点P在直线l：y=x-1上，若存在过P的直线交抛物线y=x2于A，B两点，且PA＝AB，则称点P为“λ点”，那么直线l
\u70b9P\u5728\u76f4\u7ebfl\uff1ay=x-l\u4e0a\uff0c\u82e5\u5b58\u5728\u8fc7P\u7684\u76f4\u7ebf\u4ea4\u629b\u7269\u7ebfy=x 2 \u4e8eA\u3001B\u4e24\u70b9\uff0c\u4e14|PA|=|AB|\uff0c\u5219\u79f0\u70b9P\u4e3a\u201c \u70b9\u201d\u3002\u90a3\u4e48A
\uff081\uff09\u2235\u70b9A\u3001B\u662f\u629b\u7269\u7ebfy=x2\u4e0e\u76f4\u7ebfy=-12x+32\u7684\u4ea4\u70b9\uff0c\u2234x2=-12x+32\uff0c\u89e3\u5f97x=1\u6216x=-32\uff0e\u5f53x=1\u65f6\uff0cy=1\uff1b\u5f53x=-32\u65f6\uff0cy=94\uff0c\u2234A\uff08-32\uff0c94\uff09\uff0cB\uff081\uff0c1\uff09\uff0e\uff082\uff09\u2460\u2235\u70b9P\uff08-2\uff0ct\uff09\u5728\u76f4\u7ebfy=-2x-2\u4e0a\uff0c\u2234t=2\uff0c\u2234P\uff08-2\uff0c2\uff09\uff0e\u8bbeA\uff08m\uff0cm2\uff09\uff0c\u5982\u7b54\u56fe1\u6240\u793a\uff0c\u5206\u522b\u8fc7\u70b9P\u3001A\u3001B\u4f5cx\u8f74\u7684\u5782\u7ebf\uff0c\u5782\u8db3\u5206\u522b\u4e3a\u70b9G\u3001E\u3001F\uff0e\u2235PA=AB\uff0c\u2234AE\u662f\u68af\u5f62PGFB\u7684\u4e2d\u4f4d\u7ebf\uff0c\u2234GE=EF\uff0cAE=12\uff08PG+BF\uff09\uff0e\u2235OF=|EF-OE|\uff0cGE=EF\uff0c\u2234OF=|GE-EO|\u2235GE=GO-EO=2+m\uff0cEO=-m\u2234OF=|2+m-\uff08-m\uff09|=|2+2m|\u2234OF=2m+2\uff0c\u2235AE=12\uff08PG+BF\uff09\uff0c\u2234BF=2AE-PG=2m2-2\uff0e\u2234B\uff082+2m\uff0c2m2-2\uff09\uff0e\u2235\u70b9B\u5728\u629b\u7269\u7ebfy=x2\u4e0a\uff0c\u22342m2-2=\uff082+2m\uff092\u89e3\u5f97\uff1am=-1\u6216-3\uff0c\u5f53m=-1\u65f6\uff0cm2=1\uff1b\u5f53m=-3\u65f6\uff0cm2=9\u2234\u70b9A\u7684\u5750\u6807\u4e3a\uff08-1\uff0c1\uff09\u6216\uff08-3\uff0c9\uff09\uff0e\u2461\u8bbeP\uff08a\uff0c-2a-2\uff09\uff0cA\uff08m\uff0cm2\uff09\uff0e\u5982\u7b54\u56fe1\u6240\u793a\uff0c\u5206\u522b\u8fc7\u70b9P\u3001A\u3001B\u4f5cx\u8f74\u7684\u5782\u7ebf\uff0c\u5782\u8db3\u5206\u522b\u4e3a\u70b9G\u3001E\u3001F\uff0e\u4e0e\u2460\u540c\u7406\u53ef\u6c42\u5f97\uff1aB\uff082m-a\uff0c2m2+2a+2\uff09\uff0e\u2235\u70b9B\u5728\u629b\u7269\u7ebfy=x2\u4e0a\uff0c\u22342m2+2a+2=\uff082m-a\uff092\u6574\u7406\u5f97\uff1a2m2-4am+a2-2a-2=0\uff0e\u25b3=16a2-8\uff08a2-2a-2\uff09=8a2+16a+16=8\uff08a+1\uff092+8\uff1e0\uff0c\u2234\u65e0\u8bbaa\u4e3a\u4f55\u503c\u65f6\uff0c\u5173\u4e8em\u7684\u65b9\u7a0b\u603b\u6709\u4e24\u4e2a\u4e0d\u76f8\u7b49\u7684\u5b9e\u6570\u6839\uff0e\u5373\u5bf9\u4e8e\u4efb\u610f\u7ed9\u5b9a\u7684\u70b9P\uff0c\u629b\u7269\u7ebf\u4e0a\u603b\u80fd\u627e\u5230\u4e24\u4e2a\u6ee1\u8db3\u6761\u4ef6\u7684\u70b9A\uff0c\u4f7f\u5f97PA=AB\u6210\u7acb\uff0e\uff083\uff09\u2235\u25b3AOB\u7684\u5916\u5fc3\u5728\u8fb9AB\u4e0a\uff0c\u2234AB\u4e3a\u25b3AOB\u5916\u63a5\u5706\u7684\u76f4\u5f84\uff0c\u2234\u2220AOB=90\u00b0\uff0e\u8bbeA\uff08m\uff0cm2\uff09\uff0cB\uff08n\uff0cn2\uff09\uff0c\u5982\u7b54\u56fe2\u6240\u793a\uff0c\u8fc7\u70b9A\u3001B\u5206\u522b\u4f5cx\u8f74\u7684\u5782\u7ebf\uff0c\u5782\u8db3\u4e3aE\u3001F\uff0c\u5219\u6613\u8bc1\u25b3AEO\u223d\u25b3OFB\uff0e\u2234AEOF\uff1dOEBF\uff0c\u5373m2n\uff1d?mn2\uff0c\u6574\u7406\u5f97\uff1amn\uff08mn+1\uff09=0\uff0c\u2235mn\u22600\uff0c\u2234mn+1=0\uff0c\u5373mn=-1\uff0e\u8bbe\u76f4\u7ebfm\u7684\u89e3\u6790\u5f0f\u4e3ay=kx+b\uff0c\u8054\u7acby\uff1dkx+by\uff1dx2<
解：本题采作数形结合法易于求解，如图，设A（m，n），P（x，x-1）
则B（2m-x，2n-x+1），
∵A，B在y=x2上，
∴n=m2，2n-x+1=（2m-x）2
消去n，整理得关于x的方程x2-（4m-1）x+2m2-1=0（1）
∵△=（4m-1）2-4（2m2-1）=8m2-8m+5＞0恒成立，
∴方程（1）恒有实数解，
∴有无穷多解．
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