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设椭圆方程是x^2/a^2+y^2/b^2=1e=c/a=根号3/24/a^2+8/b^2=1c^2=a^2-b^2解得a^2=4,b^2=1即椭圆方程是x^2/4+y^2=1.2.令P(x1,y1),Q(x2,y2).直线PQ方程为:y=kx+m(其中k=(y2-y1)/(x2-x1)) ①代入x^2/4+y^2=1②并整理得:(1+4k^2)x^2+8kmx+4m^2-4=0. ③依题意有Δ=16(1+4k^2-m^2)>0. x1+x2=-(8km)/ (1+4k^2).| x2-x1|=√Δ]/(1+4k^2), | y2-y1|=|k( x2-x1)|= |k|√Δ]/(1+4k^2), x1x2=4(m^2-1) /(1+4k^... |
