Question

已知动点M在抛物线y^2=2px(p>0)上,当M在何处时,到定点P（p,p)的距离最短.

Answer 1

设M(y0^2/(2p),y0),

对y^2=2px两边对x求导得

2yy'=2p,得y'=p/y

故过点M的切线斜率为p/y0

要使M到定点P(p,p)的距离最短,需MP垂直于过点M的切线.也即

(y0-p)/[y0^2/(2p)-p]=-1/(p/y0)

化简得y0^3=2p^3

于是得y0=2^(1/3)*p

定点M为M(2^(-1/3)*p,2^(1/3)*p)