Let p (x, y) = (x, x 2/4) and a (0, a) be on the curve, then the distance u from p to a is satisfied.

s=u^2=x^2+(x^2/4-a)^2=x^4/ 16+[ 1-a/2]x^2+a^4

∫s is a continuous function, and x tends to infinity.

∴s has a minimum, and this minimum is an extreme doubt.

S' = x 3/4+(2-a) x = x (x 2/4+2-a), so that s'=0,

(1) when a≥2, we get x 1=0, X2 = 2 √ (A-2) (extreme doubt).

The corresponding function values are s1= a 2, S2 = 4a-8+(a-2-a) 2 = 4a-4.

∫a2-(4a-4)≥0, ∴ S2 is the minimum value, and the shortest distance Usmall = √ S2 = √ (4a-4) = 2 √ (a-1).

(2) When a < 2, x=0, and there is only one extreme doubt, it is the minimum point.

The corresponding function value is s 1 = a 2,

The shortest distance is usmall = ∴ s1= │ a.

Note: This problem can also be solved directly by finding the maximum and minimum values, and then determining the minimum value. It feels more troublesome.