抛物线的焦点弦经典性质及其证明过程精

1、有关抛物线焦点弦问题的探讨过抛物线pxy22=(p>0的焦点F作一条直线L和此抛物线相交于B,(22yx两点A,(11yx、结论1:pxxAB+=21pxxpxpxBFAFAB+=+=+=21212(2(结论2:若直线L的倾斜角为8则弦长92sin2pAB=证：(1若27t9=寸，直线L的斜率不存在，此时AB为抛物线的通径，结论得证.,=.pAB2(2若2冗9丰时,设直线L的方程为：9tan2(py-=即2cotpyx+?=8代入抛物线方程得0cot222=-?-ppyy曲韦达定理9cot2,21221pyypyy=+-=由弦长公式得99922212sin2cot1(2cot1ppyyA

2、B=+=-+=结论3:过焦点的弦中通径长最小pp2sin21sin22a><99-AB的最小值为p2,即过焦点的弦长中通径长最短.结论4:(82为定值pABSoAB=?(8sin2sinsin2221sin21sin21sin21sin2132220PABSpppABOFBFAFOFAFOFBFOFSSSOABAFOBFOAB=?=?=+?=?+?=+=?9999结论05:(1221pyy-=(2x1x2=42p证44(,2,22222121222211PPyyxxpyxpyx=.=结论6:以AB为直径的圆与抛物线的准线相切证:设M为AB的中点，过A点作准线的垂线AA1,过B点作准

3、线的垂线BB1,过M点作准线的垂线MM1,由梯形的中位线性质和抛物线的定义知22111ABBFAFBBAAMM=+=+=故结论得证结论7:连接A1F、B1F则A1F±B1FFAAFOAFOAFAAOFAAAFAFAAAFAA11111111,/=/./=/./=/=同理？=/./=/901111FBAFBBFOB.A1F±B1F结论8:(1AM1IBM1(2M1F±AB(3BFAFFM?=21(4设AM1与A1F相交于H,M1B与FB1相交于Q则M1,Q,F,H四点共圆(52121214MMBMAM=+证:由结论（6知M1在以AB为直径的圆上AM1IBM111FB

4、A?为直角三角形,M1是斜边A1B1的中点111111111AFAFAAFAMFAMFMMA/=/=/.=?=/=/+/9011111MAAMFAFAA?=/+/.90111FMAAFA.M1F±ABBFAFFM?=.21AM1IBM1FBFA90111±?=/.又BAM?=/90FBA11所以M1,Q,F,H四点共圆,22121ABBMAM=+(2121211242MMMMBBAABFAF=+=+=结论9:(1、AO、B1三点共线(2B,O,A1三点共线(3设直线AO与抛物线的准线的交点为B1,则BB1平行于X轴(4设直线BO与抛物线的准线的交点为A1,则AA1平行于X轴

5、证:因为pypykyppyyxykoBoA2212111122,221-=-=，而221pyy-=所以122222oBoAkpyypplx一一一k所以三点共线。同理可征(2(3(4结论10:pFBFA211=+证:过A点作AR垂直X轴于点R，过B点作BS垂直X轴于点S，设准线与x轴交点为E,由勺倾斜角为因为直线L贝U88cos1cos=+=+=PAFAFAFPFREFERPAF9cos11-=同理可得PBF8cos11+=pFBFA211=+结论11:证:AABBEAEBAAFABBBFFABFEAEBAAEFBB1111111111,/=EBBEAAEBB90111111/？.？=/=/=相

6、似于EAAEBBEAAPEQEFBEFAEF90EBBBEFEAAAEF11/；?/平分角即=+=+0KKXBEAEBEAEBFAFBEAE=+轴对称关于和直线直线=(490AEBFBEFAF2?/=时，当冗92pxy2p-xkyL22=?=w将其代入方程的方程为时，设直线当冗0(k2kpxxy,B(x,y,A(x04pk2xp(k-xk2221221122222+=+=+贝U设得x1x2=4p2假设122y1KKBEAE2211BEAE-=+?+.?!pxypx=-则AEBEAFAE(1PEQ(2(3KK0BFBE(4AEBE,AEBE22EF兀7t0豆=+=线段平分角当时当时不垂直于p21

7、|CD|1|AB|1=+?+?+=?.?+?+=2px2px-2p-xk2p-xk2px2px-yy21212121即(222222222212122121k2p01k4p1kxx2pxx1kkkkp-+=+.=+-+-+结论得证假设错误不可能.=-02结论12:过抛物线的焦点作两条互相垂直的弦AB、CD,则推广与深化：深化1:性质5中，把弦AB过焦点改为AB过对称轴上一点E(a,0,则有pa2yy21-=.证:设AB方程为my=x-a，代入px2y2=.得:0ap2pmy2y2=-,.pa2yy21-=.深化2:性质12中的条件改为焦点弦AB不垂直于x轴,AB的中垂线交x轴于点R，则21|A

8、B|FR|=证明:设AB的倾斜角为a直线AB的方程为：2px(tgay-=，代入px2y2=得:px24ppxx(atg222二+-,即:04papctg2p(xx222=+-.由性质1得asinp2apctg2p2pxx|AB|2221=+=+=,又设AB的中点为M，则|acosapctg|acos2p2xx|FM|221=-+=asinp|acosapctg|acos|FM|FE|222=,.,.21|AB|FR|=深化3:过抛物线的焦点F作n条弦nn2211BABABA?、，且它们等分周角2冗则有(1E=?n1iii|FB|FA|1为定值；(2£=n1|BA|1为定值.证明:(1设抛物线方程为aFxA,cos1p1=/8=.由题意乃+=/?冗+=冗+=/n1naFxAn2aFxA,naFxAn32,所以222211pasinpacos1pacos(1pacos1|FB|FA|1=+ti-?-=?,同理22nn2222pn1na(sin|FB|FA|1,pna(sin|FB|FA|1-+=?兀+?冗易知2nn1na(sinn2a(sinna(sinasin2222=乃+?+冗+冗+；222n1i2222iip2npn1na(sinpna(sinpasin|FB|FA|1=-+?+兀+=