Question

圆锥曲线问题已知椭圆C：x²/a² +y²/b²=1（a＞b＞0）的离心率为√6

Answer 1

c/a=√6/3,b^2=a^2-c^2=1/3a^2

∴椭圆C:x²/a² +3y²/a²=1

x^2+3y^2=a²

1

AB:x=y+c代入 x^2+3y^2=a²

（y+c)^2+3y^2=a^2

4y^2+2cy+c^2-a^2=0

设A（x1,y1),B(x2,y2).N(x0,y0)

2y0=y1+y2=-c/2,y1y2=(c^2-a^2)/4

y0=-c/4,x0=3/4c

KoN=y0/x0=-1/3

2

M(x,y)

OM=mOA+nOB

=(mx1+nx2,my1+ny2)

∴(mx1+nx2)^2+3(my1+ny2)^2-a²=0

∴m^2(x1²+3y1²)+n^2(x2²+3y^2)+2mnx1x2+6mny1y2-a^2=0

m^2a^2 +n^2a^2+2mn(x1x2+3y1y2)-a^2=0

3y1y2=-3a^2/4,

x1x2=(y1+c)(y2+c)=y1y2+c(y1+y2)+c^2

∴x1x2+3y1y2=4y1y2+c(y1+y2)+1

= 4(c^2-a^2)/4+(-c^2/4)+c^2

=-a^2/3-c^2/6+c^2=0

∴ m^2a^2 +n^2a^2=a^2

∴m²+n²=1,m=cosθ,n=sinθ

即总存在角θ∈R使等式：

向量OM=cosθ向量OA+sinθ向量OB成立