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

Question

If the eccentric angle of a point lying in the first quadrant on the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) be \(\alpha\) and the line joining the centre to the point makes an angle \(\beta\) with \(x\)-axis then \(\alpha-\beta\) will be maximum when \(\alpha=\) (A) 0 (B) \(\cot ^{-1} \sqrt{\frac{a}{b}}\) (C) \(\tan ^{-1} \sqrt{\frac{a}{b}}\) (D) \(\pi / 4\)

Step-by-Step Solution

Answer

(C) \( \tan^{-1} \sqrt{\frac{a}{b}} \)

Step 1: Understand the Problem

We are given an ellipse with the equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) and need to analyze a point on it located in the first quadrant. The eccentric angle is \( \alpha \) and the angle that the line from the center to this point makes with the x-axis is \( \beta \). We are to find when \( \alpha - \beta \) is maximized.

Step 2: Express Coordinates with Eccentric Angle

The parameterization of a point on an ellipse using the eccentric angle \( \alpha \) is \( (a \cos \alpha, b \sin \alpha) \). This represents the coordinates of the point on the ellipse in terms of \( \alpha \).

Step 3: Calculate Angle with x-axis

The angle \( \beta \) that the line from the center of the ellipse to the point makes with the x-axis can be expressed as: \( \beta = \tan^{-1} \left( \frac{b \sin \alpha}{a \cos \alpha} \right) = \tan^{-1} \left( \frac{b}{a} \tan \alpha \right) \).

Step 4: Determine Expression for \( \alpha - \beta \)

The difference \( \alpha - \beta \) becomes \( \alpha - \tan^{-1} \left( \frac{b}{a} \tan \alpha \right) \). This will be our function of interest for maximizing.

Step 5: Differentiate to Find Critical Points

To maximize \( \alpha - \beta \), take the derivative of \( \alpha - \tan^{-1} \left( \frac{b}{a} \tan \alpha \right) \) with respect to \( \alpha \) and set it to zero:\[ \frac{d}{d\alpha} \left( \alpha - \tan^{-1} \left( \frac{b}{a} \tan \alpha \right) \right) = 1 - \frac{\frac{b}{a} \sec^2 \alpha}{1 + \left( \frac{b}{a} \tan \alpha \right)^2} = 0 \]Simplifying this expression leads to the critical point condition.

Step 6: Solve for \( \alpha \)

Solving the condition obtained from differentiation gives:\[ \frac{b/a}{1 + \frac{b^2}{a^2} \tan^2 \alpha} = 1 \]This simplifies to:\[ \tan^2 \alpha = \frac{a}{b} \]Thus, \( \alpha = \tan^{-1} \sqrt{\frac{a}{b}} \).

Step 7: Conclude by Analyzing Options

The solution \( \alpha = \tan^{-1} \sqrt{\frac{a}{b}} \) matches option (C). Therefore, \( \alpha - \beta \) is maximized when \( \alpha = \tan^{-1} \sqrt{\frac{a}{b}} \).

Key Concepts

Ellipse

An ellipse is a conic section that looks like an elongated circle. It can be thought of as a circle stretched along two perpendicular axes. It's described by a standard equation: \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\]where \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes, respectively. This equation represents the collection of all points whose distances to two focal points have a constant sum.
Ellipses have several interesting properties and are found abundantly in natural phenomena, such as planetary orbits.

• The center of the ellipse is located at the origin \((0,0)\).
• When \(a > b\), the ellipse is stretched more along the x-axis, making it appear wider.
• Conversely, if \(b > a\), it's longer along the y-axis, making it taller.

Understanding ellipses is crucial in calculus and physics and helpful in fields like astronomy and engineering.

Eccentric Angle

The eccentric angle \(\alpha\) is an angle used in the parametric representation of the ellipse. It essentially helps us locate a point on the ellipse via trigonometric functions. The coordinates of a point on the ellipse in terms of its eccentric angle \(\alpha\) are given by:

• \(x = a \cos \alpha\)
• \(y = b \sin \alpha\)

This angular representation is pivotal when dealing with trigonometric functions and understanding the geometry of shapes like ellipses. Using the eccentric angle makes it easy to express complex curved paths in simpler algebraic forms.
The eccentric angle is an elegant way to relate the shape of the ellipse—which could appear as a complicated structure—to simple angles that we encounter in trigonometry, making calculation and representation much easier.Understanding the eccentric angle is fundamental for tasks involving parametric equations and a variety of problem-solving contexts, like the one from the original exercise.

Trigonometric Functions

Trigonometric functions, such as sine, cosine, and tangent, are essential in describing the relationship between angles and sides of triangles. In the context of ellipses, they help in describing points on the ellipse using eccentric angles. These functions are key to solving problems involving periodic patterns and rotations. The core trigonometric functions are:

• \(\sin \theta\)
• \(\cos \theta\)
• \(\tan \theta\)

These functions help in expressing the coordinates of a point on an ellipse as \( (a \cos \alpha, b \sin \alpha) \), where \(\alpha\) is the eccentric angle.
Another important trigonometric concept used in our exercise is the inverse tangent or arc tangent, denoted \(\tan^{-1}\) or \(\arctan\). This function returns the angle whose tangent is a given number. For example, the angle \(\beta\) in our solution is expressed as:\[\beta = \tan^{-1} \left( \frac{b}{a} \tan \alpha \right)\]Trigonometric functions provide us an accessible way of solving complex geometric problems by translating them into angles or simpler function evaluations. Mastery of these functions aids not just in academic pursuits but also real-world applications like signal processing, architecture, and physics.