Problem 228 - A Complete Resource Book in Mathematics for JEE Main

Question

The area (in sq. units) of the region described by \(\left\\{(x, y): y^{2} \leq 2 x\right.\) and \(\left.y \geq 4 x-1\right\\}\) is: \(\quad\) (A) \(\frac{5}{64}\) (B) \(\frac{15}{64}\) (C) \(\frac{9}{32}\) (D) \(\frac{7}{32}\)

Step-by-Step Solution

Answer

The area is \( \frac{7}{32} \).

Step 1: Analyze the First Inequality

The inequality given is \( y^2 \leq 2x \). This can be recognized as a parabola opening to the right with the vertex at the origin.

Step 2: Analyze the Second Inequality

The inequality given is \( y \geq 4x - 1 \). This describes a line with a slope of 4 and a y-intercept at -1. This line forms a boundary above which the region of interest lies.

Step 3: Find Intersection Points

Set the equations equal to find their intersection: \( y^2 = 2x \) and \( y = 4x - 1 \). Substituting \( y = 4x - 1 \) into \( y^2 = 2x \), gives \((4x - 1)^2 = 2x\). Expanding and solving gives the points \((\frac{1}{2}, 1)\) and \((\frac{1}{8}, 0)\).

Step 4: Set Up the Integral for Area

The region is bounded between the curves from \( x = \frac{1}{8} \) to \( x = \frac{1}{2} \). The area, \( A \), is: \[ A = \int_{1/8}^{1/2} ((4x - 1)^2) - (0) \, dx - \int_{1/8}^{1/2} (y^2/2) \, dx \]

Step 5: Solve the Integral for Area

Evaluate both integrals: \( \int_{1/8}^{1/2} (4x - 1)^2\, dx \) simplifies using substitution and results in an area under the line, while integrating \( \int_{1/8}^{1/2} 2x \, dx \) gives the area under the parabola. Find the difference of these integrals to determine the area of the region.

Key Concepts

Parabola

In coordinate geometry, a parabola is a curve described by a quadratic equation. In this exercise, we have the inequality \( y^2 \leq 2x \), which defines one side of a parabolic region. This particular parabola opens to the right, with its vertex at the origin \((0,0)\). This means that the shape of the parabola is symmetric along the x-axis.

When analyzing parabolas, it's important to determine their orientation and vertex position. Here:

• Equation: \( y^2 = 2x \)
• Opens: Right
• Vertex: (0, 0)

The inequality \( y^2 \leq 2x \) indicates that we're interested in the region inside or on the parabola. This inequality bounds our area on the left side of the region under consideration.

Inequalities

Inequalities are essential in defining regions in coordinate geometry. They tell us about the boundaries and the areas we are interested in, which could be inside, outside, or on a particular curve. Here, we have two inequalities forming the region:

• \(y^2 \leq 2x\): Defines a parabola.
• \(y \geq 4x - 1\): Describes a line.

The second inequality, \( y \geq 4x - 1 \), denotes a line. The slope of this line is 4, meaning it rises sharply, and it crosses the y-axis at -1.

For solving such problems, understanding how these inequalities interact to form boundaries is key. The region specified will be where both inequalities hold true. Think of the inequalities as constraints on a graph, limiting which points (or region) satisfy both equations.

Integral Calculus

Integral calculus is used here to find the area of the region that fits within the specified inequalities. Once we establish the boundaries formed by the intersecting equations of the curve and line,we use integrals to calculate the exact area.

To set up the integral for this region:

• Identify limits of integration from intersection points: \(x = \frac{1}{8}\) to \(x = \frac{1}{2}\).
• Set up two integrals:
  • The first covers the area under the line, \((4x - 1)^2\).
  • The second accounts for the area under the parabola, \( y^2/2 \).

The outcome is the net area, representing the overlap of the parabolic and linear areas. This process is a classic use case of integration in calculating areas between curves.

Intersection Points

Intersection points are where two curves or lines meet. They're crucial for defining limits of integration or finding bounds in problems involving multiple curves. Here, the line \( y = 4x - 1 \) intersects the parabola \( y^2 = 2x \).

To find these points, solve the system of equations:- Substitute the linear equation into the parabolic equation.- Solve: \((4x - 1)^2 = 2x\).

This gives solutions \((x, y)\):

• \((\frac{1}{2}, 1)\)
• \((\frac{1}{8}, 0)\)

These points are critical as they form the range for integration and delineate the exact region of interest in our geometry, helping to precisely compute the area.