Question

(理)已知點A(4m,0)、B(m,0)(m是大于0的常數(shù)),動點P滿足=6m||.

(1)求點P的軌跡C的方程;

(2)點Q是軌跡C上一點,過點Q的直線l交x軸于點F(-m,0),交y軸于點M,若||=2||,求直線l的斜率.

(文)已知橢圓的中心在坐標原點O,焦點在x軸上,左焦點為F,左準線與x軸的交點為M,.

(1)求橢圓的離心率e;

(2)過左焦點F且斜率為的直線與橢圓交于A、B兩點,若=-2,求橢圓的方程.

Answer 1

(理)解：(1)設(shè)P(x,y),則=(-3m,0),=(x-4m,y),=(m-x,-y).               

∵,

∴-3m(x-4m)=6m.

則點P的軌跡C的方程為=1.                                    

(2)設(shè)Q(x_Q,y_Q),直線l:y=k(x+m),則點M(0,km).

當時,由于F(-m,0),M(0,km),得

(x_Q,y_Q-km)=2(-m-x_Q,-y_Q).

x_Q=,y_Q=km.                                                       

又點Q()在橢圓上,所以=1.

解之,得k=±2.                                                          

當時,x_Q=-2m,y_Q=-km.                                          

于是=1,解得k=0.                                             

故直線l的斜率是0,±2.                                                  

(文)解：(1)設(shè)橢圓方程為,F(-c,0),M(,0).

由,有(,0)=4(-c,0).                                         

則有=4c,即.

∴e=.                                                              

(2)設(shè)直線AB的方程為y=(x+c),直線AB與橢圓的交點為A(x₁,y₁),B(x₂,y₂).

由(1)可得a²=4c²,b²=3c².

由

消去y,得11x²+16cx-4c²=0.                                                   

x₁+x₂=,x₁x₂=c².

=(x₁,y₁)·(x₂,y₂)=x₁x₂+y₁y₂,

且y₁·y₂=2(x₁+c)(x₂+c)=2x₁x₂+2c(x₁+x₂)+2c².

∴3x₁x₂+2c(x₁+x₂)+2c²=-2,                                                   

即c²+2c²=-2.

∴c²=1.則a²=4,b²=3.

橢圓的方程為=1.