读后续写Possible version lI stared at the paper and then looked around. I was cold. Although I could not see my sistersnor the new house, which was hidden from my sight by a grey wall, I got the feeling that I was notalone. Walking along the shore was a tall familiar figure who looked lost in his thoughts. I lookedgain at the paper in my hand, closed my eyes and wishedAs /opened my eyes, I found my father was standing right beside me. Happy birthday, sonhe said, patting my shoulder gently. I smiled up at him and took his hand. Looking back, life fromthat day on marked the start of a close bond and a strong friendship with my father. What was mybirthday surprise? Wishes do come truePossible version 2I stared at the paper and then looked around. On the backside, there was a map. I noticed thatit was clearly leading me in a direction, but who could have sent this? I stood there in silence andstudied it. I decided to follow the map and it led me down the beach until I reached a staircase. Iooked back and could no longer see my siblings. I continued to follow the point on the map, untilI unexpectedly lost my footing and fell zksq down the stepsAs I opened my eyes, I found my father was standing right beside me. Happy birthday, sonhe said, patting my shoulder gently. Then he asked how I had come upon this place? I explained tohim that I had found the bottle in the ocean and when I opened it had the words on it. Shocked bythe message, my father realized what I had found I threw this bottle into the ocean many years agowhen I was alone. he said. I used to come to this exact place to get away from everyone else. NowI get to share this experience with you. What a birthday surprise
2.2解:(1)双曲线x22=1的离心率为得椭圆的离心率HB-4,即2a=4,即a=2,b=√3,椭圆的方程 y=(2)过左顶点A的直线/的斜率显然存在,设为k,方程设为y=k(x 2),可得E(0,2k)IL A(-2B(2,0),设P(ax24,可得(3 4)2 662-12=0.则-2=16/y(x 2即x6-8k3 4k即有D在平面内假设存在一定点P,使得E·B=0成立可得B=(=m)(3 4km 24k2-12n3 4k由于上式恒成立,可得k(4m 6)-3n=0,即有4m 6=0,且·3n=0.可得m=·号n=0则存在P(3,0),使得B=0h成立kL=3k.当k=0时,SAy=0此时SA=w223 4k23 4k12kL=3kL.当k=0时,5:.mr=0此时SADP=PD=3 4k23 4k,当且仅当=3取得等号当k≠0时,S:AD2时4|kkTIkI.综上可得,5△ADP的最大值为
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