高考数学复习 名师原创理科数学专题卷：专题十三《圆锥曲线与方程》

由题意可设直线AB:x?my?1?m?0?，代入y?4x，得y?4my?4?0，

222?y12??y2?,y1?,B?,y2?，则y1?y2?4m,y1y2??4； 设A??4??4?又H?1,2?，设直线AH,BH的斜率分别为k1,k2， 则k1?y1?2y2?244?,k??， 222y1y2y1?2y2?2?1?144设M??1,yM?,N??1,yN?， 令x??1，得yM?2?2?y1?2?8， ?y1?2y1?22?y2?2?8同理，得yN?2?， ?y2?2y2?2从而

yMyN?y1y2?2?y1?y2??4?2?y1?2?2?y2?2?4???4??4?2?4m?4???4； ·??y1?2y2?2y1y2?2?y1?y2??4?4?2?4m?4?8??8?yM?yN??2????2??

y?2y?212?????11??4?8???

?y1?2y2?2??4?y1y2?2?y1?y2??48???y1?y2??4??

?4?8?4m?4??4?2?4m?4

A

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??4m.

又以MN为直径的圆的方程为： ?x?1?2??y?yM??y?yN??0， 即y2??y2M?yN?y?yM·yN??x?1??0，即x2?2x?3?y2?4my?0，令{y?0x2?2x?3?y2?0，解得x??3或x?1，

从而以MN为直径的圆恒过定点??3,0?和?1,0?.

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