" ^2+1),则f(x)的最小值为     .

解析f(x)可看作是平面直角坐标系中x轴上的一点(x,0)到两定点(1,1)和(-1,1)的距离之和,结合图形可得,f(x)的最小值为2√2.