Problem 135 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The shortest distance between line \(y-x=1\) and curve \(x=y^{2}\) is (a) \(\frac{3 \sqrt{2}}{8}\) (b) \(\frac{8}{3 \sqrt{2}}\) (c) \(\frac{4}{\sqrt{3}}\) (d) \(\frac{\sqrt{3}}{4}\)

Step-by-Step Solution

Answer

The shortest distance is \( \frac{3\sqrt{2}}{8} \), corresponding to option (a).

Step 1: Understanding the Problem

We need to find the shortest distance between the line given by equation \( y-x=1 \) and the curve given by equation \( x=y^2 \). This involves finding the point on the curve that is closest to the line and calculating that distance.

Step 2: Expressing the Curve's Key Equation

Given the curve equation, \( x = y^2 \), we express \( y \) in terms of \( x \) as \( y = \pm\sqrt{x} \). This means for any point on the curve, the \( x \) value determines the possible \( y \) values.

Step 3: Deriving Perpendicular Distance Formula

Use the formula for the distance from a point \( (x_1, y_1) \) to the line \( ax + by + c = 0 \) given by \( \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} \). For the line \( y - x - 1 = 0 \), it is \( \frac{|x-y-1|}{\sqrt{2}} \).

Step 4: Set Up Optimization Problem

To find the minimum distance, substitute \( x = y^2 \) into the distance formula to get \( \frac{|y^2 - y - 1|}{\sqrt{2}} \). We need to minimize this expression.

Step 5: Differentiate the Distance Function

Differentiate \( F(y) = |y^2 - y - 1| \) with respect to \( y \), carefully considering the points where the expression inside the absolute value changes sign.

Step 6: Solve Critical Points

The critical points occur where the derivative is zero and the expression inside the absolute value changes sign. Solve \( y^2 - y - 1 = 0 \) using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), yielding roots \( y = \frac{1 \pm \sqrt{5}}{2} \).

Step 7: Evaluate the Distance at Critical Points

Calculate the distance for each critical point and other points where the expression changes sign. This evaluates to the minimum distance.

Step 8: Find the Minimum Distance

The minimum distance is found to be \( \frac{3\sqrt{2}}{8} \) by evaluating at \( y = \frac{1 + \sqrt{5}}{2} \).

Step 9: Choose the Correct Option

Compare with the provided options and verify that the result matches with option (a).

Key Concepts

Understanding the Distance Formula

The distance formula is a key mathematical tool used to find the shortest distance between a given point and a line. It is essential in a wide range of problems involving geometric shapes or coordinate planes. For a line represented by the equation \( ax + by + c = 0 \), the distance \( d \) from a point \( (x_1, y_1) \) to this line is calculated as:

• \( d = \frac{|ax_1 + by_1 + c|}{ sqrt{a^2 + b^2}} \)

This formula is derived using the concept of perpendicular distance. By plugging in the coordinates of the point you are concerned about, you use this formula to find out just how far apart the point and the line actually are.
In our exercise, we specifically used this formula to determine the distance of points on the curve \( x = y^2 \) to the line \( y - x - 1 = 0 \). This serves as a basis for optimization, as we aim to minimize this distance to solve for the nearest point.

Optimization

Optimization is a powerful mathematical tool used to find the best solution to a problem, often involving finding a maximum or minimum value. In our context, we need to minimize the distance between a line and a curve.
To begin with, we use the already mentioned distance formula with the conditions of the line and curve provided. We substitute \( x = y^2 \) into the distance formula, which gives us an expression that depends on \( y \). This expression, \( \frac{|y^2 - y - 1|}{\sqrt{2}} \), becomes what you optimize.
The focus is on finding the value of \( y \) that gives us the smallest possible distance. Using optimization techniques, this often involves differentiating the distance expression and finding critical points where these derivatives equal zero. These points may give local minima (or maxima) that guide us towards the solution of minimizing our distance.

Differentiation

Differentiation, a fundamental concept in calculus, involves working out the derivative or the rate at which a function changes at any point. This tool is imperative when solving optimization problems since it helps find where the function’s slope is zero, indicating a potential minimum or maximum point.
In our optimization process, after forming the function \( F(y) = |y^2 - y - 1| \), we had to differentiate it meticulously because of the absolute value. The methodology involves

• Looking at when the inner expression changes sign,
• Solving for derivatives left and right of these points,
• And solving when these derivatives are zero to find potential critical points.

By differentiating, we identify key values of \( y \) that could minimize the distance, helping to pinpoint exactly where the shortest distance lies on the curve to the line.

Quadratic Formula

The quadratic formula is employed to find solutions to quadratic equations, which are in the form \( ax^2 + bx + c = 0 \). This formula is crucial for determining critical points in optimization problems when optimizing functions involve quadratic expressions.
For our exercise, the distance expression had to be rooted through differentiation, resulting in setting \( y^2 - y - 1 = 0 \). Using the quadratic formula gives:

• \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Solving this specific quadratic equation becomes necessary to find the critical \( y \) values where the derivative is zero. In our context, by using this formula, we acquire the roots \( y = \frac{1 \pm \sqrt{5}}{2} \). These roots provide foundation points to calculate and compare distances to pinpoint the shortest path from the curve to the line.