...△AOB的位置如图所示,已知∠AOB=90°,AO=BO,点A的坐标为(-3,1...
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发布时间：2024-10-22 22:10
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时间：2024-11-13 08:06
则∠ACO=∠ODB=90°，
∴∠AOC+∠OAC=90°．
又∵∠AOB=90°，
∴∠AOC+∠BOD=90°
∴∠OAC=∠BOD．
在△ACO和△ODB中，
∠ACO＝∠ODB∠OAC＝∠BODAO＝BO
∴△ACO≌△ODB（AAS）．
∴OD=AC=1，DB=OC=3．
∴点B的坐标为（1，3）．
（2）因抛物线过原点，
故可设所求抛物线的解析式为y=ax2+bx．
将A（-3，1），B（1，3）两点代入，
得a+b＝39a?3b＝1，
解得：a=56，b=136
故所求抛物线的解析式为y=56x2+136x．
（3）在抛物线y=56x2+136x中，对称轴l的方程是x=-b2a=-1310
点B1是B关于抛物线的对称轴l的对称点，
故B1坐标（-185，3）
在△AB1B中，底边B1B=235，高的长为2．
故S△AB1B=12×235×2=235已赞过已踩过你对这个回答的评价是？评论收起 ._1uevpeq{zoom:1;background-color:#fff;border:0;margin-bottom:10px;padding:30px 0 20px 42px;position:relative}._1uevpeq.ec-1841{padding:20px 0}._1uevpeq.ec-2246{padding:20px 0 10px}.ec-1841 .y7we4hu{font-size:16px;margin-bottom:-5px}.y7we4hu{color:#7a8f9a;height:25px;line-height:25px;overflow:hidden;position:relative}.y7we4hu h2{margin:0;padding:0}.y7we4hu:after{clear:both;content:" ";display:block;height:0;visibility:hidden}a.tycfu7u{color:#666;float:right;font-size:12px;margin-left:8px;text-decoration:none}.hhhv6ex{color:#666;font-size:13px;line-height:normal;line-height:20px;margin-top:10px}.vnsdjzp{margin-top:15px;position:relative}.vnsdjzp h3{font-weight:400;padding:0}.vnsdjzp a{text-decoration:none}.vnsdjzp em{color:#d81419;font-style:normal}.ec-2246 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