Question

已知P是椭圆C:=1(a＞b＞0)上异于长轴端点的任意一点,A为长轴的左端点,F为椭圆的右焦点,椭圆的右准线与x轴、直线AP分别交于点K、M,=3 .

(1)若椭圆的焦距为6,求椭圆C的方程;

(2)若·=0,且=λ,求实数λ的值.

Answer 1

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

∴b²=a²-c²=7,                                                            

从而椭圆方程是=1.                                               

解法二:记c=,由=3,得a+c=3(-c)=,

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

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

从而椭圆方程是=1.                                            

(2)解法一:点P(x_P,y_P)同时满足=1和(x+a)(x-c)+y²=0.

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

此方程必有两实根,一根是点A的横坐标-a,另一根是点P的横坐标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,

可设a=4t,c=3t,则b=t,

椭圆方程可为=1,

即7x²+16y²=112t².                                                 

设直线AM的方程为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,                            

此方程两根为A、P两点的横坐标,

由韦达定理有-4t·x_P=,x_P=,

∴x_P=,

从而y_P=.

由于k_PF=,k²=,                             

=||==.

∴=2,λ=2.