A curve 'C' with negative slope through the point(0,1) lies in the I Quadrant. The tangent at any point 'P' on it meets the x-axis at 'Q'. Such that `PQ=1`. Then
The curve in parametric form is

Options

A. `x = cos theta + In tan (theta//2), y = sin theta`
B. `x = -cos theta + In tan (theta//2), y = sin theta`
C. `x = -cos theta - In tan (theta//2), y = sin theta`
D. None of these

Explanation

To find the curve \( C \) in parametric form, we start with the given conditions and derive the necessary equations step by step.

### Step 1: Understand the problem
We know that the curve \( C \) passes through the point \( (0, 1) \) and has a negative slope. The tangent at any point \( P \) on the curve meets the x-axis at point \( Q \) such that the length \( PQ = 1 \).

### Step 2: Use the formula for the length of the tangent
The length of the tangent from a point \( P(x, y) \) on the curve to the x-axis can be expressed using the formula:
\[ PQ = \frac{y}{\sqrt{1 + \left(\frac{dy}{dx}\right)^2}} \]
Given that \( PQ = 1 \), we can set up the equation:
\[ 1 = \frac{y}{\sqrt{1 + \left(\frac{dy}{dx}\right)^2}} \]

### Step 3: Rearranging the equation
Squaring both sides gives:
\[ 1 = \frac{y^2}{1 + \left(\frac{dy}{dx}\right)^2} \]
Rearranging this, we find:
\[ 1 + \left(\frac{dy}{dx}\right)^2 = y^2 \]
Thus,
\[ \left(\frac{dy}{dx}\right)^2 = y^2 - 1 \]

### Step 4: Taking the square root
Taking the square root of both sides, we have:
\[ \frac{dy}{dx} = \pm \sqrt{y^2 - 1} \]
Since the curve has a negative slope, we choose the negative root:
\[ \frac{dy}{dx} = -\sqrt{y^2 - 1} \]

### Step 5: Change of variables
To solve this differential equation, we can use the substitution \( y = \sin(\theta) \). Then, we have:
\[ \frac{dy}{dx} = \cos(\theta) \frac{d\theta}{dx} \]
Substituting this into our equation gives:
\[ \cos(\theta) \frac{d\theta}{dx} = -\sqrt{\sin^2(\theta) - 1} \]
Since \( \sin^2(\theta) - 1 = -\cos^2(\theta) \), we can rewrite this as:
\[ \cos(\theta) \frac{d\theta}{dx} = -\cos(\theta) \]
Dividing both sides by \( \cos(\theta) \) (assuming \( \cos(\theta) \neq 0 \)):
\[ \frac{d\theta}{dx} = -1 \]

### Step 6: Integrate
Integrating both sides gives:
\[ \theta = -x + C \]

### Step 7: Express \( y \) in terms of \( x \)
From our substitution \( y = \sin(\theta) \), we have:
\[ y = \sin(-x + C) \]

### Step 8: Determine the constant \( C \)
Since the curve passes through the point \( (0, 1) \):
\[ 1 = \sin(-0 + C) \implies C = \frac{\pi}{2} \]
Thus, we have:
\[ y = \sin\left(-x + \frac{\pi}{2}\right) = \cos(x) \]

### Step 9: Parametric form of the curve
The parametric equations for the curve can be expressed as:
\[ x = t, \quad y = \cos(t) \]

### Final Answer
The curve in parametric form is:
\[ \begin{cases} x = t \\ y = \cos(t) \end{cases} \]