The point on the curve x^2+2y=0 that is nearest the point (0, -1/2) occurs where y is? (Answer: 0)

Question

loser66 · Answer

from the curve x^2 +2y =0 --> y = -1/2x^2
                                                           y' = ?

anonymous · Answer

-x.

loser66 · Answer

Oh, sorry sweetie, we have to come up at other way, let me think more, I am so sorry

loser66 · Answer

@jim_thompson5910  help me please, I don't know how to explain

jim_thompson5910 · Answer

any point on x^2 + 2y = 0 is of the form (x, -1/2x^2) and this is shown when Loser66 solved for y

jim_thompson5910 · Answer

what you need to do is use the distance formula 

d = sqrt( (x2 - x1)^2 + (y2 - y1)^2 )

to find a function (in terms of x) of the distance from (x,-1/2x^2) to (0, -1/2)

jim_thompson5910 · Answer

once you have that distance function, you can use calculus to find the min

loser66 · Answer

I have it here for the graph but don't know how to put it in logic http://www.wolframalpha.com/input/?i=x^2+%2B2y+%3D0 thanks @jim_thompson5910

jim_thompson5910 · Answer

you're welcome