Find min/max distance between (0,4) and y=x^2/4

[Poppynz]

I am trying to find the minimum and maximum distance between point (0,4) and the line y=(x^2)/4 in the region of 0<=x<=2√3.

This is what i have so far

y(distance)=√((x-0)^2+((x^2)/4)^2)
=(x^2+(X^4)/16+16)^0.5
then calculate y'
then make y'=0
then determine if max or min using 2nd derivative test

The problem i am having is that the answers i get for x when y' = 0 are imaginary. i have computed this problem using MATLAB and get the same answers except that they are real and have isolated the problem to my simplification of y.

matlab gives it as y=0.25(-16x^2+x^4+256)^0.5

Could some one please explain how to simplify y=(x^2+(X^4)/16+16)^0.5 to y=0.25(-16x^2+x^4+256)^0.5 or suggest a better way to approach this problem

Thanks in advance

 

[ideasrule]

y(distance)=√((x-0)^2+((x^2)/4)^2)
=(x^2+(X^4)/16+16)^0.5

I think you meant [(x^2)/4-4]^2, not [(x^2)/4]^2. Also, you might find it easier to calculate the derivative of y^2 instead of y. Squaring the distance eliminates the square root, which is a major pain in the neck.