一道抛物线的题

圆x2+y2-9x=0 ,与顶点在原点O，焦点在X轴上的抛物线C交A，B两点。三角形OAB的垂心恰为抛物线的焦点，求 抛物线C的解析式？

OAB是以O为顶点的等腰三角形,垂心恰为抛物线的焦点(P/2,0),
即高为(P/2)*3/2=(3/4)P ,(亦即A的X坐标)
圆x2+y2-9x=0 与抛物线Y=2PX ,解得X=9-2P
即 9-2P=(3/4)P ,得P=36/11, 于是
抛物线C: Y^2=(72/11)X

设抛物线的解析式为y^2=2px,(p>0) ，焦点F(p/2,0)

将y^2=2px代入圆的方程有：x^2+(2p-9)x=0,x=0或x=9-2p

y=±√(18p-4p^2)

即A(9-2p,√(18p-4p^2)),B(9-2p,-√(18p-4p^2))

OA的斜率k1=√(18p-4p^2)/(9-2p)

BF的斜率k2=-√(18p-4p^2)/(9-2p-p/2)=√(18p-4p^2)/(5p/2-9)

k1k2=-1,即[√(18p-4p^2)/(9-2p)]*[√(18p-4p^2)/(5p/2-9)]=-1

亦即18p^2-117p+162=0

p=2或p=9/2 (舍去，因为判别式为0）

即抛物线的解析式为y^2=4x

设抛物线的解析式为y^2=2px,(p>0) ，焦点F(p/2,0)

将y^2=2px代入圆的方程有：x^2+(2p-9)x=0,x=0或x=9-2p

y=±√(18p-4p^2)

即A(9-2p,√(18p-4p^2)),B(9-2p,-√(18p-4p^2))

OA的斜率k1=√(18p-4p^2)/(9-2p)

BF的斜率k2=-√(18p-4p^2)/(9-2p-p/2)=√(18p-4p^2)/(5p/2-9)

k1k2=-1,即[√(18p-4p^2)/(9-2p)]*[√(18p-4p^2)/(5p/2-9)]=-1

亦即18p^2-117p+162=0

p=2或p=9/2

即抛物线的解析式为y^2=4x或y^2=9x