Problem 68 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The shortest distance between line \(y-x=1\) and curve \(x=y^{2}\) is (A) \(\frac{4}{\sqrt{3}}\) (B) \(\frac{\sqrt{3}}{4}\) (C) \(\frac{3 \sqrt{2}}{8}\) (D) \(\frac{8}{3 \sqrt{2}}\)

Step-by-Step Solution

Answer

The shortest distance is \( \frac{3\sqrt{2}}{8} \) (Option C).

Step 1: Understand the Problem

We are tasked with finding the shortest distance between the line described by the equation \( y - x = 1 \) and the curve \( x = y^2 \). The shortest distance will typically occur between the line and a point on the curve where the tangent is perpendicular to the line.

Step 2: Represent the Line

Rearranging the equation \( y - x = 1 \) gives the line equation \( y = x + 1 \). This line has a slope of \( m=-1 \).

Step 3: Find the Point on the Curve with a Perpendicular Tangent

For the curve \( x = y^2 \), find the slope of the tangent. The derivative is \( \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \). To be perpendicular to the line \( y = x + 1 \), the slope of the tangent should be 1 (negative reciprocal of -1). We set \( \frac{1}{2\sqrt{x}} = 1 \), which gives \( x = \frac{1}{4} \). The corresponding \( y \) is \( \sqrt{\frac{1}{4}} = \frac{1}{2} \) or \(-\frac{1}{2} \).

Step 4: Calculate Distances

Calculate the distance from the point \( \left( \frac{1}{4}, \frac{1}{2} \right) \) to the line and \( \left( \frac{1}{4}, -\frac{1}{2} \right) \) to the line. For point \( \left( \frac{1}{4}, \frac{1}{2} \right) \):The point-line distance formula is used: \( \frac{|1/2 - 1/4 - 1|}{\sqrt{1+1}} = \frac{|1/2 - 1/4 - 1|}{\sqrt{2}} \). Calculate this to get \( \frac{|-3/4|}{\sqrt{2}} = \frac{3}{4\sqrt{2}} \).For \( \left( \frac{1}{4}, -\frac{1}{2} \right) \): Using the point-line distance formula: \( \frac{|-1/2 - 1/4 - 1|}{\sqrt{2}} \). Calculate:\( \frac{|-7/4|}{\sqrt{2}} = \frac{7}{4\sqrt{2}} \).

Step 5: Conclude Minimum Distance

The shortest distance occurs at the smaller of the two calculated distances, which is \( \frac{3}{4\sqrt{2}} \). Simplifying gives \( \frac{3\sqrt{2}}{8} \).

Key Concepts

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, is a method of representing and analyzing geometric figures using a coordinate system. In the context of finding the shortest distance between a line and a curve, it plays a crucial role. A coordinate system allows us to express lines and curves using equations.

For example, the line can be written as a linear equation like \( y - x = 1 \), which rearranges to \( y = x + 1 \). Here:

• "\(y\)" and "\(x\)" represent the respective coordinates on a standard Cartesian plane.
• "\(1\)" represents the y-intercept of the line.
• "\(x + 1\)" indicates the slope or angle of the line relative to the x-axis.

Coordinate geometry enables us to visualize these relationships and perform various calculations, such as finding intersections and calculating distances. These calculations involve

• Identifying points on the curves or lines.
• Analyzing slopes for perpendicularly intersecting lines or tangents.

It facilitates understanding of spatial relationships in a two-dimensional plane, which is essential for solving problems involving distances between different geometric entities.

Derivatives in Calculus

Derivatives in calculus measure how a function changes as its input changes. It is a core concept used to determine the slope of a tangent line at any point on a curve. In our context, the derivative helps find where a tangent to a curve is perpendicular to a given line.

For the curve \( x = y^2 \), we need to calculate its derivative to determine the slope of the tangent:

• The expression \( \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \) represents the derivative of the curve.
• This derivative shows how quickly or slowly \(y\) changes with \(x\).

The concept of perpendicular lines requires us to find a slope that is the negative reciprocal of the line’s slope. For a line with a slope of \(-1\), the tangent slope becomes \(1\). By setting the derivative equal to \(1\), we find the relevant point on the curve, leading to

• Resolving it to get \(x = \frac{1}{4}\),
• This particular calculation helps us target exactly where on the curve the shortest distance to the line occurs.

Understanding derivatives is crucial for solving such geometric problems and understanding the dynamics of changing systems in calculus.

Tangent Lines

A tangent line to a curve at a given point represents a line that just "touches" the curve at that point, without intersecting it. It possesses the same slope as the curve at that exact point. This understanding of tangents allows us to tackle problems related to distances:

• To find a tangent, we first compute the curve's derivative to ascertain the slope of the tangent.
• The tangent's slope helps determine if it is perpendicular to any other geometrical figure, such as a line.

In our exercise, the shortest distance between the curve and line is crucially marked by a tangent line that is perpendicular to the given line, since the shortest distance would be represented by the perpendicular path from the curve to the line.

• Once the point is located at \( \left( \frac{1}{4}, \pm\frac{1}{2} \right) \), the tangent’s slope was verified to be perpendicular to the line \(y - x = 1\).
• Such points are critical as they represent potential minima where the distance can indeed be shortest.

Recognizing such points and lines ensures that approaches and calculations are handled efficiently and correctly.