Find the point on the parabola y=x² nearest to the point (-3,0)?

2 Answers
Sep 9, 2015

#(x,y) = (-1,1)# is the closest point on #y=x^2# to #(-3,0)#

Explanation:

The distance from any point #(x,y)# to a point #(hatx,haty)# is
#color(white)("XXX")##sqrt((x-hatx)^2+(y-haty)^2)#

For points on #y=x^2# this becomes
#color(white)("XXX")d(x)=sqrt((x-hatx)^2+(x^2-haty)^2)#
and
more specifically for the point #(hatx,haty)=(-3,0)# this becomes
#color(white)("XXX")sqrt((x+3)^2+(x^2-0)^2)#

#color(white)("XX") = sqrt(x^4+x^2+6x+9)#

The problem is to minimize #d(x)#
or equivalently (but slightly simpler) to minimize
#color(white)("XXX")f(x)=x^4+x^2+6x+9#

The minimum occurs when #f'(x)= 0#
That is when
#color(white)("XXX")4x^3+2x+6=0#

An obvious (by inspection) root is #x=-1#
(and in fact there are no other real roots)

If #x=-1#
then #y=x^2 = (-1)^2 =1#

May 21, 2018

#(-1,1)#

Explanation:

The squared distance from #(x,x^2)# to #(-3,0)# is

#q= (x- -3)^2 + (x^2 - 0)^2 = x^4 + x^2 + 6x + 9#

We want the minimum #q#,

# 0 = {dq}/{dx} = 4x^3 + 2x + 6 #

That's a cubic but an easy one. By the time honored method of trying small numbers, we find #x=-1# is a solution so #x+1# is a factor.

#0 = (x+1)(4x^2 -4x + 6)#

The quadratic has no real roots, so #(-1,1)# is the sole solution.