Question

已知圓C的方程為x²+y²=4.

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

(2)過(guò)圓C上一動(dòng)點(diǎn)M作平行于x軸的直線(xiàn)m,設(shè)m與y軸的交點(diǎn)為N,若向量,求動(dòng)點(diǎn)Q的軌跡方程,并說(shuō)明此軌跡是什么曲線(xiàn).

(文)(本小題共13分)已知圓C的方程為x²+y²=4.

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

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

Answer 1

解:(1)①直線(xiàn)l垂直于x軸時(shí),直線(xiàn)方程為x=1,l與圓的兩個(gè)交點(diǎn)坐標(biāo)為(1,)和(1,),其距離為滿(mǎn)足題意.                                            

②若直線(xiàn)l不垂直于x軸,設(shè)其方程為y-2=k(x-1),

即kx-y-k+2=0.                                                           

設(shè)圓心到此直線(xiàn)的距離為d,

則=,得d=1,                                                 

∴1=,k=.                                                       

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

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

(2)設(shè)點(diǎn)M的坐標(biāo)為(x₀,y₀)(y₀≠0),Q點(diǎn)坐標(biāo)為(x,y),

則N點(diǎn)坐標(biāo)是(0,y₀).                                                        

∵,

∴(x,y)=(x₀,2y₀),

即x₀=x,y₀=.                                                             

又∵x₀²+y₀²=4,

∴x²+=4(y≠0).                                                          

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

軌跡是一個(gè)焦點(diǎn)在y軸上的橢圓,除去短軸端點(diǎn).                               

注:多端點(diǎn)時(shí),合計(jì)扣1分.

(文)解:(1)①若直線(xiàn)l垂直于x軸,則此時(shí)直線(xiàn)方程為x=1,l與圓的兩個(gè)交點(diǎn)坐標(biāo)分別為(1,)和(1,),這兩點(diǎn)間的距離為2,滿(mǎn)足題意.                                

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

設(shè)圓心到此直線(xiàn)的距離為d,

∵2=,得d=1.                                                 

∴1=,解得k=.                                                 

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

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

(2)設(shè)Q點(diǎn)坐標(biāo)為(x,y),∵M(jìn)點(diǎn)坐標(biāo)是(x₀,y₀),=(0,y₀),,

∴(x,y)=(x₀,2y₀).

∴x=x₀,y=2y₀.                                                            

∵x₀²+y₀²=4,

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

∴Q點(diǎn)的軌跡方程是=1.                                           

軌跡是一個(gè)焦點(diǎn)在y軸上的橢圓.