英语周报八年级下册2021-2022第32期答案

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!

21.【考查目标】必备知识:本題主要考查拋物线与圆的方程、点到直线的距离、直线与圆的位置关系等知识关键能力:逻辑思維能力、运算求解能力学科素养:理性思维、数学应用、数学探索【解題思路】(1)根据题意得出P,Q的坐标,由P在C上得到P的值,即可得到C的方程,由⊙M与l相切得到⊙M的半径,即可得到⊙M的方程;(2)设出A1,A2,A3的坐标,分两种情况判断直线A2A3与⊙M的位置关系解:(1)由题意,直线x=1与C交于P,Q两点,且OP⊥0Q,设C的焦点为F,P在第一象限则根据抛物线的对称性,∠POF=∠QOF=45°,所以P(1,1),Q(1,-1)设C的方程为y=2px(p>0),则1=2,p=2,所以C的方程为y2=x由题意,圆心M(2,0)到l的距离即⊙M的半径,且距离为1,所以⊙M:(x-2)2+y2=1(2)设A1(x1,y1),A2(x2,y2),A3(x3,y3)当A1,A2,A3中有一个为坐标原点,另外两个点的横坐标均为3时,AA2,A1A3均与⊙M相切此时直线A2A3与⊙M相切当x1≠x2≠x3时,直线A1A2:x-(y1+y2)y+yy2=0则12+21=1,即(3-1)2+2y,y+3√(y1+y)同理可得(y2-1)y3+2yy+3-y=0所以y2,y是方程(y2-1)2+2y1t+3-y2=0的两个根,则y2323=直线A2A3的方程为x-(y2+y3)y+y2y3=0,直线A2A3的方程为x-(y2+y3)y+y2y3=0设M到直线A4的距离为d则=,(2+为y)2(2+31+(y2+y3)22即d=1,所以直线A2A3与⊙M相切综上,直线A2A3与⊙M相切

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