Problem 30 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The points on the curve \(x y^{2}=1\) which are nearest to the origin are (A) \(\left[\left(\frac{1}{2}\right)^{1 / 3}, \pm\left(\frac{1}{2}\right)^{-1 / 6}\right]\) (B) \(\left[\left(\frac{1}{2}\right)^{1 / 3}, 2^{-1 / 6}\right]\) (C) \(\left(2^{1 / 3}, \pm\left(\frac{1}{2}\right)^{-1 / 6}\right)\) (D) None of these

Step-by-Step Solution

Answer

The correct answer is (A).

Step 1: Define the Problem

We need to find the points on the curve \(xy^2 = 1\) that are closest to the origin \((0, 0)\). This is equivalent to minimizing the distance between a point \((x, y)\) on the curve and the origin.

Step 2: Set Up the Distance Formula

The distance \(D\) from a point \((x, y)\) to the origin is given by the formula \(D = \sqrt{x^2 + y^2}\). To simplify, we aim to minimize \(D^2 = x^2 + y^2\), as it avoids the square root and is equivalent to minimizing \(D\).

Step 3: Use the Constraint

We have the constraint from the curve equation \(xy^2 = 1\). Express \(y^2\) in terms of \(x\): \(y^2 = \frac{1}{x}\). Substitute \(y^2\) into the distance equation to get \(D^2(x) = x^2 + \frac{1}{x}\).

Step 4: Differentiate to Find Critical Points

Differentiate \(D^2(x) = x^2 + \frac{1}{x}\) with respect to \(x\): \[\frac{d}{dx}(D^2) = \frac{d}{dx}(x^2) + \frac{d}{dx}(\frac{1}{x}) = 2x - \frac{1}{x^2}\]Set the derivative equal to zero to find critical points:\[2x - \frac{1}{x^2} = 0\]

Step 5: Solve for Critical Points

Solve the equation \(2x = \frac{1}{x^2}\) to find the critical points:Multiply both sides by \(x^2\):\[2x^3 = 1\]Solve for \(x\):\[x = \left(\frac{1}{2}\right)^{1/3}\]

Step 6: Find Corresponding\(y\) Values

Use the curve equation \(xy^2 = 1\) to find \(y\). With \(x = \left(\frac{1}{2}\right)^{1/3}\), solve for \(y^2\):\[\left(\frac{1}{2}\right)^{1/3}y^2 = 1\Rightarrow y^2 = \left(\frac{1}{2}\right)^{-1/3}\]So, \(y = \pm \left(\left(\frac{1}{2}\right)^{-1/6}\right)\).

Step 7: Determine the Closest Points

The points closest to the origin are \(\left(\left(\frac{1}{2}\right)^{1/3}, \pm \left(\frac{1}{2}\right)^{-1/6}\right)\).

Step 8: Select the Answer Choice

Match the solution to the given choices. The correct choice is (A): \[\left[\left(\frac{1}{2}\right)^{1/3}, \pm \left(\frac{1}{2}\right)^{-1/6}\right]\].

Key Concepts

Distance Minimization

Understanding how to minimize distance is essential in calculus. To find the point on a curve that is closest to a particular point, such as the origin, you need to find the shortest distance between these two entities.
Minimizing the distance is much simpler if you consider the squared distance. For a point \((x, y)\) and another \((0, 0)\), the distance \(D\) can be given by \(\sqrt{x^2 + y^2}\). However, minimizing \(D\) directly is challenging due to the square root.
Instead, we look at \(D^2 = x^2 + y^2\). This avoids complications while still leading to the same solution. By considering \(D^2\) instead of \(D\), we simplify our approach to finding the minimal distance on the curve defined by constraints.

Curve Analysis

Analyzing a curve involves understanding its equation, constraints, and how points behave along it. In our example, we analyze the curve given by \(xy^2 = 1\). This equation describes a relationship between \(x\) and \(y\) that must hold true for all points on the curve.
By using this relationship, we can express one variable in terms of the other. For instance, from \(xy^2 = 1\), we derive \(y^2 = \frac{1}{x}\). This transformation helps us substitute into the distance function to work solely with a single variable.

• This substitution simplifies our computations significantly, converting the problem to a single-variable calculus problem.
• Such analysis often involves rearranging and substituting equations to find relationships that allow easier computations.
• Keep in mind that these transformations do not change the curve's properties; they simply allow us to analyze it more effectively.

Critical Points

Critical points on a curve are where the derivative of a function equals zero or is undefined, giving us clues about the function's behavior. To find the closest points to the origin on our curve, we must find where the derivative of \(D^2 = x^2 + \frac{1}{x}\) equals zero.
First, differentiate \(D^2\) with respect to \(x\):\[\frac{d}{dx}(D^2) = 2x - \frac{1}{x^2}\]. Setting the derivative equal to zero, \(2x - \frac{1}{x^2} = 0\), allows us to solve for critical points.
This simplifies down to an equation we can solve: \(2x^3 = 1\), leading to \(x = (\frac{1}{2})^{1/3}\). Critical points are essential because they often indicate maxima, minima, or points of inflection.

• By solving for these points, we discover where distance achieves its minimum or maximum.
• It involves setting derivatives to zero to explore potential changes in the function's behavior.
• Critical points may yield the solution, but always substitute back to check it satisfies the original equation and conditions.