抛物线x^2=(1/4)y上点M到y=4x-5的距离最短,则过抛物线x^2=(1/4)y点M的切线,与直线y=4x-5平行,设过M的直线为:
y=4x+b
代入x^2=(1/4)y,得
x^2=(1/4)y=(1/4)*(4x+b)
x^2-x-(5/4)b=0
过点M的直线y=4x+b与抛物线x^2=(1/4)y相切,则上方程的判别式△=0,即
△=(-1)^2-4*1*[-(5/4)b]=0
b=-1/5
过点M的直线y=4x-1/5
现在问题变为求两直线y=4x-5与y=4x-1/5的距离:
y=4x-5
x=0,y=-5,N(0,-5)
点N(0,-5)到y=4x-1/5的距离
L=|4*0-(-5)-1/5|/√17 =4.8/√17
M到y=4x-5的最短距离=4.8/√17