In the curve represented parametrically by the equations `x=2ln cott+1 and y=tant+cott,` (A) tangent and normal intersect at the point (2,1).(B) normal at t=π4 is parallel to the y-axis. (C) tangent at t=π4 is parallel to the line y=x .(D) tangent at t=π4 is parallel to the x-axis.

Options

A. tangent and normal intersect at the point (2,1)
B. normal at `t=pi//4` is parallel to the y-axis
C. tangent at `t=pi//4` is parallel to the line y = x
D. tangent at `t=pi//4` is parallel to the x-axis

Answer

To solve the problem, we will analyze the given parametric equations and find the required derivatives and conditions step by step.

### Given Parametric Equations:
1. \( x = 2 \ln(\cot t) + 1 \)
2. \( y = \tan t + \cot t \)

### Step 1: Find \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \)

**Differentiating \( y \) with respect to \( t \):**
\[ y = \tan t + \cot t \]
Using the derivatives:
- \( \frac{d}{dt}(\tan t) = \sec^2 t \)
- \( \frac{d}{dt}(\cot t) = -\csc^2 t \)

Thus,
\[ \frac{dy}{dt} = \sec^2 t - \csc^2 t \]

**Differentiating \( x \) with respect to \( t \):**
\[ x = 2 \ln(\cot t) + 1 \]
Using the chain rule:
\[ \frac{d}{dt}(\ln(\cot t)) = \frac{1}{\cot t} \cdot (-\csc^2 t) = -\frac{\csc^2 t}{\cot t} \]
Thus,
\[ \frac{dx}{dt} = 2 \cdot \left(-\frac{\csc^2 t}{\cot t}\right) = -2 \frac{\csc^2 t}{\cot t} \]

### Step 2: Find \( \frac{dy}{dx} \)

Using the chain rule:
\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\sec^2 t - \csc^2 t}{-2 \frac{\csc^2 t}{\cot t}} = -\frac{(\sec^2 t - \csc^2 t) \cot t}{2 \csc^2 t} \]

### Step 3: Evaluate at \( t = \frac{\pi}{4} \)

At \( t = \frac{\pi}{4} \):
- \( \tan\left(\frac{\pi}{4}\right) = 1 \)
- \( \cot\left(\frac{\pi}{4}\right) = 1 \)
- \( \sec^2\left(\frac{\pi}{4}\right) = 2 \)
- \( \csc^2\left(\frac{\pi}{4}\right) = 2 \)

Substituting these values:
\[ \frac{dy}{dt} = 2 - 2 = 0 \]
\[ \frac{dx}{dt} = -2 \cdot \frac{2}{1} = -4 \]

Thus,
\[ \frac{dy}{dx} = \frac{0}{-4} = 0 \]

### Step 4: Analyze the Tangent and Normal

Since \( \frac{dy}{dx} = 0 \) at \( t = \frac{\pi}{4} \), the tangent line at this point is horizontal (parallel to the x-axis).

### Step 5: Check the Intersection of Tangent and Normal

The coordinates at \( t = \frac{\pi}{4} \):
\[ x = 2 \ln(1) + 1 = 1 \]
\[ y = 1 + 1 = 2 \]
So the point is \( (1, 2) \).

The equation of the tangent line at this point is:
\[ y - 2 = 0 \quad \text{(horizontal line)} \]

The normal line is vertical (since the tangent is horizontal) and has the equation:
\[ x = 1 \]

Thus, the tangent and normal intersect at \( (1, 2) \).

### Conclusion

Now we can summarize the findings regarding the options:

(A) Tangent and normal intersect at the point (2, 1). **(False)**
(B) Normal at \( t = \frac{\pi}{4} \) is parallel to the y-axis. **(True)**
(C) Tangent at \( t = \frac{\pi}{4} \) is parallel to the line \( y = x \). **(False)**
(D) Tangent at \( t = \frac{\pi}{4} \) is parallel to the x-axis. **(True)**