Question

已知圆C的方程为x²+y²=4.

(1)直线l过点P(1,2),且与圆C交于A、B两点,若|AB|=,求直线l的方程;

(2)过圆C上一动点M作平行于x轴的直线m,设m与y轴的交点为N,若向量,求动点Q的轨迹方程,并说明此轨迹是什么曲线.

(文)(本小题共13分)已知圆C的方程为x²+y²=4.

(1)直线l过点P(1,2),且与圆C交于A、B两点,若|AB|=,求直线l的方程;

(2)圆C上一动点M(x₀,y₀),=(0,y₀),若向量,求动点Q的轨迹方程,并说明此轨迹是什么曲线.

Answer 1

解:(1)①直线l垂直于x轴时,直线方程为x=1,l与圆的两个交点坐标为(1,)和(1,),其距离为满足题意.                                            

②若直线l不垂直于x轴,设其方程为y-2=k(x-1),

即kx-y-k+2=0.                                                           

设圆心到此直线的距离为d,

则=,得d=1,                                                 

∴1=,k=.                                                       

故所求直线方程为3x-4y+5=0.                                              

综上所述,所求直线方程为3x-4y+5=0或x=1.                                  

(2)设点M的坐标为(x₀,y₀)(y₀≠0),Q点坐标为(x,y),

则N点坐标是(0,y₀).                                                        

∵,

∴(x,y)=(x₀,2y₀),

即x₀=x,y₀=.                                                             

又∵x₀²+y₀²=4,

∴x²+=4(y≠0).                                                          

∴Q点的轨迹方程是=1(y≠0).                                       

轨迹是一个焦点在y轴上的椭圆,除去短轴端点.                               

注:多端点时,合计扣1分.

(文)解:(1)①若直线l垂直于x轴,则此时直线方程为x=1,l与圆的两个交点坐标分别为(1,)和(1,),这两点间的距离为2,满足题意.                                

②若直线l不垂直于x轴,设其方程为y-2=k(x-1),即kx-y-k+2=0.                  

设圆心到此直线的距离为d,

∵2=,得d=1.                                                 

∴1=,解得k=.                                                 

故所求直线方程为3x-4y+5=0.                                              

综上所述,所求直线方程为3x-4y+5=0或x=1.                                  

(2)设Q点坐标为(x,y),∵M点坐标是(x₀,y₀),=(0,y₀),,

∴(x,y)=(x₀,2y₀).

∴x=x₀,y=2y₀.                                                            

∵x₀²+y₀²=4,

∴x²+()²=4,即=1.                                                

∴Q点的轨迹方程是=1.                                           

轨迹是一个焦点在y轴上的椭圆.