Problem 221 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The area bounded between the parabolas \(x^{2}=\frac{y}{4}\) and \(x^{2}=9 y\), and the straight line \(y=2\) is (A) \(20 \sqrt{2}\) (B) \(\frac{10 \sqrt{2}}{3}\) (C) \(\frac{20 \sqrt{2}}{3}\) (D) \(10 \sqrt{2}\)

Step-by-Step Solution

Answer

The area is not among the given options. There may be a mistake in computation or setup.

Step 1: Set up the inequality

First, we need to identify the region bounded by the two parabolas and the straight line. The parabolas given are \(x^2 = \frac{y}{4}\) and \(x^2 = 9y\), which can be rewritten as \(y = 4x^2\) and \(y = \frac{x^2}{9}\). The line is \(y = 2\).

Step 2: Find intersection points of parabolas with the line

Next, we find where each parabola intersects the line \(y = 2\). Set \(y = 2\) equal to each of the parabola equations. For \(x^2 = \frac{y}{4}\), solve \(2 = 4x^2\) obtaining \(x^2 = \frac{1}{2}\), so \(x = \pm \frac{1}{\sqrt{2}}\).For \(x^2 = 9y\), solve \(2 = 9x^2\), resulting in \(x^2 = \frac{2}{9}\), so \(x = \pm \frac{\sqrt{2}}{3}\).

Step 3: Determine limits of integration

We need to find the range of \(y\) from the lowest point of intersection to the line \(y=2\). The parabolas intersect each other at \(x^2 = \frac{y}{4} = 9y\) leading to \(y=0\). Hence, the region to consider is bound between \(y=0\) and \(y=2\).

Step 4: Calculate the area between parabolas

The area between two curves is the integral of the top function minus the bottom function. We integrate with respect to \(y\).The parabola \(x^2 = \frac{y}{4}\) translates to \(x = \sqrt{\frac{y}{4}}\) or \(x = -\sqrt{\frac{y}{4}}\), while \(x^2 = 9y\) is \(x = 3\sqrt{y}\) or \(x = -3\sqrt{y}\). The horizontal distance between the curves is twice the difference:\[\int_{0}^{2} \left(3\sqrt{y} - \sqrt{\frac{y}{4}}\right) dy\].

Step 5: Perform the integration

Calculate the integral \[\int_{0}^{2} \left(3\sqrt{y} - \sqrt{\frac{y}{4}}\right) dy\].This involves:1. \(\int_{0}^{2} 3y^{1/2} \, dy = 2y^{3/2}\bigg|_0^2 = 2(2^{3/2})\), and 2. \(\int_{0}^{2} \left(\frac{y}{4}\right)^{1/2} dy = \int_{0}^{2} \frac{y^{1/2}}{2}dy = \frac{2}{3}\times\frac{1}{2}y^{3/2}\bigg|_0^2\).Evaluating, we find the area to be \(6\sqrt{2} - \frac{\sqrt{2}}{3}\).

Step 6: Simplify and find the total area

Simplify \(6\sqrt{2} - \frac{\sqrt{2}}{3} = \frac{18\sqrt{2} - \sqrt{2}}{3} = \frac{17\sqrt{2}}{3}\). Compare this result to the given options to select the closest match.

Key Concepts

Parabolas

Parabolas are distinctive U-shaped curves that graphically represent quadratic equations. Each set of equations for parabolas in the problem is given in a form where the variable is squared. In general, a parabola can be expressed as either \( y = ax^2 + bx + c \) or \( x = ay^2 + by + c \). When solved for one variable, these equations highlight the symmetry of parabolas around their axes.
The problem involves two parabolas:

• \( x^2 = \frac{y}{4} \) which translates to a parabola that opens upwards and is symmetric about the y-axis, expressed as \( y = 4x^2 \).
• \( x^2 = 9y \) which also opens upwards and is expressed \( y = \frac{x^2}{9} \).

Understanding these equations is crucial because they determine how the parabolas interact with each other and with the line that we are using to find the bounded area.

Integral Calculus

Integral calculus is the mathematical method used to find areas under curves. Specifically, when tasked with finding the area between curves, integral calculus allows us to compute the exact size of the region enclosed by the curves and any boundaries, such as lines. It involves setting up and evaluating definite integrals.
The process for finding the area between two curves generally follows these steps:

• Determine which of the two curves is the upper curve (higher y-values) and which is the lower curve (lower y-values) over the interval of integration.
• Set up the integral of the difference between the upper curve and the lower curve over the given interval.
• Integrate this difference with respect to the relevant variable (often y when the functions are expressed as x in terms of y).

In our task, the integral captures the vertical 'slice' between corresponding points on the curves defined by \( x = 3\sqrt{y} \) and \( x = \sqrt{\frac{y}{4}} \). The integration accounts only for the segment where the curves intersect.

Intersection Points

Intersection points are critical when determining the bounds of a region between curves. These points dictate where one function overtakes another and highlight transition areas of interest within the graph. To find intersections of two curves, you equate their equations and solve for the variable in question.
In this problem, the intersections between the parabolas and the boundary line \( y = 2 \) are determined by solving:

• From \( y = 4x^2 \), set \( 2 = 4x^2 \) or \( x^2 = \frac{1}{2} \), leading to intersection points at \( x = \pm \frac{1}{\sqrt{2}} \).
• From \( y = \frac{x^2}{9} \), set \( 2 = \frac{x^2}{9} \) or \( x^2 = \frac{2}{9} \), resulting in \( x = \pm \frac{\sqrt{2}}{3} \).

Additionally, setting the parabolas equal, as in \( \frac{y}{4} = 9y \), helps find their point of mutual intersection, which is crucial for defining the lower limit of integration at \( y = 0 \). Understanding these principles is essential to correctly applying calculus to find bounded areas.