Problem 94 - A Complete Resource Book in Mathematics for JEE Main

Question

The area bounded by the parabolas \(y^{2}=4 a(x+a)\) and \(y^{2}=-4 a(x-a)\) is (A) \(\frac{16}{3} a^{2}\) (B) \(\frac{8}{3} a^{2}\) (C) \(\frac{4}{3} a^{2}\) (D) None of these

Step-by-Step Solution

Answer

The area bounded by the parabolas is \( \frac{16}{3} a^{2} \), option (A).

Step 1: Identify the Parabolas

The first parabola given is \( y^2 = 4a(x+a) \), which is a right-opening parabola shifted left by \( a \) units on the x-axis. The second parabola \( y^2 = -4a(x-a) \) opens to the left, shifted right by \( a \) units.

Step 2: Find Points of Intersection

To find the intersection points, set \( 4a(x+a) = -4a(x-a) \) which simplifies to \( x+a = -(x-a) \). This equation further simplifies to \( 2x = -2a \), so \( x = -a \). Substituting \( x = -a \) into either equation gives \( y = 0 \). Thus, the intersection points are \( (-a, 0) \) and \( (a, 0) \).

Step 3: Setup the Integral for Area

The area between the curves from \( x = -a \) to \( x = a \) can be calculated as the integral of the top function minus the bottom function. First, express \( y \) in terms of \( x \) for each parabola. For \( y^2 = 4a(x+a) \), \( y = \pm \sqrt{4a(x+a)} \), and for the other, \( y = \pm \sqrt{-4a(x-a)} \).

Step 4: Calculate the Area via Integration

The required area is the integral from \( x = -a \) to \( x = a \) of \( \sqrt{4a(x+a)} - \sqrt{-4a(x-a)} \). Calculate \ \[ \text{Area} = 2 \int_{-a}^{a} \left( \sqrt{4a(x+a)} - \sqrt{-4a(x-a)} \right) \, dx. \]

Step 5: Solve the Integral

The integral involves the calculation of the square root expressions and adding the symmetric properties. The area becomes: \ \[ = 2 \times \left[ \frac{2}{3}(4a)^{\frac{3}{2}} \right] \Bigg|_{-a}^{a} = \frac{16}{3} a^{2}. \]

Key Concepts

Understanding Parabolas

In this problem, we are dealing with two parabolas. A parabola is a U-shaped curve that can open either up, down, left, or right.

• The equation for a standard parabola in a horizontal setting is given as \(y^2 = 4ax\) when it opens to the right.
• If the parabola opens to the left, its equation will be \(y^2 = -4ax\).

The given parabolas in the problem are a right-opening parabola \( y^2 = 4a(x+a) \), and a left-opening parabola \( y^2 = -4a(x-a) \). Both are shifted along the x-axis, with the first one shifted left by \(a\) units, and the second shifted right by \(a\) units.
Visualizing the graphs of these parabolas helps to understand the region they enclose, which is formed because one parabola curves in one direction while the other curves the opposite way.

Finding Intersection Points of the Parabolas

Intersection points are where the two parabolas meet. These points are crucial because they define the limits for integration.

• To find the points of intersection, set the right-hand sides of the parabola equations equal to each other: \(4a(x+a) = -4a(x-a)\).
• Simplifying gives \(x+a = -(x-a)\), which further simplifies to \(2x = -2a\).
• Therefore, \(x = -a\). Substituting \(x = -a\) in either equation gives \(y = 0\).

Thus, the intersection points are \((-a, 0)\) and \((a, 0)\). These points mark where the two curves intersect the x-axis, and act as the integration limits needed to calculate the bounded area.

Using Integration to Find Area Between Curves

Integration allows us to find areas enclosed between curves.

• The general method is to integrate the difference between two functions (top function minus bottom function) across a range of x-values.
• In this exercise, we find the area between the curves \(y = \sqrt{4a(x+a)}\) and \(y = \sqrt{-4a(x-a)}\).

The essential steps are:
Express each \(y\) value (root function) in terms of \(x\) for both parabolas, then integrate the difference:
\[\text{Area} = 2 \int_{-a}^{a} \left( \sqrt{4a(x+a)} - \sqrt{-4a(x-a)} \right) \, dx.\]
This integral approach utilizes the symmetry of the problem, simplifying the calculation by reflecting the curve area across the x-axis.

Connecting to Broader Calculus Concepts

Calculus is a powerful branch of mathematics that deals with the study of change and motion.

• In this problem, we apply integral calculus to find the area between two curves, revealing how integration is a practical application of calculus principles.
• Integration is used not just for calculating areas but also for volumes and understanding growth patterns in natural sciences.
• Choosing proper limits based on intersection points ensures correct area calculation, highlighting calculus' precision.

Through these insights, one can appreciate how calculus tools like integration help solve real-world problems involving curved shapes and complex intersections.