如图在平面直角坐标系中,四边形OABC是矩形,且B点的坐标是(2,5),抛物线y=ax2随顶点P沿折线O-A-B-C运动

⑴顶点在C(0,5)时,抛物线Y=aX^2+5,又过A(2,0),
∴0=4a+5,a=-5/4,
∴Y=-5/4X^2+5.
⑵①BC中点(1,5),抛物线：Y=-5/4(X-1)^2+5
令X=0得Y=15/4,∴M(0,15/4),N(2,15/4),
设对称轴交MN于Q,则PQ=5/4,MQ=1,
∴tan∠PMN=PQ/MQ=5/4.
②设顶点为(m,5),
抛物线为Y=-5/4(X-m)^2+5,
令X=0,Y=-5/4m^2+5,
令X=2,Y=-5/4(2-m)^2+5,
∴AM=5/4m^2,BN=5/4(2-m)^2,
∵∠MPN=90°,
∴ΔPAM∽ΔNBP,
∴PA/AM=BN/PB,
m/(5/4m^2)=［5/4(2-m)^2］/(2-m),
(m-1)^2=9/25,
m=8/5或2/5,
∴P(8/5,5)或(2/5,5).