How to find minimal distant from one point to a parabola? Find the minimal distance from the point (2,0) to the parabola $y^2=2x$.
Any point on the parabola can be written as $(2m^2,2m)$

So, we have $f^2(m)=(2-2m^2)^2+(0-2m)^2=4m^4-4m^2+4=(2m^2-1)^2+3$
$$(2,0)to (x,y)\\dis=\sqrt{(x1-x2)^2+(y1-y2)^2}\\dis=\sqrt{(x-2)^2+(y-0)^2}\\dis=\sqrt{(x-2)^2+(y-0)^2}=\\dis=\sqrt{(x-2)^2+2x} \\dis=\sqrt{x^2-2x+4}=\\dis=\sqrt{x^2-2x+4}=\\dis=\sqrt{(x-1)^2+3}=\\min((x-1)^2+3)=0+3\\so\\min(dis=\sqrt{(x-1)^2+3} )=\sqrt{3}$$when put x=1