Problem 163 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

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

Step-by-Step Solution

Answer

The correct answer is (C) \((0,0)\).

Step 1: Parameterization of the Curve

Identify the parameterized form of the curve given: \( x = a(1 + \cos \theta) \) and \( y = a \sin \theta \). These equations describe the curve in terms of parameter \( \theta \).

Step 2: Differentiate to Find the Tangent

Differentiate \( x \) and \( y \) with respect to \( \theta \):\[ \frac{dx}{d\theta} = -a \sin \theta \] and \[ \frac{dy}{d\theta} = a \cos \theta \].Thus, the slope of the tangent to the curve at \( \theta \) is given by \( \frac{dy}{dx} = \frac{a \cos \theta}{-a \sin \theta} = -\cot \theta \).

Step 3: Find the Slope of the Normal

Since the slope of the tangent line is \(-\cot \theta\), the slope of the normal line is the negative reciprocal: \( \tan \theta \).

Step 4: Equation of the Normal

The equation of the normal line passing through the point \((x_1, y_1) = (a(1 + \cos \theta), a \sin \theta)\) is given by:\[ y - y_1 = \tan \theta (x - x_1) \]This can be written as:\[ y - a \sin \theta = \tan \theta (x - a(1 + \cos \theta)) \].

Step 5: Simplify the Equation

Substitute \( y = 0 \) and \( x = 0 \) to check if it satisfies the normal equation. The equation simplifies to check fixed point:Substitute \((x, y) = (0, 0)\) and verify:\[ -a \sin \theta = \tan \theta (-a(1 + \cos \theta)) \] Simplifying,\[ -a \sin \theta = \frac{a \sin \theta}{a \cos \theta} (-a(1 + \cos \theta)) \] which holds true, confirming our point is \((0, 0)\).

Step 6: Conclusion

The normal to the curve \( x = a(1 + \cos \theta), \ y = a \sin \theta \) at any \( \theta \) passes through the point \((0, 0)\).

Key Concepts

Parametric equations

Parametric equations provide an efficient way to represent curves in the plane. Instead of defining a curve solely in terms of \( x \) and \( y \), parametric equations express both \( x \) and \( y \) as functions of a third variable, often labeled as \( t \) or \( \theta \).
In the given problem, we have:

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

Here, \( \theta \) is the parameter that varies, generating different points \((x, y)\) on the curve for each \( \theta \) value. This approach allows us to describe more complex curves that might be difficult to express as a single function of \( x \) or \( y \).
For example, the curve could represent parts of a circle, ellipse, or other shapes that aren't necessarily functions since they can double back on themselves or create loops.

Slope of the tangent

The slope of the tangent line to a curve at any given point is crucial because it provides us with the direction in which the curve is heading at that point.
To find the tangent's slope in parametric equations, we differentiate both \( x \) and \( y \) with respect to the parameter, \( \theta \).
For the curve described by \( x = a(1 + \cos \theta) \) and \( y = a \sin \theta \), we compute:

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

The slope of the tangent, \( \frac{dy}{dx} \), is found by dividing these derivatives:\[ \frac{dy}{dx} = \frac{a \cos \theta}{-a \sin \theta} = -\cot \theta \]
The maple housed within \( -\cot \theta \) gives us a straightforward way to determine the angle of the tangent line at any point defined by \( \theta \).

Equation of the normal

The normal line to a curve at a given point is perpendicular to the tangent at that point. Finding the equation of the normal involves determining its slope and point of contact.
Since the tangent's slope is \( -\cot \theta \), the slope of the normal, being the negative reciprocal, is \( \tan \theta \).
The equation for a line is typically expressed in point-slope form: \( y - y_1 = m(x - x_1) \). For the normal line:

• Point \((x_1, y_1)\) is where the curve is intersected, here \( (a(1+\cos \theta), a \sin \theta) \).
• Slope \( m \) is \( \tan \theta \).

Thus, the normal's equation becomes:\[ y - a \sin \theta = \tan \theta (x - a(1 + \cos \theta)) \]
To check if this line passes through a fixed point, substitute potential coordinates like \( (0, 0) \) into the equation. If it holds true, as simplification shows it does, then \((0, 0)\) is indeed a fixed point through which the normal always passes, confirming the problem conclusion.