Hello StroodleWe still seem to be having problems with LaTex at present, so I'll write it without. The distance, D, of the point (x, y) from (3, 0) is given by

D^2 = (x-3)^2 + y^2Now replace y by x^2, and expand to get:

D^2 = x^4 + x^2 - 6x + 9Differentiate and put the result equal to zero:

4x^3 + 2x - 6 = 0which has a solution x = 1.

The second derivative is positive at this point, giving a minimum value of D^2 (and hence D) at the point (1, 1).

Grandad