Question

已知P是橢圓C:=1(a＞b＞0)上異于長軸端點(diǎn)的任意一點(diǎn),A為長軸的左端點(diǎn),F為橢圓的右焦點(diǎn),橢圓的右準(zhǔn)線與x軸、直線AP分別交于點(diǎn)K、M,=3 .

(1)若橢圓的焦距為6,求橢圓C的方程;

(2)若·=0,且=λ,求實(shí)數(shù)λ的值.

Answer 1

(1)解法一:由=3,得a+3=3(-3),a=4.                        

∴b²=a²-c²=7,                                                            

從而橢圓方程是=1.                                               

解法二:記c=,由=3,得a+c=3(-c)=,

∵a+c＞0,∴3a=4c.                                                     

又2c=6,c=3,∴b²=a²-c²=7,                                              

從而橢圓方程是=1.                                            

(2)解法一:點(diǎn)P(x_P,y_P)同時(shí)滿足=1和(x+a)(x-c)+y²=0.

消去y²并整理得c²x²+a²(a-c)x-a³c+a²b²=0,                               

此方程必有兩實(shí)根,一根是點(diǎn)A的橫坐標(biāo)-a,另一根是點(diǎn)P的橫坐標(biāo)x_P,

-a·x_P=,x_P=.                                  

∴x_P-x_A=-(-a)=,x_m-x_P=-=.

∴=||=||=||,                  

由=代入上式可得=2.

∴=,λ=2.                                              

解法二:由(1) =3,3a=4c,

可設(shè)a=4t,c=3t,則b=t,

橢圓方程可為=1,

即7x²+16y²=112t².                                                 

設(shè)直線AM的方程為y=k(x+4t)(k存在且k≠0),

代入7x²+16y²=112t²,

整理得(16k²+7)x²+128k²tx+256k²t²-112t²=0,                            

此方程兩根為A、P兩點(diǎn)的橫坐標(biāo),

由韋達(dá)定理有-4t·x_P=,x_P=,

∴x_P=,

從而y_P=.

由于k_PF=,k²=,                             

=||==.

∴=2,λ=2.