0)的半焦距为c,直线l过(a,0),(0,b)两点,且点(1,0）到直线l的距离与点（-1,0）到

0)的半焦距为c,直线l过(a,0),(0,b)两点,且点(1,0）到直线l的距离与点（-1,0）到

∵a>1>0,b>0∴ab>b>0∴∣-b-ab∣=b+ab ∣b-ab∣=-(b-ab)=ab-b∴∣b-ab∣+∣-b-ab∣=2ab即2ab/c≥4c/5∴5ab≥2c2将b=(c^2-a^2)^(-1/2)带入上式,可得(c^2-a^2)^(-1/2)×5a≥2c2两边同除以5a2,得(e^2-1)^(1/2)≥2e2/5解得5/2≤e2≤5又e>1,解得(5/2)^(1/2)≤e≤5^(1/2)======以下答案可供参考======供参考答案1:对的供参考答案2:x = 0, y = 3, C(0, 3)线段BC的方程: x/(-3) y/3 = 1, y = x 3 ( -3 设E(e, e 3), (-3 经过B、E、O三点的圆的圆心G显然在OB的中垂线x = -3/2上；G还在OE的中垂线上。OE的斜率=(e 3)/e, OE的中垂线斜率= -e/(e 3), OE的中点H(e/2, (e 3)/2)OE的中垂线方程：y - (e 3)/2 = -[e/(e 3)](x - e/2)取x = -3/2, y = (2e 3)/2G(-3/2, (2e 3)/2)为了简单起见，令G(-3/2, b), b = (2e 3)/2圆的方程：(x 3/2)² (y - b)² = r²圆过点O: 9/4 b² = r²(x 3/2)² (y - b)² = 9/4 b² (i)BC斜率 = 1BF斜率= -1, BF的方程: y - 0 = -(x 3), y = -(x 3) (ii)(i)(ii)联立; 2x² (2b 9)x 6b 9 = 0(x 3)(2x 2b 3) = 0x = -3 (点B)x = -(2b 3)/2 = -(e 3)代入(ii), y = eF(-e-3, e)OE = √[e² (e 3)²]OE的方程: y = (e 3)x/e, (e 3)x - ey = 0F与OE的距离h = |(e 3)(-e-3) - e²|/√[(e 3)² (-e)²]= |2e² 6e 9|/√[e² (e 3)²]△OEF面积S = (1/2)*OE*h = |e² 3e 9/2|= (e 3/2)² 9/4e = -3/2时，S最小， 此时E(-3/2, 3/2)

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