(Ⅱ)由①②知,C的方程为:3x2?y2?3a2,
4?3a2A(a,0),F(2a,0),x1?x2?2,x1x2???0,
2故不妨设x1剠?a,x2(步骤5) a,
BF?FD??x1?2a??y12??x2?2a?22?x1?2a??3x12?3a2?a?2x1, ?x2?2a?22?y22??3x22?3a2?2x2?a,
BF?FD??a?2x1??2x2?a?
??4x1x2?2a?x1?x2??a2 ?5a2?4a?8.(步骤6)
又BF?FD?17,故5a?4a?8?17, 解得a?1,或a??故BD?29(舍去), 52x1?x2?2??x1?x2?2?4x1x2?6.(步骤7)
连结MA,则由A(1,0),M(1,3)知MA?3,从而MA?MB?MD,且MA?x轴,因此以M为圆心,MA为半径的圆经过A、B、D三点,且在点A处与x轴相切,所以过A、
B、D三点的圆与x轴相切.(步骤8)
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