The parabola is given as:

$$y = x^{2} - 2ax + 1$$

Solution:

The turning point occurs at: $$x = a, \quad y = 1 - a^{2}.$$ So, the turning point is: $$(a, \, 1 - a^{2}).$$

The distance of this point from the origin is: $$D^{2} = a^{2} + (1 - a^{2})^{2}.$$

Expanding: $$D^{2} = a^{2} + 1 - 2a^{2} + a^{4} = a^{4} - a^{2} + 1.$$

Let $u = a^{2} \geq 0$. Then: $$D^{2} = u^{2} - u + 1.$$

This quadratic in $u$ can be written as: $$D^{2} = \left(u - \tfrac{1}{2}\right)^{2} + \tfrac{3}{4}.$$

The minimum occurs when $u = \tfrac{1}{2}$, i.e. $$a^{2} = \tfrac{1}{2} \quad \implies \quad a = \pm \tfrac{1}{\sqrt{2}}.$$

Answer: (d) $a = \pm \tfrac{1}{\sqrt{2}}$