22解:(1)由ctn=4,得m=生coso sing,即psn2l4pcose=0因为{x=p9y=pine,得曲线C的直角坐标方程为y2=4x,由直线l的参数方程是2+te(3分y-tsing(为参数)得sina〓tana即y=(tan)x-2tana,可得直线l的普通方程cOSa(tana)x5分)(2)不妨设点A,B对应的参数公s轴,2·tana0把直线4的参数方程/x=2+c0(t为参y-isin数)代人y2=4x得:t2sina-4cosa·t-8=0,则△=16csa+32sna>0,t+4co7分)n2所以|AB|=|41-t2|=1+t2)2-414=4gy+3216+16sisin asIn得6sina-sin2a-1=0,所以sina=2(负舍),所以si=(9分)所以a=兀或a=3(10分)

书面表达Dear sir or madamI am Li Hua, a middle school student from China who is crazy about differentcultures around the world. Recently, I learned on your official website that therell be a summer camp in my city. I'm writing to participate in it.First, having learned English for almost ten years, I can speak English flu-ently. Moreover, with the development of China, there are more and more young-sters who show great interest in Chinese culture and this is really a wonderfulportunity for us to communicate the different cultures with each other, I will he a-hle to learn from other cultures to broaden my horizons and spread Chinese cultureI hope I can he accepted as a member in your summer caLooking forward to your reply.RegardsLi H

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