The equation of tangent to the curve `x=a(theta+ sin theta), y=a (1+ cos theta)" at "theta=(pi)/(2)` is

Options

A. `2x-2y=a(pi+2)`
B. `2x+2y=a(pi+2)`
C. `2x-2y=a(pi+4)`
D. `2x+2y=a(pi+4)`

Answer

To find the equation of the tangent to the curve given by the parametric equations \( x = a(\theta + \sin \theta) \) and \( y = a(1 + \cos \theta) \) at \( \theta = \frac{\pi}{2} \), we will follow these steps:

### Step 1: Find the coordinates of the point on the curve at \( \theta = \frac{\pi}{2} \).

We will substitute \( \theta = \frac{\pi}{2} \) into the equations for \( x \) and \( y \):

\[ x = a\left(\frac{\pi}{2} + \sin\left(\frac{\pi}{2}\right)\right) = a\left(\frac{\pi}{2} + 1\right) \]

\[ y = a\left(1 + \cos\left(\frac{\pi}{2}\right)\right) = a(1 + 0) = a \]

Thus, the point on the curve at \( \theta = \frac{\pi}{2} \) is:

\[ \left(x_1, y_1\right) = \left(a\left(\frac{\pi}{2} + 1\right), a\right) \]

### Step 2: Calculate \(\frac{dy}{dx}\) using parametric differentiation.

The derivatives of \( x \) and \( y \) with respect to \( \theta \) are:

\[ \frac{dx}{d\theta} = a\left(1 + \cos \theta\right) \]
\[ \frac{dy}{d\theta} = -a\sin \theta \]

Now, we can find \(\frac{dy}{dx}\):

\[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{-a\sin \theta}{a(1 + \cos \theta)} = \frac{-\sin \theta}{1 + \cos \theta} \]

### Step 3: Evaluate \(\frac{dy}{dx}\) at \( \theta = \frac{\pi}{2} \).

Substituting \( \theta = \frac{\pi}{2} \):

\[ \frac{dy}{dx}\bigg|_{\theta = \frac{\pi}{2}} = \frac{-\sin\left(\frac{\pi}{2}\right)}{1 + \cos\left(\frac{\pi}{2}\right)} = \frac{-1}{1 + 0} = -1 \]

### Step 4: Write the equation of the tangent line.

The equation of the tangent line in point-slope form is given by:

\[ y - y_1 = m(x - x_1) \]

Where \( m = \frac{dy}{dx} \) at the point \( (x_1, y_1) \):

Substituting the values:

\[ y - a = -1\left(x - a\left(\frac{\pi}{2} + 1\right)\right) \]

Simplifying this gives:

\[ y - a = -x + a\left(\frac{\pi}{2} + 1\right) \]

Rearranging the equation:

\[ y = -x + a\left(\frac{\pi}{2} + 2\right) \]

### Final Answer:

Thus, the equation of the tangent to the curve at \( \theta = \frac{\pi}{2} \) is:

\[ y = -x + a\left(\frac{\pi}{2} + 2\right) \]