The tangents to the curve `x=a(theta - sin theta), y=a(1+cos theta)` at the points `theta = (2k+1)pi, k in Z` are parallel to :

Options

A. `y=x`
B. `y= -x`
C. `y=0`
D. `x=0`

Answer

To solve the problem step by step, we will find the slope of the tangent to the given parametric curve at the specified points and determine the line that is parallel to that tangent.

### Step 1: Define the Parametric Equations
The curve is given by the parametric equations:
- \( x = a(\theta - \sin \theta) \)
- \( y = a(1 + \cos \theta) \)

### Step 2: Differentiate \( y \) and \( x \) with respect to \( \theta \)
To find the slope of the tangent, we need to compute \( \frac{dy}{d\theta} \) and \( \frac{dx}{d\theta} \).

- Differentiate \( y \):
\[ \frac{dy}{d\theta} = a \cdot \frac{d}{d\theta}(1 + \cos \theta) = a(-\sin \theta) \]

- Differentiate \( x \):
\[ \frac{dx}{d\theta} = a \cdot \frac{d}{d\theta}(\theta - \sin \theta) = a(1 - \cos \theta) \]

### Step 3: Find the Slope \( \frac{dy}{dx} \)
The slope of the tangent line is given by:
\[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{-a \sin \theta}{a(1 - \cos \theta)} = \frac{-\sin \theta}{1 - \cos \theta} \]

### Step 4: Evaluate the Slope at \( \theta = (2k + 1)\pi \)
Now we need to evaluate the slope at the points where \( \theta = (2k + 1)\pi \):
- At \( \theta = (2k + 1)\pi \):
- \( \sin((2k + 1)\pi) = 0 \)
- \( \cos((2k + 1)\pi) = -1 \)

Substituting these values into the slope equation:
\[ \frac{dy}{dx} = \frac{-0}{1 - (-1)} = \frac{0}{2} = 0 \]

### Step 5: Conclusion
The slope of the tangent line at the points \( \theta = (2k + 1)\pi \) is \( 0 \). This means the tangent line is horizontal.

### Step 6: Identify the Line Parallel to the Tangent
A line with a slope of \( 0 \) is parallel to the x-axis. The equation of such a line can be expressed as:
\[ y = c \quad \text{(where c is a constant)} \]
In the context of the options provided, the line that is parallel to the tangent at these points is \( y = 0 \).

### Final Answer
The tangents to the curve at the points \( \theta = (2k + 1)\pi \) are parallel to the line \( y = 0 \).

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