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Answer 1 · 2018-04-24T09:17:32

Closest point is #(1,1)#

Let us consider the parametric equation of parabola #y=x^2#

It is #(t,t^2)# and its distance #D# from #(3,0)# is

#D=sqrt((t-3)^2+(t^2-0)^2)#

= #sqrt(t^4+t^2-6t+9)#

and #(dD)/(dt)=1/(2sqrt(t^4+t^2-6t+9))*(4t^3+2t-6)#

= #(4t^3+2t-6)/(2sqrt(t^4+t^2-6t+9))#

This will be #0# when #4t^3+2t-6=0# or #2t^3+t-3=0#

or #(t-1)(2t^2+2t+3)=0#

As #2t^2+2t+3=2(t+1/2)^2+5/2#, this is always greater than #0#

Hence #(dD)/(dt)=0# when #t=1# i.e. at #(1,1^2)# or #(1,1)#

graph{(y-x^2)((x-1)^2+(y-1)^2-0.002)((x-3)^2+y^2-0.002)=0 [-1.29, 3.71, -0.28, 2.22]}

Answer 2 · 2018-04-24T09:32:30

See below

Let that closest point be #(alpha, beta)#

The (square of the) distance to #(3,0)# is:

#D = (alpha - 3)^2 + (beta - 0 )^2#, with condition #beta = alpha ^2# so, #beta^2 = alpha ^4#

#implies D = alpha ^4 + alpha^2 - 6 alpha + 9 #

Optimising the square of the distance will produce the same result as optimising the distance itself, but with simplified workings:

#D_alpha = 4 alpha ^3 + 2 alpha - 6 = 0 implies 2 alpha ^3 + alpha - 3 = 0 #

#alpha = 1# factors out on inspection

#implies (alpha - 1) (2 alpha^2 + 2 alpha + 3) = 0 #

The second expression does not have real roots, check the discriminant.

Optimisation #implies (alpha, beta) = (1,1)#

#D_(alpha alpha)(1,1) = 12 alpha ^2 + 2 = 14 # so this is a min.