不妨設a=3,c=√6,則b^2=3,橢圓方程為x^2/9+y^2/3=1,
右焦點F(√6,0）,AB:y=x-√6,代入上式得
x^2+3(x^2-2x√6+6）=9,
4x^2-6x√6+9=0,
x1=(3√6+3√2）/4,x2=(3√6-3√2）/4,
y1=(-√6+3√2）/4,y2=(-√6-3√2）/4.
向量OM=cosx向量OA+sinx向量OB,
(3cost,√3sint)=cosx*((3√6+3√2）/4,(-√6+3√2）/4)+sinx*((3√6-3√2）/4,(-√6-3√2）/4)
12cost=cosx(3√6+3√2）+sinx(3√6-3√2）,
4√3sint=cosx(-√6+3√2）+sinx(-√6-3√2）,
解得cosx=[(√6+√2）cost+(√6-√2）sint]/4,①
sinx=[(√6-√2）cost-(√6+√2）sint]/4,②
①^2+②^2,得（cosx)^2+(sinx)^2=1,
∴命題成立.
不設a=3,c=√6,而設a=3k,c=k√6,k>0,計算方法類似.