20.本题考査抛物线的定义和方程,直线与抛物线的位置关系,弦长,面积等基础知识,考查运算求解能力、推理论证能力,考査数形结合思想,化归与转化思想.满分1分解:(1)∵动圆C过点F(1,0)且与直线l:x=-1相切点C到F(,0)的距离等于C到l的距离点C的轨迹是以F为焦点,为准线的抛物线,其方程为y2=4x分分(2)①证法一:设A(x1,y),B(x2,y2),则A(-,y)B(-1,y2)为线段AB的中点,∴M(-1,当+)依题意可设直线AB的方程为x=y+1得y2-4y-4=0△=16r2+16>0,y1+y2=4,yy2=5分M(-12),ku=(-1)-1当=0时,A,B关于x轴对称,点M恰为与x轴的交点,满足AB⊥FMAB⊥FM综上,AB⊥FM7分证法二:连接AF,BF,设直线l与x轴的交点为TAA∥x轴,AA=AF,∴∠AFA=∠AAF=∠AFT同理,∠BFB=∠BBF=∠BFT∠AFB1=∠AFT+∠BFT5分△AAM≌△AFM6分∴∠AFM=∠AAM=90°,即AB⊥FM7分=4F②法一:由AM AM得△AM≌△AFM8分∠AAM=∠AFM=90°同理△BBM≌△BFMSaAM 2SBum BF ya由yy2=-4知y,y2异号,故y1=-2y20分y2=2,yx+x2+22分法二:由AMAM得△AAM≌△AFM∠AAM=∠AFM=90°理△BBM≌△BFMS2S由对称性,不妨设点A在x轴上方,直线AB的倾斜角为a由定义易得|4F=A4|=14F1cosa+|F7=|4 FIcosa+2同理BF210分1 +cos a2,即cosaFl a-pr1+o-212分

书面表达One possible version:NOTICEA sports meeting will be held in the playground of our school from next Thursday to FridayAs you know, the pressure of study is very heavy now, especially for those senior 3. So the purpose of the sports meeting is to letevery student get relaxed, as a result of which we students can live happily and heal thilyEveryone is welcome to take part in it. Those who perform excellently at the sports meeting will get prizes. But don' t take theresults so serously because taking part is more important than the result. Good luck to everyone!

以上就是2022英语周报九年级HBE版第四期答案，更多英语周报答案请关注本网站。

本文地址： http://www.ncneedu.cn/post/8843.html

文章来源：admin

版权声明：除非特别标注，否则均为本站原创文章，转载时请以链接形式注明文章出处。