Question

The points on the curve \(xy^2 = 1\) which are nearest to the origin, are

• A. \(\left(\frac{1}{2^{1/3}}, \pm 2^{1/6}\right)\)
• B. \(\left(\frac{1}{2^{1/3}}, \pm 2^{-1/6}\right)\)
• C. \(\left(2^{1/3}, \pm \frac{1}{2^{1/6}}\right)\)
• D. None of these

Hint:

For distance minimization problems, minimize the square of the distance to avoid square roots.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Minimize distance \(d^2 = x^2 + y^2\) subject to \(xy^2 = 1\).

Step 2: Detailed Explanation:
Let \(S = x^2 + y^2\). From \(y^2 = 1/x\), we have \(S = x^2 + \frac{1}{x}\) for \(x>0\). \(\frac{dS}{dx} = 2x - \frac{1}{x^2} = 0 \Rightarrow 2x = \frac{1}{x^2} \Rightarrow x^3 = \frac{1}{2} \Rightarrow x = \frac{1}{2^{1/3}}\). Then \(y^2 = \frac{1}{x} = 2^{1/3} \Rightarrow y = \pm 2^{1/6}\). Check second derivative: \(\frac{d^2S}{dx^2} = 2 + \frac{2}{x^3} > 0\) so minimum.

Step 3: Final Answer:
Points are \(\left(\frac{1}{2^{1/3}}, \pm 2^{1/6}\right)\), which corresponds to option (A).