2022英语周报八年级新课程答案

书面表达One posssible versionDear Mr. Smith,I'm Li Hua, a senior 3 student. Learning that your City Culture department is looking for an English editorthis summer, I'm eager to apply for the position.As far as I'm concerned, city culture refers to a citys history, geography, customs, future, and so on. Asa native, I am quite familiar with our citys traditional culture. Besides, I have a good command of ErI' m especially good at English writing. Additionally, Im cooperative and creative, which will give me anadvantage for the position.I'll be greatly honored if I' m chosen for the job. If I'm admitted, please email me at lihua @16com.Yours sincerely,Li Hu

20.0解:(1)由题意得2a=邻c=-3-22a<0c+3(舍去),或c=-3+2√2a(2分)又c=2,所以a=2,c=1,(3分)所以b2=a2-c2=1,所以椭圆C的方程为+y2=1(4分(2)当直线l的斜率存在时,设l:y=kx+2,A(x1,31),B(232).+y2=1,联立2y=kx+2,消去y,得(1+2k2)x2+8kx+6=0由△=64k2-24(1+2k2)=16k2-24>0,解得k<-6或k2&k6则x1+x2,1C2(6分)1+2k1+2k2若在y轴上存在异于点P的定点Q,使向量QQAiBQ与OF共线I BQ则∠AQB的平分线与x轴平行,即kaA+kaB=0,(7分)设Q(0,m),则k÷y1-my2-my1x2+y2x1-m(x1+x2)所以kaA+kQB=122kx1x2+(2-m)(x1+x2)C1.26k-4k(2-m)2k(2m-1)=0(米)3要使(*)式恒成立,只有2m-1=0,解得m=2,即Q(0,2(10分)QABQ当直线l的斜率不存在时=0,显然向QAI BQQA BQ量一与OF共线(11分)IQAI IBQBQ综上所述,存在定点Q(0,。),使向量IQAI IBQI与OF共线(12分)

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