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
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![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](/uploads/image/z/8689183-7-3.jpg?t=F1F2%E4%B8%BA%E6%A4%AD%E5%9C%86C%E4%B8%A4%E7%84%A6%E7%82%B9%2CP%E4%B8%BAC%E4%B8%8A%E5%8A%A8%E7%82%B9%2CQ%E6%BB%A1%E8%B6%B3%28PQ%E5%90%91%E9%87%8F%29%3D%CE%BB%28%28PF1%E5%90%91%E9%87%8F%2F%7CPF1%7C%29-%28PF2%E5%90%91%E9%87%8F%2F%7CPF2%7C%29%29%7B%E6%B3%A8%E6%84%8F%E6%98%AF%E4%B8%AD%E9%97%B4%E6%98%AF%E5%87%8F%7D%2C%E4%B8%94PQ%E5%9E%82%E7%9B%B4%E4%BA%8EPF2%2C%E6%B1%82Q%E7%82%B9%E8%BD%A8%E8%BF%B9%E6%96%B9%E7%A8%8B.%E6%98%AFPQ%E5%9E%82%E7%9B%B4%E4%BA%8EQF2)
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).