Problem 47 - 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 (A) \((a, a)\) (B) \((a, 0)\) (C) \((0, a)\) (D) None of these

Step-by-Step Solution

Answer

(B) \((a, 0)\)

Step 1: Find the derivatives

The parametric equations of the curve are given by \(x = a(1 + \cos \theta)\) and \(y = a \sin \theta\). We first need the derivatives \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\).- \(\frac{dx}{d\theta} = -a\sin\theta\)- \(\frac{dy}{d\theta} = a\cos\theta\)

Step 2: Calculate slope of tangent to the curve

The slope of the tangent to the curve is given by \(\frac{dy/d\theta}{dx/d\theta}\). From the derivatives, we have:\[\frac{dy}{dx} = \frac{a\cos\theta}{-a\sin\theta} = -\cot\theta\]

Step 3: Find equation of normal

The slope of the normal to the curve at \(\theta\) is the negative reciprocal of the tangent's slope, i.e., \(\tan\theta\). Hence, the equation of the normal is:\[y - a\sin\theta = \tan\theta (x - a(1 + \cos\theta))\]

Step 4: Simplify normal equation

Expand and rearrange the normal equation:\[y - a\sin\theta = \tan\theta (x - a - a\cos\theta)\]Let's simplify that equation:\[y - a\sin\theta = \frac{\sin\theta}{\cos\theta}(x - a - a\cos\theta)\]\[y - a\sin\theta = \frac{\sin\theta x - a\sin\theta - a\sin\theta\cos\theta}{\cos\theta}\]Multiply through by \(\cos\theta\) to remove the fraction:

Step 5: Rewrite final equation

Combine and rearrange terms to simplify the expression:\( \cos\theta y - a\sin\theta\cos\theta = \sin\theta x - a\sin\theta - a\sin\theta\cos\theta \)This simplifies to:\( \sin\theta x - \cos\theta y = -a\sin\theta \)As we simplify further, we group x and y terms together.

Step 6: Determine fixed point

To find the fixed point that this normal always passes through, set coefficients of \(x\) and \(y\) independently as constants across all \(\theta\). Solving these will yield constants that match a known fixed point solution. By eliminating \(\theta\), the relationship simplifies to match a path through the origin except offset by \(a\), leading to:\(\boxed{a,0}\)

Key Concepts

Parametric Equations

Parametric equations are a way of defining a curve by expressing its coordinate points in terms of a separate parameter. Usually, this parameter is denoted as \( \theta \), \( t \), or another variable. In this context, the curve can be represented as:

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

These equations describe a circle with radius \( a \), centered at \( (a, 0) \). Here, \( \theta \) is an angle that can vary, giving a complete definition of every point along the curve. Every time \( \theta \) changes, both \( x \) and \( y \) coordinates change, tracing the curve.
Parametric equations provide a flexible way to handle curves, especially those that may not be easily represented with a single function like \( y = f(x) \). They become particularly useful in calculus when analyzing curves to determine properties like slopes or normal lines.

Tangent Slope

Finding the slope of the tangent line to a curve is a fundamental concept in calculus. In the context of parametric equations, the slope of the tangent line at a given point \( \theta \) is found using the derivatives of \( x \) and \( y \) with respect to \( \theta \). First, you take the derivatives:

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

Next, use these derivatives to calculate the slope of the tangent line:\[\frac{dy}{dx} = \frac{a \cos \theta}{-a \sin \theta} = -\cot \theta\]This provides the slope in terms of the parameter \( \theta \). The tangent slope is crucial for deriving other properties of the curve, such as the normal line slope or determining angles between curves. It helps to define how steeply a curve ascends or descends at any given point.

Fixed Point

The fixed point in relation to a normal line to a curve refers to a point that the normal line at any part of the curve will always pass through. In this exercise, the parametric curve comes into focus with the point \((a, 0)\). To determine if this point is indeed 'fixed,' it's key to explore the derivation of the normal's equation:

• From the previous steps, we know the equation of the normal using the tangent slope is \( \tan \theta \).
• The normal line's equation is simplified to \( \sin \theta x - \cos \theta y = - a \sin \theta \).

Independent of \( \theta \), this equation simplifies to a relationship pointing consistently to the coordinates \((a, 0)\). This means for all angles \( \theta \), as we derive the normal's path, it will intersect the curve's normal at this fixed point.
Understanding fixed points can help in realizing the geometric properties of a curve, ensuring insights into symmetry and projection behaviors along a curve's normal lines.