The minimum distance of a point on the curve y = x² from the origin is:

Let take (x,y) be the point on the curve y = x²
Another point is (0,0)
distance between (x,y) and (0,0) is given by
D = √[(x-0)² + (y-0)²] = √(x² + y²)
Since y = x², we have
D = √(x² + x⁴)
To minimize D, we minimize D² = x² + x⁴
Let f(x) = x² + x⁴
f'(x) = 2x + 4x³ = 0
2x(1 + 2x²) = 0
x = 0 or 1 + 2x² = 0 (No real solution)
Thus x = 0
If x = 0, y = 0, D = 0
However, we need to check the second derivative
f''(x) = 2 + 12x²
f''(0) = 2 > 0 (minimum)
If x = 0, the distance is 0
Let's consider the boundary conditions
If x = 1, y = 1, D = √(1² + 1²) = √2
If x = 2, y = 4, D = √(4 + 16) = √20
Let's find the minimum distance
f'(x) = 2x + 4x³ = 0
2x(1 + 2x²) = 0
x = 0
Let's check the second derivative
f''(x) = 2 + 12x²
f''(0) = 2 > 0, so it's a minimum
At x = 0, y = 0, D = 0
However, this is not a point on the curve y=x², except for the origin
Let's use Lagrange multipliers
Let L(x,y,λ) = (x² + y²) + λ(y - x²)
∂L/∂x = 2x - 2λx = 0
∂L/∂y = 2y + λ = 0
∂L/∂λ = y - x² = 0
From 2x - 2λx = 0, 2x(1-λ) = 0, so x = 0 or λ = 1
If x = 0, y = 0
If λ = 1, 2y + 1 = 0, y = -1/2
But y = x², so x² = -1/2 which is impossible
Thus, the closest point is (1/√2, 1/2)
distance = √[(1/√2)² + (1/2)²] = √(1/2 + 1/4) = √(3/4) = √3/2 = √(1.5) ≈ 1.22
√15/2 ≈ 1.936
√19/2 ≈ 2.179