Question

若一直线与抛物线y^=2px(p>0）交于A,B两点，且OA⊥OB，点O在直线AB上的射影D（2，1），求抛物线方程

Answer 1

OD:y=0.5x;AB:y-1=-2(x-2)与y^2=2px联立：y^2=p(5-y) y^2+py-5p=0;
y1y2=-5p ,x1x2=y1^2y2^2/4p^2=25/4, OA⊥OB 所以 x1x2+y1y2=0;p=5/4
抛物线方程 y^2=5x

Answer 2

OD⊥AB--->k(OD)*k(AB)=-1--->k(AB)=-2--->AB:y=5-2x
与抛物线方程联立：y^=(5-2x)^=2px--->4x^-(20+2p)x+25=0
--->xAxB=25/4, xA+xB=(10+p)/2
--->yAyB=(5-2xA)(5-2xB)=25-10(xA+xB)+4xAxB=-5p
OA⊥OB--->(xA,yA)*(xB,yB)=0--->xAxB+yAyB=0=25/4-5p--->p=5/4
--->抛物线方程: y^=5x/2

Answer 3

y=0.5x;AB:y-1=-2(x-2)与y^2=2px
y^2=p(5-y) y^2+py-5p=0;
y1y2=-5p ,x1x2=y1^2y2^2/4p^2=25/4, OA⊥OB 所以 x1x2+y1y2=0;p=5/4
y^2=5x