设点A、B为抛物线Y^2=2PX上的点,角AOB=90度,(O为原点),则AB必过点的坐标为
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A、B为抛物线Y^2=2PX上的点,角AOB=90度,则A、B必在X轴两侧

令A:(m,-√(2pm)),B((n,√(2pn)),其中m,n>0

角AOB=90度即kOA*kOB=-1

-√(2pm)/m * √(2pn)/n = -1

2p√(mn) = mn

√mn = 2p

A:(m,-√(2pm)),B((n,√(2pn))所在直线方程:

[y+√(2pm)]/(x-m) = [y-√(2pn)]/(x-n)

xy - √(2pn) x - my + m√(2pn)= xy +√(2pm) x - ny -n(√2pm)

x+ (m-n)y - [ m√(2pn)+n(√2pm)] = 0

当y=0时,x=[ m√(2pn)+n(√2pm)] /[√(2pm)+√(2pn) ] = (m√n+n√m)/(√m+√n) = √mn = 2p

∴AB恒过(2p,0)