Problem 111 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The angle between the tangents at any point \(P\) and the line joining \(P\) to the origin \(O\), where \(P\) is a point on the curve \(\ln \left(x^{2}+y^{2}\right)=c \tan ^{-1} \frac{y}{x}, c\) is a constant (A) varies as \(\tan ^{-1} x\) (B) varies as \(\tan ^{-1} y\) (C) is a constant (D) None of these

Step-by-Step Solution

Answer

The angle is constant, so the answer is (C).

Step 1: Understand the Given Equation and Curve

The given curve is \( \ln\left(x^{2}+y^{2}\right) = c \tan^{-1}\left(\frac{y}{x}\right) \). This is a polar equation similar to a spiral line where \( c \) is a constant.

Step 2: Implement Polar Coordinates

To simplify the problem, convert the Cartesian coordinates to polar coordinates where \( x = r \cos \theta \) and \( y = r \sin \theta \). Substituting \( x^2 + y^2 = r^2 \) and \( \frac{y}{x} = \tan \theta \), rewrite the equation as \[ \ln r^2 = c \theta \]. This simplifies to \( 2 \ln r = c \theta \).

Step 3: Differentiate in Polar Form

Differentiate \( 2 \ln r = c \theta \) with respect to \( \theta \). We get \[ \frac{d}{d\theta}(2 \ln r) = \frac{d}{d\theta}(c \theta) \]. Thus, \[ \frac{2}{r} \frac{dr}{d\theta} = c \].

Step 4: Find the Expression for the Tangent

The expression for \( \frac{dy}{dx} \) at any point \( P (r, \theta) \) involves taking derivatives in polar coordinates (here using \( \theta \) as the independent variable). We have \[ \frac{dy}{dx} = \frac{r \cos \theta \cdot \frac{dy}{d\theta} - r \sin \theta \cdot \frac{dx}{d\theta}}{r \cos^2 \theta + r \sin^2 \theta} \]. Substitute for \( \frac{dr}{d\theta} \) from the differential equation obtained, simplifying the expression.

Step 5: Calculate the Angle Between Tangents

The angle \( \phi \) between the tangent at \( P \) and the line \( OP \) is given by \[ \tan \phi = \left| \frac{r \, d\theta/dr}{1} \right| \]. Substituting \( \frac{dr}{d\theta} = \frac{rc}{2} \) simplifies the angle calculation. Due to the formula: \( \tan \phi \) becomes \( \frac{c}{2} \), a constant value.

Key Concepts

Polar Coordinates

Polar coordinates are an alternative way to describe locations on a plane, different from the traditional Cartesian coordinate system. In polar coordinates, a point is represented by a distance from a reference point and an angle from a reference direction.

• Distance (r): This is the distance from the origin to the point in question.
• Angle (\(\theta\)): This is the angle made with a reference direction, usually the positive x-axis.

To convert Cartesian coordinates (\(x, y\)) to polar coordinates (\(r, \theta\)), you use the formulas:\[x = r \cos \theta \y = r \sin \theta \r = \sqrt{x^2 + y^2} \\theta = \tan^{-1}\left(\frac{y}{x}\right) \\]These formulas were essential in simplifying the given curve, allowing it to be expressed in terms of polar equations like those often seen in geometry problems.

Differential Calculus

Differential calculus is a branch of mathematics focused on the concept of the derivative, which represents the rate of change of a function.In our exercise, we used differential calculus to find how 'r', the radial distance, changes with respect to \(\theta\). This was done by differentiating the polar form of the given equation:2 \(\ln r = c \theta\)Through differentiation, we derived that:\[ \frac{2}{r} \frac{dr}{d\theta} = c\]This relationship allows us to track how the curve grows or shrinks as you move along the angle \(\theta\). The derivative \(\frac{dr}{d\theta}\), therefore, gives us crucial information needed to determine properties of the curve like tangents, which are vital for solving geometry problems like this.

Tangent Lines

In geometry problems, tangent lines are important because they touch a curve at exactly one point, describing the curve's direction at that point.To find the tangent line in polar coordinates, we first need its slope \(\frac{dy}{dx}\). In our exercise, this involved using the derivatives we found earlier:\[ \frac{dy}{dx} = \frac{r \cos \theta \cdot \frac{dy}{d\theta} - r \sin \theta \cdot \frac{dx}{d\theta}}{r \cos^2 \theta + r \sin^2 \theta}\]Then, we use the formula provided:\[ \tan \phi = \left| \frac{r \cdot d\theta/dr}{1} \right|\]This helps calculate the angle \(\phi\) between the tangent line at point \(P\) and the line joining the point \(P\) to the origin \(O\). With our given equation, this angle turned out to be a constant, \(\frac{c}{2}\), emphasizing how tangent lines and differential calculus are intertwined in understanding the behavior of curves.