Calculus 6 : Deviated again

$xy = a^2 \\ y = \dfrac{a^2}{x} \\ f(x) =x^2 + y^2 = x^2 + (\dfrac{a^2}{x})^2 = x^2 + \dfrac{a^4}{x^2} = x^2+ a^4x^{-2}\\ f'(x) = 0\text{, gives you the least value of x, y} f'(x) = \dfrac{d}{dx} f(x) = \dfrac{d}{dx} (x^2 + a^4x^{-2}) \\ = \dfrac{d}{dx} x^4 + \dfrac{d}{dx} a^4x^{-2} \\ = 2x + a^4(-2x^{-3}) = 2x - \dfrac{2a^4}{x^3} \\ f'(x) = 0 \\ 2x - \dfrac{2a^4}{x^3} = 0 \\ 2x = \dfrac{2a^4}{x^3} \\ 2x \times x^3 = 2a^4 \\ x^4 = a^4 \\ x = \pm a \\ y = \dfrac{a^2}{x} = \dfrac{a^2}{\pm a}\\ y = \pm a \\ \boxed{x^2 + y^2 = (\pm a)^2 + (\pm a)^2 = 2a^2}$

The equations are perfectly symmetrical with respect to $x$ and $y$ . If there is indeed only one $(x,y)$ pair which minimizes the sum of squares, x and y must both be equal to $a$ . The sum of squares is therefore $2a^{2}$