数学高手们来做这道高二的解析几何题在抛物线x^2=2py上有两点A（x1,y1)、B(x2,y2),|AB|=y1+y2+p，求证：A 一道高二抛物线解析几何题，没算出来。。求解
\u9ad8\u4e8c\u6570\u5b66\u89e3\u6790\u51e0\u4f55\uff0c\u6c42\u9ad8\u624b\uff01\uff01\uff011. \u89e3\uff1aOD\u7684\u659c\u7387\u4e3a2/1=2\uff0c\u76f4\u7ebfAB\u4e0eOD\u5782\u76f4\uff0c\u6240\u4ee5AB\u7684\u659c\u7387\u4e3a-1/2\uff0c\u53c8AB\u8fc7\u70b9D\uff0c\u6240\u4ee5\u76f4\u7ebfAB\u7684\u65b9\u7a0b\uff08\u70b9\u659c\u5f0f\uff09\u4e3a\uff1a(y-2) = -1/2(x-1)\uff0c\u5373y = -1/2x+5/2;
\u8bb2\u76f4\u7ebf\u65b9\u7a0b\uff0c\u4e0e\u66f2\u7ebf\u65b9\u7a0bx^2 = 2py\u8054\u7acb\uff0c\u6d88\u53bby\uff0c\u53ef\u5f97\uff1ax^2+px-5p = 0\uff0c \u5219\u7531\u97e6\u8fbe\u5b9a\u7406x1x2=-5p;
\u8bbeA(x1,y1)\uff0cB(x2,y2)\uff0c\u5219\u7531OA\u4e0eOB\u5782\u76f4\uff0c\u53ef\u77e5y1/x1 * y2/x2=-1, \u5c06y1 = 1/(2p) * x1^2 \u548c y2 = 1/(2p) * x2^2\u5e26\u5165\u4e0a\u5f0f\u53ef\u5f97\uff1ax1x2=-4p^2;
\u6240\u4ee5 -5p=-4p^2\uff0c \u89e3\u5f97p=0(\u820d\u53bb)\uff0c p=5/4(\u6b63\u89e3)\u3002
2.\u89e3\uff1a\u629b\u7269\u7ebf\u7126\u70b9F(1,0)\uff0c\u51c6\u7ebfx=-1\uff0c\u629b\u7269\u7ebf\u4e0a\u7684\u70b9\u5230\u7126\u70b9\u4e0e\u51c6\u7ebf\u7684\u8ddd\u79bb\u76f8\u7b49\uff0c\u6240\u4ee5A\u548cB\u5230\u51c6\u7ebfx=-1\u7684\u8ddd\u79bb\u5206\u522b\u4e3a2\u548c5\uff0c\u6240\u4ee5A(1,y1)\u548cB(4,y2)\uff0c\u5e26\u5165\u629b\u7269\u7ebf\u65b9\u7a0b\uff0c\u53c8A\u548cB\u5206\u522b\u5728\u5bf9\u79f0\u8f74x\u8f74\u7684\u4e0a\u65b9\u548c\u4e0b\u65b9\uff0c\u6240\u4ee5\u53ef\u6c42A\u548cB\u7684\u5750\u6807\u5206\u522b\u662fA(1,2)\u548cB(4,-4);
AB\u7684\u65b9\u7a0b\u4e3a\uff1a2x+y-4=0;
\u4e09\u89d2\u5f62PAB\u7684\u5e95\u8fb9AB\u957f\u5ea6\u56fa\u5b9a\uff0c\u4e3a \u6839\u53f7\u4e0b45\u3002
\u4e3a\u4f7f\u4e09\u89d2\u5f62PAB\u7684\u9762\u79ef\u6700\u5927\uff0c\u53ea\u9700\u5728\u629b\u7269\u7ebf\u4e0a\u627e\u4e00\u70b9P(x,y)\uff0c\u4f7f\u5176\u5230\u76f4\u7ebfAB\u7684\u8ddd\u79bb\u6700\u5927\u5373\u53ef,\u5373
|2x+y-4| / \u6839\u53f75 \u6700\u5927\uff0c\u4e14x=y^2/4\uff0c\u5f00\u53e3\u5411\u4e0a\uff0c\u6240\u4ee5\u6700\u5927\u503c\u4e3a\u65e0\u7a77\u5927\uff0c\u5373\u4e09\u89d2\u5f62PAB\u7684\u9762\u79ef\u53ef\u4ee5\u4efb\u610f\u5927\uff0c\u9898\u76ee\u5e94\u8be5\u6709\u8bef\u3002
\u8bbeM\u5750\u6807\u4e3a\uff08x\u3002,y\u3002\uff09AB\u65b9\u7a0b\u4e3ay=kx+b
AB\u65b9\u7a0b\u4e0e\u629b\u7269\u7ebf\u8054\u7acb\u5f97k^2x^2+(2kb-2p)x+b^2=0
x1+x2= x1x2=
y1y2=(kx1+b)(kx2+b)+k^2x1x2+kb(x1+x2)+b^2=
|AB|=\u6839\u53f7\u4e0b(1+k^2)*|x1-x2|=\u6839\u53f7\u4e0b{(1+k^2)[(x1+x2)^2-4x1x2]}
x\u3002=(x1+x2)/2=
C\u70b9\u5750\u6807y=y\u3002x=y^2/(2p)
C\u5230AB\u7684\u8ddd\u79bb|kx+b-y|/\u6839\u53f7\u4e0b(1+k^2)
则AF=AA',BF=BB'(抛物线第二定义）A'和B'分别是A,B在准线上的投影。
AF=AA'=y1+p/2,BF=BB'=y2+p/2,则AF+BF=y1+y2+p。
由AF+BF>=AB（当且仅当A,B,F在同一直线上成立）
所以由AF+BF=AB=y1+y2+p知A,B,F在同一直线上，即A、B和抛物线的焦点共线。
证毕
希望能帮到你！
【参数法】证明：∵点A,B均在抛物线x²=2py上，∴可设A(2pa,2pa²),B(2pb,2pb²).(a≠b).易知，点A,B到准线y=-p/2的距离分别为2pa²+(p/2).和2pb²+(p/2).由题设可知，|AB|=2pa²+2pb²+p.又由抛物线定义可知，|AF|=2pa²+(p/2).|BF|=2pb²+(p/2).∴|AF|+|BF|=2pa²+2pb²+p=|AB|.在⊿ABF中，由三角形三边关系可知，|AF|+|BF|＞|AB|.但这里|AF|+|BF|=|AB|,∴三点A,F,B共线。
证明： 由已知 x^2=2py 得焦点坐标F（0，p/2) ，准线y=-p/2
A,B点纵坐标y1,y2>0,由抛物线定义知 |AF|=y1+p/2 |BF|=y2+p/2
所以|AF|+|BF| =y1+y2+p/2+p/2=y1+y2+p
而题目已知|AB|=y1+y2+p 所以|AB|=y1+y2+p=|AF|+|BF|
即A、F、B三点共线 证毕。
解：设焦点为F，过A、B分别作准线的垂线，垂足分别为G、H，
很容易写出准线方程为x= -3/2，由抛物线第二定义有
AF=AG=X1-(-3/2)= X1+(3/2)
BF=BH=X2-(-3/2)= X2+(3/2)
所以AB=AF+BF=[X1+(3/2)]+[X2+(3/2)]= X1+X2+3=4+3=7
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