Question

在抛物线y=4X平方上求一点,使这点到直线Y=4X-5的距离最短

Answer 1

设该点为A(a,b),b = 4a^2

A(a,4a^2)

y = 4x -5,4x -y -5 = 0

A到直线的距离d = |4a -4a^2 -5|/√[4^2 +(-1)^2] = |4a^2 - 4a +5|/√17 = |(2a-1)^2 +4|/√17

a=1/2时,A到直线的距离最短,A(1/2,1)