d = √[(0-x)2 + (1-y)2]
Since y = x2, x = ±√y
So, d = √[y + 1 - 2y + y2]
d = √(y2 - y + 1)
d is minimized when y2 - y + 1 is minimized
Minimum occurs at the vertex. So, y = -(-1) / (2(1)) = 1/2
Minimum distance d = √[(1/2)2 - (1/2) + 1] = √(3/4) = √3/2