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