?当P为AE与x=1的交点时?ACP的周长最小令AE的直线解析式为:y=kx+b(k?0) 第11页
(3)?要使?ACP的周长最小?只需使AP?CP最小?点P在x?1上且由(2)中图可知点C与点E关于x=1对称1539?A(-3,0),E(2,)?y?x??xP?1?yP?3?P(1,3)444令经过点P(1,3)的直线为y=mx+3-m?y=mx+3-m,y??14x?212x?154?x?(4m?2)x?4m?3?02?x1?x2?2?4m,x1x2??4m?3?y1?mx1?3?m,y2?mx2?3?m?y1?y2?m(x1?x2)?M1M2??2(x1?x2)?(y1?y2)?222(x1?x2)?m(x1?x2)?2221?m?(x1?x2)221?m?(x1?x2)?4x1x21?m?22?M1M2??M1P?(x1?x2)?4x1x2?2221?m?2(2?4m)?4(?4m?3)?4(1?m)222(x1?1)?(y1?3)?22(x1?1)?(mx1?3?m?3)?21?m?(x1?1)2222同理:M2P?1?m?(x2?1)?M1P?M2P?22221?m?(x1?1)?1?m?(x2?1)2?M1P?M2P?(1?m)(x1?1)(x2?1)?(1?m)?(1?m)?2?x1x2?(x1?x2)?1?2??4m?3?(2?4m)?1??4(1?m)4(1?m)222?4(1?m)为定值2M1P?M2PM1M2?1?M1P?M2PM1M2
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