Problem 40 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The maximum value of radius vector where \(\frac{c^{4}}{r^{2}}=\frac{a^{2}}{\sin ^{2} t}+\frac{b^{2}}{\cos ^{2} t} ;(a, b>0)\) is (A) \((a+b)^{2}\) (B) \(\frac{c^{4}}{(a+b)^{2}}\) (C) \(\frac{c^{2}}{a+b}\) (D) \(c^{2}(a+b)\)

Step-by-Step Solution

Answer

The correct answer is (C) \( \frac{c^{2}}{a+b} \).

Step 1: Interpret the Given Equation

The equation given is \( \frac{c^{4}}{r^{2}} = \frac{a^{2}}{\sin ^{2} t} + \frac{b^{2}}{\cos ^{2} t} \). We need to determine the value of \( r \) that maximizes this expression. Rearranging, we have \( r^2 = \frac{c^{4}}{\left( \frac{a^{2}}{\sin ^{2} t} + \frac{b^{2}}{\cos ^{2} t} \right)} \). The goal is to maximize \( r \).

Step 2: Apply Trigonometric Identity

Using the identity \( \sin^2 t + \cos^2 t = 1 \), we can express the equation in terms of a single trigonometric function. Set \( x = \sin^2 t \), thus \( \cos^2 t = 1 - x \). Now substitute these into the given equation, resulting in \( r^2 = \frac{c^{4}}{\frac{a^2}{x} + \frac{b^2}{1-x}} \).

Step 3: Simplify the Expression

Simplify the expression to find \( r^2 \) in terms of \( x \):\[r^2 = \frac{c^4 x(1-x)}{a^2(1-x) + b^2 x}\]Further simplification gives:\[r^2 = \frac{c^4 x (1-x)}{(a^2 - a^2 x + b^2 x)}\]

Step 4: Maximize with Respect to x

To maximize \( r^2 \) with respect to \( x \), take its derivative with respect to \( x \), set it to zero, and solve for \( x \). Calculating this derivative leads us to the critical point \( x = \frac{a}{a+b} \) by ensuring the numerator of the derivative equates to zero.

Step 5: Substitute Critical Point Back

Substitute \( x = \frac{a}{a+b} \) back into the equation for \( r^2 \):\[r^2 = \frac{c^4 \left(\frac{a}{a+b}\right)\left(1-\frac{a}{a+b}\right)}{a^2\left(1-\frac{a}{a+b}\right) + b^2\left(\frac{a}{a+b}\right)}\]

Step 6: Simplify and Identify Maximum Value

Simplifying further:\[r^2 = \frac{c^4 \frac{a(a+b-a)}{(a+b)^2}}{a^2 \frac{b}{a+b} + b^2 \frac{a}{a+b}}\]\[r^2 = \frac{c^4 \frac{ab}{(a+b)^2}}{\frac{ab}{a+b}} = \frac{c^4}{(a+b)^2}\]Thus, the maximum possible value of \( r \) is \( \frac{c^2}{a+b} \).

Step 7: Identify the Correct Answer Choice

The maximum value of the radius vector, given \( r^2 = \frac{c^4}{(a+b)^2} \), corresponds to the square root: \( \frac{c^2}{a+b} \). Hence, the correct answer is option C.

Key Concepts

Radius Vector

The radius vector, often seen in polar coordinates, represents the distance from the origin to a specific point. In this exercise, we're given a rather complex equation that involves finding the maximum value of this radius vector. This involves understanding how the equation is set up with respect to the radius vector, denoted here as \( r \).

• The radius vector \( r \) is found in the denominator, indicating we must maximize its square, \( r^2 \).
• In physical terms, the radius vector connects the origin of a coordinate system with any arbitrary point in a plane.
• Your aim is to rearrange the terms and utilize mathematical identities to solve for \( r \).

Understanding the physical meaning of the radius vector helps to grasp not just the computational steps but the reason behind maximizing \( r \) within the constraints provided by the equation.

Maximization Problems

Maximization problems in mathematics involve finding the largest possible value of a function. In our given equation, which we've rewritten as \( r^2 \), we look to maximize this by manipulating the variables involved.First, we introduce new variables to simplify our problem:- Set \( x = \sin^2 t \), and with this identity we know \( \cos^2 t = 1 - x \).Next, express \( r^2 \) in terms of \( x \):- We simplify the given equation to find \( r^2 = \frac{c^4 x(1-x)}{a^2(1-x) + b^2 x} \).To find the maximum of \( r^2 \), derivative calculus plays a crucial role:- By taking the derivative of the expression with respect to \( x \) and setting it to zero, we locate critical points that could represent maxima.- Here, solving leads us to \( x = \frac{a}{a+b} \) as a critical value where the function obtains its maximum value.The challenge is not just in computation but understanding why this approach finds the largest possible \( r \) under the conditions given, balancing the influence of \( a \) and \( b \).

Trigonometric Functions

Trigonometric functions, key players in many mathematical problems, relate angles to sides in right triangles. They're pivotal in our problem:- We're dealing with \( \sin \) and \( \cos \), fundamental trigonometric functions, to help redefine the original equation for ease of maximization.- Thanks to the identity \( \sin^2 t + \cos^2 t = 1 \), we simplify the problem by using a single variable \( x \) in place of two trigonometric expressions.Key steps when dealing with such functions:

• Identify which identities simplify the process best. In this case, substituting \( \sin^2 t \) and \( \cos^2 t \) helps reduce complexity.
• Understand how changes in \( t \) affect \( \sin^2 t \) and \( \cos^2 t \), and consequently \( r^2 \).

Mastering these functions and their identities not only streamlines calculations but also enhances your ability to tackle more complex trigonometric problems in the future.