To determine which points on the curve \(xy^2 = 1\) are nearest to the origin, we need to minimize the distance from a point \((x, y)\) on this curve to the origin \((0, 0)\). The distance \(D\) from a point \((x, y)\) to the origin is given by the formula:
\(D = \sqrt{x^2 + y^2}\)
Since we are minimizing, it is often easier to minimize the square of the distance to avoid dealing with the square root. So we minimize:
\(D^2 = x^2 + y^2\)
Given the constraint \(xy^2 = 1\), we express \(y\) in terms of \(x\):
\(y^2 = \frac{1}{x}\)
Hence, \(y = \pm \sqrt{\frac{1}{x}}\).
Substitute \(y^2\) into the equation to express \(D^2\) entirely in terms of \(x\):
\(D^2 = x^2 + \frac{1}{x}\)
To find the minimum value of \(D^2\), we differentiate it with respect to \(x\) and set the derivative to zero.
\(\frac{d}{dx}(D^2) = 2x - \frac{1}{x^2}\)
Setting \(\frac{d}{dx}(D^2) = 0\), we have:
\(2x - \frac{1}{x^2} = 0\)
This implies:
\(2x^3 = 1\)
Solving for \(x\), we get:
\(x = \frac{1}{2^{1/3}}\)
Substitute \(x = \frac{1}{2^{1/3}}\) back into the expression for \(y\):
\(y = \pm \sqrt{\frac{1}{x}} = \pm \sqrt{2^{1/3}} = \pm 2^{1/6}\)
Hence, the points on the curve closest to the origin are:
\(\left(\frac{1}{2^{1/3}}, \pm 2^{1/6}\right)\)
Thus, the correct answer is:
\(\left(\frac{1}{2^{1/3}}, \pm 2^{1/6}\right)\)