The point on the parabola y=(x-3)^2 that is near to the origin?

Question

1369 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-20T08:19:19

#P(3/2,9/4)#

Let:

#h(x) = x^2+y^2#

be the squared distance of a point of the plane from the origin.

For points laying on the parabola we have #y=(x-3)^2#, so:

#h(x) = x^2+(x-3)^2 = x^2+x^2-6x+9 = 2x^2-6x+9#

We can look for the local minimum by solving the equation:

#(dh)/dx = 0#

#4x-6 = 0#

#x= 3/2#

As #(d^2h)/dx^2 = 4 > 0# this critical point is a minimum and we can conclude that the closest point of the parabola to the origin is:

#x = 3/2#

#y = (3/2-3)^2 = 9/4#