Question

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

Answer 1

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).