抛物线中OA垂直于OB,求证AB垂直于X轴

首先,有两个焦点k>>>>>>>>>>>>>>【神机易数】团队

设kOA=kkOB=-1/k则A(2P/k^2,2P/k)B(2Pk^2,-2Pk)kAB=k/(1-k^2)AB:y+2Pk=[k/(1-k^2)](x-2Pk^2)即y=[k/(1-k^2)](x

你是高二的吧,这道题我曾做过具体如下.y^2=2px设A（x1,y1),B(x2,y2)OA垂直于OB所以x1x2+y1y2=0而y1^2=2px1y2^2=2px2所以（y1y2)^2=4p^2x1

设A(x1,y1)B(x2,y2)直线AB方程为x=my+b与抛物线联立得y1*y2=-2pbx1*x2=b^2又因为OA垂直与OB所以OAOB的向量积等於0所以x1*x2+y1*y2=0所以b^2-

在OA上取OE=OC；在OD上取OF=OB,连接BE、EF、FC,连接AF、ED交于GAG+GD>AD；EG+GF>EFAG+GD+EG+GF>AD+EF即AF+ED>AD+EF可知：AF=AB、DE

∵y^2=4x=2px,∴p=2.设OA的斜率是K,则OB的斜率是-1/K.OA方程:y=kxOB方程:y=-1/kx代入y^2=4x得:A(4/k^2,4/k),B(4k^2,-4k)AB的斜率是K

两式联立k^2x^2+(2k+1)x+k^2=0x1+x2=2k^2+1/-k=1x1x2+y1y2=1+k^2-2k^2-1+k^2=0得证1/2(根x1^2+y1^2*根x2^2+yx^2)=根1

在空间直角坐标系中记向量OA=a,向量OB=b,向量OC=c则向量BC=向量OC-向量OB=c-b向量AC=向量OC-向量OA=c-a因为OA垂直BC,OB垂直AC所以a（c-b）=0b（c-a）=0

设：y=kx(∵过点4,0）由：y^2=4xy=kx即：k^2x^2-4x=0△=0(因为有二个交点）、求出k接下直线ab方程出来了就不用说了吧

x=y+2=y²/2y²-2y-4=0y1+y2=2y1y2=-4x=y+2则x1x2=(y1+2)(y2+2)=y1y2+2(y1+y2)+4=4则y1y2/x1x2=-1即(y

设A(x1,y1)B(x2,y2),要证OA垂直OB,只要证kOAkOB=-1,即x1x2=-y1y2,那么联立抛物线和直线方程得k^2x^2+(2k^2+1)x+k^2=0,所以x1+x2=-(2k

设,直线L的方程为:Y=KX+b,则有Y=K(X+b/k),即直线必过定点(-b/k,0).y^2=4x,令,点A坐标为(t1^2/2p,t1),点B坐标为(t2^2/2p,t2).Koa=t1/(t

设A(X1,Y1),B(X2,Y2)则y1^2=2px1,y2^2=2px2∠AOB=90(y1*y2)/(x1*x2)=-1即y1*y2=-4P^2由直线AB得:y-y1=(y1-y2)/(x1-x

设A(2pm^2,2pm),B(2pn^2,2pn)OA⊥OB则(2pm^2)(2pn^2)+(2pm)(2pn)mn=-1直线方程为(2pm-2pn)x+(2pn^2-2pm^2)+4(p^2)(m

设直线AB与x轴交点M(m,0)那么直线AB可以写成x=ty+m由{y^2=2px{x=ty+m==>y^2=2pty+2pmy^2-2pty-2pm=0设A(x1,y1),B(x2,y2)根据韦达定

连接OC⌒　　⌒AC＝　BC∠COD＝∠COE∠ODC＝∠OEC＝90°OC＝OC△COD≌△COE所以CD=CE

设A(y1^2/2p,y1),B(y2^2/2p,y2),则由OA向量乘OB向量=0得,(y1y2)^2/4p^2+y1y2=0,即y1y2(y1y2/4p^2+1)=0,y1y2不等于0,所以y1y

1:设A点坐标为（Xa,Ya),B点坐标为(Xb,Yb)因为它们在抛物线y^2=-x上,则A:(-Ya^2,Ya),B(-Yb^2,Yb)又因为它们在直线y=k(x+1)上,则Ya=k(Xa+1)Yb

由y=kx+1与y=x^2得x^2-kx-1=ok^2+4>0恒成立设A（x1,y1),B(x2,y2)则x1+x2=k,x1x2=-1所以y1y2=(kx1+1)(kx2+1)=k^2x1x2+k(

y^2=-xy=k(x+1)联立,整理得k^2x^2+x(2k^2+1)+k^2=0x1*x2=1y1*y2=-1x1*x2+y1y2=0所以OA垂直OB