Problem 88 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Area of the greatest rectangle that can be inscribed in the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) is (A) \(2 a b\) (B) \(a b\) (C) \(\sqrt{a b}\) (D) \(\frac{a}{b}\)

Step-by-Step Solution

Answer

(A) The area of the greatest rectangle that can be inscribed in the ellipse is \(2ab\).

Step 1: Understand the Problem

We need to find the area of the largest rectangle that can fit inside an ellipse given by the equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). The rectangle's sides should be parallel to the coordinate axes.

Step 2: Parameterize the Rectangle

Since the rectangle is inscribed in the ellipse, let's use points \((x, y)\) on the ellipse as the vertices of the rectangle in the first quadrant. By symmetry, the opposite vertices will be \((-x, -y)\). The area of the rectangle is \(A = 4xy\).

Step 3: Use Substitution for the Ellipse Equation

From the ellipse equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), solve for \(y = b \sqrt{1 - \frac{x^2}{a^2}}\). Substitute this in the area formula: \(A = 4x \cdot b \sqrt{1 - \frac{x^2}{a^2}}\).

Step 4: Simplify the Area Expression

The area function becomes \(A(x) = 4bx \cdot \sqrt{1 - \frac{x^2}{a^2}}\). To find the maximum area, we need to differentiate this with respect to \(x\) and find the critical points.

Step 5: Differentiate the Area Function

Use the chain rule to find \(\frac{dA}{dx}\). Differentiate \(A(x) = 4bx \cdot \sqrt{1 - \frac{x^2}{a^2}}\). Simplify and set \(\frac{dA}{dx} = 0\) to solve for \(x\).

Step 6: Solve for Critical Points

Set the derivative \(\frac{dA}{dx} = 0\) and solve for \(x\). Solving gives \(x = \frac{a}{\sqrt{2}}\). From symmetry and the ellipse equation, \(y = \frac{b}{\sqrt{2}}\).

Step 7: Calculate the Maximum Area

Substitute \(x = \frac{a}{\sqrt{2}}\) and \(y = \frac{b}{\sqrt{2}}\) back into the area formula: \(A = 4xy = 4 \times \frac{a}{\sqrt{2}} \times \frac{b}{\sqrt{2}} = 2ab\).

Step 8: Verify the Solution

Check that \(x = \frac{a}{\sqrt{2}}\) and \(y = \frac{b}{\sqrt{2}}\) satisfy the ellipse equation. Substitute back into the ellipse equation to ensure that it holds true.

Key Concepts

Ellipse

An ellipse is a geometric shape resembling an elongated circle, which you might imagine as an oval. It has two main axes: the major axis and the minor axis. The major axis is the longest diameter, while the minor axis is the shortest. The equation introducing an ellipse is\( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) where:

• \(a\) represents the semi-major axis length.
• \(b\) denotes the semi-minor axis length.

The shape is symmetrical with respect to both axes, creating a balanced structure. When we study a rectangle inside an ellipse, we explore its projection properties and how the ellipse constrains that rectangle's size.

Inscribed Rectangle

An inscribed rectangle is a rectangle placed inside another shape, like an ellipse, so that all the rectangle's vertices touch the boundary of that shape. In our exercise, the rectangle is positioned inside the ellipse along the coordinate axes.
This geometric placement makes use of symmetric properties, allowing the rectangle to be equally partitioned and aligned. This concept not only focuses on maximal area fitting but also understanding how shapes interact within boundaries without exceeding them. The key challenge is solving for the rectangle dimensions that utilize the ellipse's boundary most efficiently.

Differentiation

Differentiation is a fundamental calculus technique that lets us find the rate at which one quantity changes in relation to another. In this problem, we use differentiation to maximize the area of the inscribed rectangle.
Given the area function of the rectangle \(A(x) = 4bx \cdot \sqrt{1 - \frac{x^2}{a^2}}\), we apply differentiation to find the points where this area reaches its maximum possible size. This involves finding the derivative of the area function with respect to \(x\). By setting this derivative to zero, we identify the possible values of \(x\) where the rate of change transitions, which are known as critical points.

Critical Points

Critical points are the values of \(x\) where the derivative of a function equals zero. These points are significant because they represent locations where the function potentially has a maximum or minimum value. In the context of this exercise, identifying these points helps establish when the rectangle's area is maximized inside the ellipse.
After differentiating the area function and equating it to zero, solving for \(x\) gives the critical point \(x = \frac{a}{\sqrt{2}}\). With symmetry considerations and the ellipse equation, the corresponding \(y\) value also becomes \(\frac{b}{\sqrt{2}}\). Testing these values back into the area equation confirms that they provide the maximum possible inscribed area, which is \(2ab\). Critical points essentially guide us to the optimal solutions within the constraints given.