橢圓C方程x^2/4+y^2/3=1,左右端點(diǎn)為A(-2,0), B(2,0)
c=√(a^2-b^2)=√(4-3)=1,右焦點(diǎn)為F(1,0)
設(shè)過(guò)右焦點(diǎn)的直線為y=k(x-1)
代入橢圓方程得 x^2/4+k^2(x-1)^2/3=1
整理得 3x^2+4k^2(x-1)^2-12=0
(3+4k^2)x^2-8k^2x+4k^2-12=0(1)
解此一元二次方程可得兩個(gè)x的解,x1,x2；
同時(shí)有x1+x2=8k^2/(3+4k^2),x1x2=(4k^2-12)/(3+4k^2)
代入y=k(x-1)可解得y的兩個(gè)解,y1=k(x1-1),y2=k(x2-1).
不妨設(shè)直線與橢圓的交點(diǎn)為M(x1,y1),N(x2,y2)
則直線AM的方程為：y=y1/(x1+2)*(x+2)
直線BN的方程為：y=y2/(x2-2)*(x-2)
由二直線可解出交點(diǎn)P為：y1/(x1+2)*(x+2)=y2/(x2-2)*(x-2)
y1/y2=(x1+2)/(x2-2)*(x-2)/(x+2)=(x1-1)/(x2-1)
(x-2)/(x+2)=(x1-1)/(x2-1)*(x2-2)/(x1+2)
1-4/(x+2)=1-(x1+3x2-4)/(x1x2-x1+2x2-2)
(x+2)/4=(x1x2-x1+2x2-2)/(x1+3x2-4)
x=4(x1x2-x1+2x2-2)/(x1+3x2-4)-2
x=4[2x1x2-3x1+x2]/(2x1+6x2-8)
∵(2x1x2-3x1+x2)-(2x1+6x2-8)
=2x1x2-5(x1+x2)+8
=2*4(k^2-3)/(3+4k^2)-5*8k^2/(3+4k^2)+8
=-8[5k^2-k^2+3]/(3+4k^2)+8
=-8+8=0
∴(2x1x2-3x1+x2)=(2x1+6x2-8)
(2x1x2-3x1+x2)/(2x1+6x2-8)=1
∴x=4
代入解得y=6y1/(x1+2)=2y2/(x2-2)
即交點(diǎn)橫坐標(biāo)為x=4,縱坐標(biāo)隨點(diǎn)M,N的變化而變化,即為直線x=4
∴交點(diǎn)P在定直線x=4上運(yùn)動(dòng)