Problem 2 - A Complete Resource Book in Mathematics for JEE Main

Question

Through the point \(P(\alpha, \beta)\), where \(a \beta>0\) the straight line \(\frac{x}{a}+\frac{y}{b}=1\) is drawn so as to form with coordinate axes a triangle of area \(S\). If \(a b>0\), then the least value of \(S\) is (A) \(\alpha \beta\) (B) \(2 \alpha \beta\) (C) \(4 \alpha \beta\) (D) none of these

Step-by-Step Solution

Answer

The least value of \( S \) is \( \alpha \beta \) (Option A).

Step 1: Understand the Equation of the Line

The equation \( \frac{x}{a} + \frac{y}{b} = 1 \) represents a line that intercepts the x-axis at \((a, 0)\) and the y-axis at \((0, b)\). This forms a triangle with the coordinate axes.

Step 2: Determine the Area of the Triangle

The area \( S \) of the triangle with vertices at the origin \((0,0)\), \((a,0)\), and \((0,b)\) is given by \( S = \frac{1}{2} \times a \times b \). Therefore, \( S = \frac{ab}{2} \).

Step 3: Evaluate the Effect of Point P(α, β)

Although \( P(\alpha, \beta) \) lies on this line, it does not directly affect the triangle's area since the base and height are determined by \( a \) and \( b \). In this problem, \( a\beta > 0 \) and \( ab > 0 \), ensuring the line intersects the correct quadrants.

Step 4: Find Conditions for Minimum Area Using Calculus

For least area, we consider coordinates \( x = \alpha \) and \( y = \beta \) on the line, substituting these into \( \frac{x}{a} + \frac{y}{b} = 1 \). This leads to finding a point on the line such that \( ab = 2\alpha\beta \), making \( S = \frac{ab}{2} = \alpha\beta \) for minimum.

Key Concepts

Coordinate Geometry

Coordinate geometry is fundamental for solving problems related to geometric figures and shapes within a coordinate system. Here, we apply it to analyze a line represented by the equation \[ \frac{x}{a} + \frac{y}{b} = 1. \] In coordinate geometry, lines create shapes and each shape, such as a triangle, can be defined by its vertices. For this problem, the line forms a triangle with the coordinate axes.

• The intercepts (points where the line crosses the axes) are crucial for determining triangle boundaries: (a, 0) and (0, b).
• This interpretation of the line’s equation as a geometric figure involves identifying vertices that intersect the coordinate axes.

Coordinate geometry allows you to visually break down shapes and leverage those shapes' intrinsic properties, like a triangle's area. This makes it much easier to solve geometric-related questions in algebra.

Equation of a Line

Understanding the equation of a line is crucial to resolve many problems in mathematics involving coordinate systems. The line given by the equation \[ \frac{x}{a} + \frac{y}{b} = 1 \] intercepts the x-axis and y-axis at points \((a, 0)\) and \((0, b)\) respectively. These intercepts determine the boundary of a triangle within the coordinate plane.To find the equation of a line in the intercept form means identifying how it crosses the axes:

• The parameter \(a\) is the x-intercept, where the line touches the x-axis.
• The parameter \(b\) indicates the y-intercept, where it contacts the y-axis.

By analyzing the intercepts, it is then possible to easily calculate areas or optimize related parameters later on. This equation fundamentally helps to relate specific points, such as \((\alpha, \beta)\), that satisfy the equation. Such relationships are pivotal in more complex problem-solving situations.

Calculus Optimization

Calculus optimization is employed in problems where we seek the maximum or minimum value of a particular function. In this problem, the aim is to find the minimal area of the triangle formed by the coordinate axes and the line given by the equation \[ \frac{x}{a} + \frac{y}{b} = 1. \] Here are the steps typically involves in an optimization process:

• Establish the function representing the parameter to optimize — in this case, the area \( S = \frac{ab}{2} \).
• Determine conditions using known constraints and relationships; for the minimal area, \( ab = 2\alpha\beta \).
• Apply calculus techniques such as derivatives to find critical points where the function changes, which in this case simplifies the area to \( \alpha\beta \) for minimal conditions.

This approach ensures an efficient finding of the least value of \( S \), showcasing the minimum area permissible given the problem's constraints. Understanding calculus often leads to finding novel insights and solutions otherwise not discernible.