F1F2为椭圆C两焦点,P为C上动点,Q满足(PQ向量)=λ((PF1向量/|PF1|)-(PF2向量/|PF2|)){注意是中间是减},且PQ垂直于PF2,求Q点轨迹方程.是PQ垂直于QF2

F1F2为椭圆C两焦点,P为C上动点,Q满足(PQ向量)=λ((PF1向量/|PF1|)-(PF2向量/|PF2|)){注意是中间是减},且PQ垂直于PF2,求Q点轨迹方程.是PQ垂直于QF2
F1F2为椭圆C两焦点,P为C上动点,Q满足(PQ向量)=λ((PF1向量/|PF1|)-(PF2向量/|PF2|)){注意是中间是减},且PQ垂直于PF2,求Q点轨迹方程.
是PQ垂直于QF2

F1F2为椭圆C两焦点,P为C上动点,Q满足(PQ向量)=λ((PF1向量/|PF1|)-(PF2向量/|PF2|)){注意是中间是减},且PQ垂直于PF2,求Q点轨迹方程.是PQ垂直于QF2
F1F2为椭圆C两焦点,P为C上动点,Q满足向量PQ=λ(PF1/|PF1|-PF2/|PF2|),且PQ垂直于QF2,求Q点轨迹方程.
设F1(-c,0),F2(c,0),P(x1,y1),Q(x,y),|PF1|+|PF2|=2a,
由向量PQ=λ(PF1/|PF1|-PF2/|PF2|),得
(x-x1,y-y1)=λ[(-c-x1,-y1)/(a+cx1/a)-(c-x1,-y1)/(a-cx1/a)],
∴x=x1+λ[(-ac-ax1)/(a^2+cx1)+(-ac+ax1)/(a^2-cx1)]
=x1+λac[x1^2+(a-c)x1-a^2]/(a^4-c^2*x1^2),
（x-x1)(a^4-c^2*x1^2)=λac[x1^2+(a-c)x1-a^2],①
y=y1+λ[-ay1/(a^2+cx1)+ay1/(a^2-cx1)]
=y1+2λacx1y1/(a^4-c^2*x1^2),②
由PQ垂直于QF2得
（x-x1)(c-x)-y(y-y1)=0,
∴y1=[y^2-(x-x1)(c-x)]/y,
代入②,y=[1+2λacx1/(a^4-c^2*x1^2)]*[y^2-(x-x1)(c-x)]/y,
∴y^2*(a^4-c^2x1^2)=(a^4-c^2*x1^2+2λacx1)*[y^2-(x-x1)(c-x)],③
①*（x-c）+③,消去x1^3项,化简得
0=[x1^2+(a-c)x1-a^2](x-c)+2x1*[y^2-(x-x1)(c-x)],
∴(c-x)x1^2+[(x-c)(2x+a-c)+2y^2]x1-a^2(x-c)=0,
解出x1,代入①（或③）,就得Q的轨迹方程（甚繁）.