let T denote a curve `y=f(x)` which is in the first quadrant and let the point (1,0) lie on it. Let the tangent to T at a point P intersect the y-axis at `Y_(P)` and `PY_P` has length 1 for each poinit P on T. then which of the following option may be correct?

Options

A. Order 1
B. `xy'+sqrt(1-x^2)=0`
C. `y =log_(e)((1+sqrt(1-x^2))/(x))+sqrt(1+x^2)`
D. `xy 'sqrt(1-x^2)=0`

Answer

To solve the problem, we need to analyze the given conditions and derive the equation of the curve \( y = f(x) \) based on the properties of the tangent lines at points on the curve.

### Step-by-Step Solution:

1. **Understanding the Tangent Line**:
Let \( P(x, y) \) be a point on the curve \( y = f(x) \). The equation of the tangent line at point \( P \) can be expressed as:
\[ y - f(x) = f'(x)(x - x_0) \]
where \( f'(x) \) is the derivative of \( f(x) \) at point \( x \).

2. **Finding the Intersection with the Y-axis**:
The tangent line intersects the y-axis when \( x = 0 \). Substituting \( x = 0 \) in the tangent equation gives:
\[ y - f(x) = f'(x)(0 - x) \implies y = f(x) - f'(x)x \]
Let \( Y_P \) denote the y-coordinate of the intersection point. Thus:
\[ Y_P = f(x) - f'(x)x \]

3. **Length of Segment \( PY_P \)**:
The length of the segment \( PY_P \) is given as 1. Using the distance formula, we have:
\[ PY_P = \sqrt{(x - 0)^2 + \left(f(x) - (f(x) - f'(x)x)\right)^2} = 1 \]
Simplifying the expression inside the square root:
\[ PY_P = \sqrt{x^2 + (f'(x)x)^2} = 1 \]

4. **Squaring Both Sides**:
Squaring both sides gives:
\[ x^2 + (f'(x)x)^2 = 1 \]
Rearranging this, we have:
\[ x^2(1 + (f'(x))^2) = 1 \]

5. **Solving for \( f'(x) \)**:
We can isolate \( f'(x) \):
\[ 1 + (f'(x))^2 = \frac{1}{x^2} \implies (f'(x))^2 = \frac{1}{x^2} - 1 \]
Thus:
\[ f'(x) = \pm \sqrt{\frac{1 - x^2}{x^2}} = \pm \frac{\sqrt{1 - x^2}}{x} \]

6. **Integrating to Find \( f(x) \)**:
Integrating \( f'(x) \):
\[ f(x) = \pm \int \frac{\sqrt{1 - x^2}}{x} \, dx \]
This integral can be solved using trigonometric substitution, let \( x = \sin(\theta) \):
\[ dx = \cos(\theta) d\theta \]
The integral becomes:
\[ f(x) = \pm \int \frac{\sqrt{1 - \sin^2(\theta)}}{\sin(\theta)} \cos(\theta) d\theta = \pm \int \frac{\cos^2(\theta)}{\sin(\theta)} d\theta \]

7. **Final Form of the Function**:
After integrating, we can express \( f(x) \) in terms of \( x \) and a constant \( C \). The exact form will depend on the integration process, but it will yield a logarithmic function or an expression involving square roots.

8. **Finding the Constant Using Given Point**:
Since the curve passes through the point \( (1, 0) \), we can substitute \( x = 1 \) and \( f(1) = 0 \) to find the constant \( C \).