How do you find a point on curve y = x^2 that is closet to the point (16, 1/2) ?

First we will draw the graph of given curve and point 

To find the closest point on curve we draw a perpendicular from point P(16, 1/2) to the curve

y = x^2.

Let the point Q(t, t^2) be the point on curve that is foot of perpendicular from point P on the curve.

Now,

Slope of line PQ = (t^2 - 1/2)/(t - 16)

Slope of tangent to the curve at Q is value of dy/dx at (t, t^2).

So,

dy/dx = 2x

dy/dx at (t, t^2) = 2t

Now, as tangent at Q and line PQ are perpendicular.

So,

(Slope of PQ)*(Slope of TANGENT at Q) = -1

=> ((t^2 - 1/2)/(t - 16))*2t = -1

=> 2t^3 - t = -t + 16

=>2t^3 = 16

=> t^3 = 8

=> t = 2

So, Q(2 , 4) is the point.