2022英语周报ojs第6期答案

20.(理科【解析】(1)证明:由题意lx=5,代入y2=4x中,解得y=±2√5,不妨取M(5,2√5),N(5,-2√5)则AM=(4,2√5-2),AN=(4,-2√5-2)所以AMAN=4×4+(2V5-2)x(-2V5-2)=0所以AM⊥AN,即△AMN为直角三角形(2)存在m=6时,使得AM⊥AN由题意可得四边形OAPB为平行四边形,则kwp=ko=2,设直线=2x-m),M(23,N(2联立=2(x-m)得y2-2y-4m=0,由题意,判别式△=4+16m>0,y1+y2=2y1y2=-4m,要使AM⊥AN,则AM·AN=0,又AM=(21-1,y1-2)AN=(-1,y2-2),即(2)(y2-2)=0化简,得(y1+2)(y2+2)+16=0,即y1y2+2(y1+y2)+20=0,代入解得m=6.故m=6时,有AM⊥AN(文科)解析】(1)设P(x,y),则kpkpx+2 x整理可得31,y≠0,故C的方程为x+y=1,y≠0,说明C是(不包含y=0)的椭圆(2)假设存在满足题意的定点Q,设Q(0,m),当直线MN的斜率存在时,设直线MN的方程为y=kx+,M(x1,y1),N(x2,y2)由消去y,得(3+4k2)x2+4kx-11=0由直线MN过椭圆内一点(0,)故△>0,由根与系数关系得:x3+42,1=3+42,由∠MQO0=∠NQO,得直线MQ与直线NQ斜率和为零.故-m12-mkx1+-m如*t +x2k+(1-m).4=4(6-m)20I 1所以m=6,存在定点Q0,6),当直线MN的斜率不存在时定点(0,6)也符合题意

写作Paragraph 1: Jatin looked up and saw the other boys rushing ahead Refusing togive up at this point, he jumped to his feet and ran as fast as he could. Hedashed across the finishing line and won the fourth prize. Thinking he would lether sister down, Jatin couldn t help sobbing. Just then he suddenly heard theannouncement from the broadcast that he won the thirdprize as the runner whocollided with Jatin was disqualified for breaking the rule. holding the shoes closeto his chest during the award ceremony Jatin felt overjoyed and relieved as heeventually won the shoes for his sister.(要点1 Jatin获奖)Paragraph 2Filled with pleasure, Jatin walked home quickly.(高分句型一) He could notwait to see his sister and share the good news. On his arrival, Jatin found Zahrawas waiting outside. Nervous and anxious, Zahra wondered whether herbrother had brought her what she was expectingZahra, see what i have gotfor you. Jatin said excitedly taking the prize out of his bag with his tremblinghands. The instant Zahra saw the shoes, her face lit up. So thrilled was shethat she threw herself at Jatin.(高分句型二)" Thank you!ltin." Tears of joywelled up in their eyes as the brother and the sister hugged tightly together (4 R2把奖品给妹妹)

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