已知抛物线y=x²与直线y=2x在第一象限内有一个交点a,在x轴上是否存在一点p,使△aop为等腰三角形?
\u5982\u56fe\u6240\u793a\uff0c\u629b\u7269\u7ebfy=x2\u4e0e\u76f4\u7ebfy=2x\u5728\u7b2c\u4e00\u8c61\u9650\u5185\u6709\u4e00\u4e2a\u4ea4\u70b9A\uff0e\uff081\uff09\u6c42A\u70b9\u5750\u6807\uff1b\uff082\uff09\u5728x\u8f74\u4e0a\u662f\u5426\u5b58\u5728\u4e00\u70b9P\uff0c\u89e3\u7b54\uff1a\u89e3\uff1a\uff081\uff09\u89e3\u65b9\u7a0b\u7ec4y=x2y=2x\u5f97x=0y=0\u6216x=2y=4\uff0c\u6240\u4ee5A\u70b9\u5750\u6807\u4e3a\uff082\uff0c4\uff09\uff1b\uff082\uff09\u5b58\u5728\uff0e\u4f5cAB\u22a5x\u8f74\u4e8eB\u70b9\uff0c\u5982\u56fe\uff0c\u5f53PB=OB\u65f6\uff0c\u25b3AOP\u662f\u4ee5OP\u4e3a\u5e95\u7684\u7b49\u8170\u4e09\u89d2\u5f62\uff0c\u800cA\uff082\uff0c4\uff09\uff0c\u6240\u4ee5P\u70b9\u5750\u6807\u4e3a\uff084\uff0c0\uff09\uff0e
\u4ee4x^2=2x \u89e3\u5f97 x=2 \u6216x=0\u3002\u7531\u4e8e\u7b2c\u4e00\u8c61\u9650\uff0c\u6240\u4ee5 x\u4e0d\u7b49\u4e8e0\u3002 x=2\u65f6\uff0c y=4 \u6240\u4ee5A\u70b9\u5750\u6807\u4e3a\uff082,4\uff09
OA\u957f\u5ea6\u4e3a2\u221a5\uff0c\u82e5AOP\u4e3a\u7b49\u8170\u4e09\u89d2\u5f62\uff0c\u6709\u4e24\u79cd\u60c5\u51b5
\uff081\uff09 AP=2\u221a5\uff0c\u4ee5A\u4e3a\u5706\u5fc3\uff0c2\u221a5\u4e3a\u534a\u5f84\u505a\u5706\uff0c\u4ea4X\u8f74\u4e8eO\uff080\uff0c0\uff09\uff0c\uff084\uff0c0\uff09\u3002\u6240\u4ee5P\u70b9\u5750\u6807\u4e3a\uff084\uff0c0\uff09
\uff082\uff09OP=2\u221a5\uff0c\u4ee5O\u4e3a\u5706\u5fc3\uff0c2\u221a5\u4e3a\u534a\u5f84\u505a\u5706\uff0c\u4ea4x\u8f74\u4e8e\uff08-2\u221a5\uff0c0\uff09\uff0c\uff082\u221a5\uff0c0\uff09\u3002\u6240\u4ee5P\u70b9\u5750\u6807\u4e3a\uff08-2\u221a5\uff0c0\uff09\uff0c\uff082\u221a5\uff0c0\uff09\u3002
Y=X²
Y=2X
得X=0,Y=0或X=2,Y=4,
∴A(2,4),OA=2√5,
①当OP=OA=2√5时,P1(2√5。,0),P2(-2√5,0),
②AO=AP,P3(4,0),
③PO=PA,P4(5,0)。
其中作O的垂直平分线PQ,交OA于Q,交X轴于P,过A作AC⊥X轴于C
则OQ=1/2OA=√5,
ΔOPQ∽ΔOCA,OP/OQ=OC/OA=2/(2√5)=1/√5,
∴OP=5。
由y=x²与y=2x得到A(2,4)
若AO=AP,则P(4,0);
若PO=PA,则P(5,0);
若OA=OP,则P(2√5,0)或(-2√5,0)。
过程有点麻烦,就直接报答案吧
P(4,0)
P(-2√5,0)
P((√10)/4,0)
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