如图,已知抛物线y=x^2-6x+9的顶点为P，与y轴焦点与B，一经过点B的直线y=-x+b 如图，已知抛物线y= x 2 -2x+1的顶点为P，A为抛...
\u5982\u56fe\uff0c\u5df2\u77e5\u629b\u7269\u7ebfy=x²-6x+9\u7684\u9876\u70b9\u4e3a\u70b9P\uff0c\u4e0e y\u8f74\u4ea4\u4e8e\u70b9B\uff0c\u4e00\u7ecf\u8fc7\u70b9B\u7684\u76f4\u7ebfy=-x+b\u4e0e\u8be5\u629b\u7269\u7ebf\u4ea4\u4e8e\u70b9A\u3002\uff081\uff09
\u629b\u7269\u7ebf\u4e0ey\u8f74\u4ea4\u70b9\u4e3a(0,9)\uff0c\u6240\u4ee5b=9
\u76f4\u7ebf\u65b9\u7a0b\u4e3ay=-x+9
\u4e0e\u629b\u7269\u7ebf\u65b9\u7a0b\u8054\u7acb\uff0c\u89e3\u5f97
x=0,5
\u6240\u4ee5\u4ea4\u70b9A\u4e3a(5,4)
\uff082\uff09
P\u70b9\u5750\u6807\u4e3a(3,0)\uff0c\u5230\u76f4\u7ebfy=-x+9\u7684\u8ddd\u79bb\u4e3a3\u221a2
AB\u957f\u5ea6\u4e3a5\u221a2
\u6240\u4ee5S\u25b3APB=3\u221a2*5\u221a2/2=15
\u89e3\uff1a\uff081\uff09\u914d\u65b9\uff0c\u5f97y= \uff08x-2\uff09 2 -1\uff0c\u2234\u629b\u7269\u7ebf\u7684\u5bf9\u79f0\u8f74\u4e3a\u76f4\u7ebfx=2\uff0c\u9876\u70b9\u4e3aP\uff082\uff0c-1\uff09\uff0c\u53d6x=0\u4ee3\u5165y= x 2 -2x+1\uff0c\u5f97y=1\uff0c\u2234\u70b9A\u7684\u5750\u6807\u662f\uff080\uff0c1\uff09\uff0c\u7531\u629b\u7269\u7ebf\u7684\u5bf9\u79f0\u6027\u77e5\uff0c\u70b9A\uff080\uff0c1\uff09\u4e0e\u70b9B\u5173\u4e8e\u76f4\u7ebfx=2\u5bf9\u79f0\uff0c\u2234\u70b9B\u7684\u5750\u6807\u662f\uff084\uff0c1\uff09\uff0c\u8bbe\u76f4\u7ebfl\u7684\u89e3\u6790\u5f0f\u4e3ay=kx+b\uff08k\u22600\uff09\uff0c\u5c06B\u3001P\u7684\u5750\u6807\u4ee3\u5165\uff0c\u6709 \uff0c\u89e3\u5f97 \uff0c\u2234\u76f4\u7ebfl\u7684\u89e3\u6790\u5f0f\u4e3ay=x-3\uff1b\uff082\uff09\u8fde\u7ed3AD\u4ea4O\u2032C\u4e8e\u70b9E\uff0c\u2235\u70b9D\u7531\u70b9A\u6cbfO\u2032C\u7ffb\u6298\u540e\u5f97\u5230\uff0c\u2234O\u2032C\u5782\u76f4\u5e73\u5206AD\uff0c\u7531\uff081\uff09\u77e5\uff0c\u70b9C\u7684\u5750\u6807\u4e3a\uff080\uff0c-3\uff09\uff0c\u2234\u5728Rt\u25b3AO\u2032C\u4e2d\uff0cO\u2032A=2\uff0cAC=4\uff0c\u2234O\u2032C=2 \uff0c\u636e\u9762\u79ef\u5173\u7cfb\uff0c\u6709 \u00d7O\u2032C\u00d7AE= \u00d7O\u2032A\u00d7CA\uff0c\u2234AE= \uff0cAD=2AE= \uff0c\u4f5cDF\u22a5AB\u4e8eF\uff0c\u6613\u8bc1Rt\u25b3ADF\u223dRt\u25b3CO\u2032A\uff0c\u2234 \uff0c\u2234 \uff0cDF= \uff0c\u53c8\u2235OA=1\uff0c\u2234\u70b9D\u7684\u7eb5\u5750\u6807\u4e3a1- \uff0c\u2234\u70b9D\u7684\u5750\u6807\u4e3a\uff08 \uff0c- \uff09\uff1b\uff083\uff09\u663e\u7136\uff0cO\u2032P\u2225AC\uff0c\u4e14O\u2032\u4e3aAB\u7684\u4e2d\u70b9\uff0c\u2234\u70b9P\u662f\u7ebf\u6bb5BC\u7684\u4e2d\u70b9\uff0c\u2234S \u25b3DPC =S \u25b3DPB \uff0c\u6545\u8981\u4f7fS \u25b3DQC =S \u25b3DPB \uff0c\u53ea\u9700S \u25b3DQC =S \u25b3DPC \uff0c\u8fc7P\u4f5c\u76f4\u7ebfm\u4e0eCD\u5e73\u884c\uff0c\u5219\u76f4\u7ebfm\u4e0a\u7684\u4efb\u610f\u4e00\u70b9\u4e0eCD\u6784\u6210\u7684\u4e09\u89d2\u5f62\u7684\u9762\u79ef\u90fd\u7b49\u4e8eS \u25b3DPC \uff0c\u6545m\u4e0e\u629b\u7269\u7ebf\u7684\u4ea4\u70b9\u5373\u7b26\u5408\u6761\u4ef6\u7684Q\u70b9\uff0c\u5bb9\u6613\u6c42\u5f97\u8fc7\u70b9C\uff080\uff0c-3\uff09\u3001D\uff08 \uff09\u7684\u76f4\u7ebf\u7684\u89e3\u6790\u5f0f\u4e3ay= x-3\uff0c\u636e\u76f4\u7ebfm\u7684\u4f5c\u6cd5\uff0c\u53ef\u4ee5\u6c42\u5f97\u76f4\u7ebfm\u7684\u89e3\u6790\u5f0f\u4e3a \uff0c\u4ee4 x 2 -2x+1= \uff0c\u89e3\u5f97x 1 =2\uff0cx 2 = \uff0c\u4ee3\u5165y= \uff0c\u5f97y 1 =-1\uff0cy 2 = \uff0c\u56e0\u6b64\uff0c\u629b\u7269\u7ebf\u4e0a\u5b58\u5728\u4e24\u70b9Q 1 \uff082\uff0c-1\uff09\uff08\u5373\u70b9P\uff09\u548cQ 2 \uff08 \uff09\uff0c\u4f7f\u5f97S \u25b3DQC =S \u25b3DPB \u3002
1，求出b，再求A的坐标。解：y=-x+b
直线y与y轴有交点B。当x=0，y=b，B点坐标为（0，b）
抛物线与y轴有交点，则x=0时，y=9，则B点坐标为（0，9）
所以b=9
直线y=-x+9
抛物线与直线有2个交点，所以
-x+9=x平方-6x+9
解出x（x-5）=0
当x=0和x=5时候，直线与抛物线有2个交点，带入x=0和x=5，A点为（5，4）
2，三角形APB的面积可以用用大三角形减除2个小三角形的面积求出
定直线与x轴交点为N，则三角形APB=S三角形BON-（S三角形BOP+S三角形ABN）
根据三角形的面积公式可以得出S（BON）=(OB)*ON/2=9×9/2=81/2=40.5
S(BOP)=BO*OP/2=9*3/2=13.5
S（ABN）=A(5,4)*(ON-OP)/2=4×6/2=12
所以，S三角形APB=40.5-13.5-12=15
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