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设b>0,椭圆方程为
,抛物线方程为x2=8(y-b),如图所示,过点F(0,b+2)作x轴的平行线,与抛物线在第一象限的交点为G,已知抛物线在点G的切线经过椭圆的右焦点F1,(1)求满足条件的椭圆方程和抛物线方程;(2)设A,B分别是椭圆长轴的左、右端点,试探究在抛物线上是否存在点P,使得△ABP为直角三角形?若存在,请指出共有几个这样的点?并说明理由(不必具体求出这些点的坐标).
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![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAACAAAAAgCAYAAABzenr0AAAA5ElEQVRYR2NkGGDAOMD2M+B0wKpVq9huv/jbxPD/fxwDA4MkhQ59zsDIuEhVgrkuLCzsF7JZOB3QOml5B8P//+UUWoyqnZGxszovsoI4B0xc9gyfzxmZmb2rcsK3IRvWMWm599///7fgcfTz6vwoKWId8B+f75kZGX0q8iK3kugAhur8KJRQxx0FE5fhdQC5UTPqgNEQGA2B0RAYDYHREBgNgaETAuTW96TqG7yt4tbRFhGRIQBqHTP9+8dIoDUMTxqk5AK8zXKYiaDWMYhNpANIaJYPdMdkwLtmpOZnctUPeDkAAP6tmiH1ggTLAAAAAElFTkSuQmCC)
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参考答案:
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参考解析:
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