高考数学问题:抛物线y=x^2+2ax+b和x轴交于A,B两点 如图，已知抛物线y=-x2+ax+b与x轴从左至右交于A、B...
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x1\uff1c0\uff0cx2\uff1e0\uff0cb\uff1e0\uff0c
\u2235x1\uff0cx2\u662f\u65b9\u7a0b-x2+ax+b=0\u7684\u4e24\u6839\uff0c
\u2234x1+x2=a\uff0cx1�6�1x2=-b\uff1b\uff081\u5206\uff09
\u5728Rt\u25b3ABC\u4e2d\uff0cOC\u22a5AB\uff0c
\u2234OC2=OA�6�1OB\uff0c
\u2235OA=-x1\uff0cOB=x2\uff0c
\u2234b2=-x1�6�1x2=b\uff0c\uff082\u5206\uff09
\u2235b\uff1e0\uff0c
\u2234b=1\uff0c
\u2234C\uff080\uff0c1\uff09\uff1b\uff083\u5206\uff09
\uff082\uff09\u5728Rt\u25b3AOC\u548cRt\u25b3BOC\u4e2d\uff0c
tan\u03b1-tan\u03b2=OCOA-OCOB=-1x1-1x2=-x1+x2x1x2=ab=2\uff0c\uff084\u5206\uff09
\u2234a=2\uff0c
\u2234\u629b\u7269\u7ebf\u89e3\u6790\u5f0f\u4e3a\uff1ay=-x2+2x+1\uff0e\uff085\u5206\uff09
\uff083\uff09\u2235y=-x2+2x+1\uff0c
\u2234\u9876\u70b9P\u7684\u5750\u6807\u4e3a\uff081\uff0c2\uff09\uff0c
\u5f53-x2+2x+1=0\u65f6\uff0cx=1\u00b12\uff0c
\u2234A\uff081-2\uff0c0\uff09\uff0cB\uff081+2\uff0c0\uff09\uff0c\uff086\u5206\uff09
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1, 抛物线y=x^2+2ax+b和x轴交于A,B两点,要使抛物线的顶点在以AB为直径的圆内抛物线:y=x^2+2ax+b=(x+a)^2+b-a^2，顶点C(-a,b-a^2),b=a^2，抛物线与X轴相切，故|b-a^2|>0
y=0,x^2+2ax+b=0
x=-a±√(a^2-b)
A[-a-√(a^2-b),0],B[-a+√(a^2-b),0]
|AB|=2√(a^2-b),b<a^2
AB为直径的圆D，r（D）=√(a^2-b)
顶点在以AB为直径的圆D内，则AC=BC,∠ACB≥90°，即0<|CD|≤rD
|CD|=|b-a^2|
0<|b-a^2|≤√(a^2-b)
b≤a^2
0<a^2-b≤√(a^2-b)
0<(a^2-b)^2≤a^2-b
a,b应满足关系式:a^2>b≥a^2-1
2,已知抛物线y=x^2-1上一定点B(-1,0)和两动点P,Q,当点P在抛物线上运动时,BP⊥PQ,则点Q的横坐标取值范围是
yP=(xP)^2-1,yQ=(xQ)^2-1
k(BP)=yP/(xP-xB)=[(xP)^2-1]/(xP+1)=xP-1
k(PQ)=[(yQ-yP)/(xQ-xP)]=[(xQ)^2-(xP)^2)]/(xQ-xP)=xQ+xP
BP⊥PQ
k(BP)*k(PQ)=-1
(xP-1)*(xQ+xP)=-1
(xP)^2+(xQ-1)xP+1-xQ=0
△≥0
(xQ-1)^2-4*(1-xQ)≥0
(xQ)^2+2xQ-3≥0
(xQ+3)*(xQ-1)≥0
Q的横坐标取值范围是:xQ≥1,xQ≤-3
设点画图即可求解啊
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