已知橢圓的兩個焦點分為F1(0,-2√2),F2(0,2√2),離心率e=(2√2)/3.(1) 求橢圓方程;(2) 一條不與坐軸平行的直線L與橢圓交于不同的兩點M,N,且線段MN中點的橫坐標為-(1/2),求直線L傾斜角的取值范圍.

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解:∵F1(0,-2√2),F2(0,2√2),c=2√2∵e=(2√2)/3.∴a=3,b=1∴橢圓方程為:x^+y/9=1設M(x1,y1),N(x2,y2),線段MN的中點T(-1/2,yo),直線L的斜率k.∵9x1^+y1^=9……①∵9x2^+y2^=9……②①-②得:9(x1^-x2^)+(y1^-y2^)=09(x1+x2)(x1-x2)+(y1+y2)(y1-y2)=0∴9×[2×(-1/2)]+(2yo)k=0,∴yo=9/(2k)∴T(-1/2,9/(2k)),中點T它在橢圓內,(-1/2)^+[9/(2k)]^＜1k^＞27∴k＜-3√3或k＞3√3∴傾斜角的取值范圍θ∈(aretan3√3,π/2)∪(π/2,π-aretan3√3)