Problem 163 - A Complete Resource Book in Mathematics for JEE Main

Question

The normal to the curve \(x=a(1+\cos \theta), y=a \sin \theta\) at \(\theta\) always passes through the fixed point [2004] (A) \((a, 0)\) (B) \((0, a)\) (C) \((0,0)\) (D) \((a, a)\)

Step-by-Step Solution

Answer

The normal to the curve always passes through the point \((0,0)\). Option (C) is correct.

Step 1: Find the Parametric Equations

The given parametric equations for the curve are:\[ x = a(1 + \cos \theta) \] and \[ y = a \sin \theta \].

Step 2: Determine the Derivatives for the Normal

Compute the derivatives \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \): \[ \frac{dx}{d\theta} = -a \sin \theta\] and \[ \frac{dy}{d\theta} = a \cos \theta \].

Step 3: Calculate the Slope of the Tangent

The slope of the tangent line to the curve is given by \( \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = -\cot \theta \).

Step 4: Determine the Slope of the Normal

The normal's slope is the negative reciprocal of the tangent's slope, so it is \( \tan \theta \).

Step 5: Find the Equation of the Normal

Using the point \((x_0, y_0) = (a(1 + \cos \theta), a \sin \theta)\) and the slope \( \tan \theta \), the equation of the normal is:\[ y - a \sin \theta = \tan \theta (x - a(1 + \cos \theta)) \].

Step 6: Simplify the Equation of the Normal

Rearrange and simplify the normal's equation: \[ y - a \sin \theta = \tan \theta \cdot x - a \cdot \tan \theta (1 + \cos \theta) \].

Step 7: Find the Fixed Point

Substituting common coordinates like the origin \((0,0)\) into the normal's equation, and confirm the fixed point by seeing if it satisfies the equation for all \( \theta \), derive that the normal always passes through the origin \((0, 0)\).

Step 8: Select the Correct Option

From the calculations, we see that the normal passes through the point \((0,0)\), which corresponds to option (C).

Key Concepts

Derivatives

In mathematics, a derivative represents the rate of change of a function. If you think of a curve as the path traced out by a moving point, then the derivative at any point on the curve gives you the direction of travel at that very point.
For parametric equations such as \[ x = a(1 + \cos \theta), \ y = a \sin \theta \], it is important to find how each parameter, \(\theta\), changes with respect to the curve.
In the given exercise, the derivatives \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \) are determined as:

• \( \frac{dx}{d\theta} = -a \sin \theta \)
• \( \frac{dy}{d\theta} = a \cos \theta \)

Once these are calculated, they can be used to find the slope of the tangent line to the curve at any given point.

Slope of Tangent

The slope of a tangent line to a curve can help you understand how steep the curve is at a particular point. For a curve given by parametric equations, the slope of the tangent line at a point is determined by the ratio of the derivatives from the previous section.

• It is given as \( \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \).
• In this problem, substituting the derivatives results in a tangent slope of \( -\cot \theta \).

This slope indicates the direction and steepness at which the curve is changing at that specific point.
Whereas, understanding the tangent slope can simplify tasks like graphing or predicting the path of the curve.

Equation of Normal

A normal line to a curve at a given point is perpendicular to the tangent line at that point. Its slope is the negative reciprocal of the slope of the tangent.
For the curve described by the parametric equations, the normal's slope is computed as \( \tan \theta \).
Using the point \((x_0, y_0) = (a(1 + \cos \theta), a \sin \theta)\) and this slope, the equation of the normal line is:

• \( y - a \sin \theta = \tan \theta (x - a(1 + \cos \theta)) \)

The simplification of this equation helps to determine if a specific point is "fixed," meaning the normal lines always pass through it, regardless of \(\theta\).
In the exercise provided, further analysis shows that the fixed point is indeed \((0,0)\), as the normal line passes through this point for all values of \(\theta\).