Question

The normal to the curve \[ x=a(1+\cos\theta), \quad y=a\sin\theta \] at \(\theta\) always passes through a fixed point, then the fixed point is

• A. \((a,0)\)
• B. \((0,a)\)
• C. \((0,0)\)
• D. \((a,a)\)

Hint:

For parametric curves, first find \(\frac{dy}{dx}\) using \(\frac{dy/d\theta}{dx/d\theta}\). Then write the normal equation and test the given fixed points.

Answer 1

The Correct Option is A

Step 1: Differentiate the parametric equations.
Given, \[ x=a(1+\cos\theta) \] and \[ y=a\sin\theta \] Differentiate with respect to \(\theta\): \[ \frac{dx}{d\theta}=-a\sin\theta \] \[ \frac{dy}{d\theta}=a\cos\theta \] Thus, \[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{a\cos\theta}{-a\sin\theta} = -\cot\theta \] Hence, slope of tangent is \[ -\cot\theta \]

Step 2: Find the slope of the normal.
Slope of normal is \[ m_n=\tan\theta \] The point on the curve is \[ \left(a(1+\cos\theta),a\sin\theta\right) \] Equation of the normal is \[ y-a\sin\theta = \tan\theta \left(x-a(1+\cos\theta)\right) \]

Step 3: Check whether the normal passes through a fixed point.
Substitute \[ x=a,\quad y=0 \] Then, \[ 0-a\sin\theta = \tan\theta \left(a-a(1+\cos\theta)\right) \] \[ -a\sin\theta = \tan\theta(-a\cos\theta) \] \[ -a\sin\theta = -a\sin\theta \] which is true for every \(\theta\).
Hence, every normal passes through the fixed point \[ (a,0) \]

Step 4: Final conclusion.
Therefore, the fixed point is \[ \boxed{(a,0)} \]