已知橢圓C:x^2/a^2+y^2/b^2=1(a>b>0)的離心率e=√2/2,左、右焦點(diǎn)分別為F1.F2,定點(diǎn)p（2,√3）,且|F1F2|=|PF2|
1.求橢圓C的方程
2.設(shè)直線l：y=kx+m與橢圓C交于M.N兩點(diǎn),直線F2M與F2N的斜率和為零,求m與k的關(guān)系
1.解析：∵橢圓C:x^2/a^2+y^2/b^2=1(a>b>0)的左、右焦點(diǎn)分別為F1.F2,定點(diǎn)p（2,√3）,且|F1F2|=|PF2|
4c^2=(2-c)^2+3==>3c^2+4c-7=0==>c1=1,c2=-7/3(舍)
又離心率e=√2/2,∴a=√2,b=1
∴橢圓C ：x^2/2+y^2=1
2.解析：設(shè)直線l：y=kx+m==> y^2=k^2x^2+2kmx+m^2
代入橢圓x^2/2+y^2=1
得(1+2k^2)x^2+4kmx+2m^2-2=0
設(shè)M(x1,y1),N(x2,y2)
則x1+x2=-4km/(1+2k^2),x1x2=(2m^2-2)/(1+2k^2)
∵直線F2M與F2N的斜率和為零
Y1/(x1-1)=-y2/(x2-1)==>y1/y2=-(x1-1)/ (x2-1)
(kx1+m)/ (kx2+m)=(1-x1)/ (x2-1)
kx2+m- kx1x2-mx1=kx1x2+mx2-kx1-m
2kx1x2+(m-k)(x1+x2)-2m=0
2k(2m^2-2)-(m-k)4km-2m(1+2k^2)=0
∴m+2k=0==>m=-2k