Find the minimum distance from a point (6,3) to the parabola y^2=8x.

Answer 1

the distance from (6,3) to (x,y)=(y^2/8,y) is

z = √((y^2/8-6)^2 + (y-3)^2)

dz/dy = (y^3-16y-96) / 4√((y^2/8-6)^2 + (y-3)^2)

the denominator is never 0, so dz/dy=0 when y^3-16y-96=0, which is at y=5.72

so, the minimum distance is 3.32