Question

已知圓C的方程為x²+y²=4.

(1)直線l過點P(1,2),且與圓C交于A、B兩點,若|AB|=,求直線l的方程;

(2)過圓C上一動點M作平行于x軸的直線m,設m與y軸的交點為N,若向量,求動點Q的軌跡方程,并說明此軌跡是什么曲線.

(文)(本小題共13分)已知圓C的方程為x²+y²=4.

(1)直線l過點P(1,2),且與圓C交于A、B兩點,若|AB|=,求直線l的方程;

(2)圓C上一動點M(x₀,y₀),=(0,y₀),若向量,求動點Q的軌跡方程,并說明此軌跡是什么曲線.

Answer 1

解:(1)①直線l垂直于x軸時,直線方程為x=1,l與圓的兩個交點坐標為(1,)和(1,),其距離為滿足題意.                                            

②若直線l不垂直于x軸,設其方程為y-2=k(x-1),

即kx-y-k+2=0.                                                           

設圓心到此直線的距離為d,

則=,得d=1,                                                 

∴1=,k=.                                                       

故所求直線方程為3x-4y+5=0.                                              

綜上所述,所求直線方程為3x-4y+5=0或x=1.                                  

(2)設點M的坐標為(x₀,y₀)(y₀≠0),Q點坐標為(x,y),

則N點坐標是(0,y₀).                                                        

∵,

∴(x,y)=(x₀,2y₀),

即x₀=x,y₀=.                                                             

又∵x₀²+y₀²=4,

∴x²+=4(y≠0).                                                          

∴Q點的軌跡方程是=1(y≠0).                                       

軌跡是一個焦點在y軸上的橢圓,除去短軸端點.                               

注:多端點時,合計扣1分.

(文)解:(1)①若直線l垂直于x軸,則此時直線方程為x=1,l與圓的兩個交點坐標分別為(1,)和(1,),這兩點間的距離為2,滿足題意.                                

②若直線l不垂直于x軸,設其方程為y-2=k(x-1),即kx-y-k+2=0.                  

設圓心到此直線的距離為d,

∵2=,得d=1.                                                 

∴1=,解得k=.                                                 

故所求直線方程為3x-4y+5=0.                                              

綜上所述,所求直線方程為3x-4y+5=0或x=1.                                  

(2)設Q點坐標為(x,y),∵M點坐標是(x₀,y₀),=(0,y₀),,

∴(x,y)=(x₀,2y₀).

∴x=x₀,y=2y₀.                                                            

∵x₀²+y₀²=4,

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

∴Q點的軌跡方程是=1.                                           

軌跡是一個焦點在y軸上的橢圓.