The equation of the curve lying in the first quadrant, such that the portion of the x - axis cut - off between the origin and the tangent at any point P is equal to the ordinate of P, is (where, c is an arbitrary constant)

Options

A. `y=ce^((x)/(y))`
B. `ye^((x)/(y))=c`
C. `ye^((y)/(x))=c`
D. `y=ce^((y)/(x))`

Answer

To solve the problem, we need to find the equation of the curve in the first quadrant such that the portion of the x-axis cut off between the origin and the tangent at any point \( P \) is equal to the ordinate of \( P \).

### Step-by-Step Solution:

1. **Understanding the Problem**:
- Let \( P(x, y) \) be a point on the curve.
- The tangent at point \( P \) can be represented by the equation of the tangent line.

2. **Equation of the Tangent**:
- The equation of the tangent line at point \( P(x, y) \) is given by:
\[ y - y_1 = m(x - x_1) \]
where \( m = \frac{dy}{dx} \) is the slope at point \( P \).
- Rearranging gives:
\[ y = \frac{dy}{dx}(x - x) + y \]

3. **Finding the x-intercept of the Tangent**:
- Set \( y = 0 \) to find the x-intercept:
\[ 0 = \frac{dy}{dx}(x - x) + y \]
- This implies:
\[ x_{intercept} = x - \frac{y}{\frac{dy}{dx}} \]

4. **Condition Given in the Problem**:
- According to the problem, the length of the x-axis cut off (from the origin to the x-intercept) is equal to the ordinate \( y \):
\[ x - \frac{y}{\frac{dy}{dx}} = y \]

5. **Rearranging the Equation**:
- Rearranging gives:
\[ x = y + \frac{y}{\frac{dy}{dx}} \]
- This can be rewritten as:
\[ x = y \left( 1 + \frac{1}{\frac{dy}{dx}} \right) \]

6. **Differential Equation Formation**:
- Rearranging leads to:
\[ \frac{dy}{dx} = \frac{y}{x - y} \]
- This can be rearranged to form a separable differential equation:
\[ \frac{dy}{y} = \frac{dx}{x - y} \]

7. **Integrating Both Sides**:
- Integrate both sides:
\[ \int \frac{dy}{y} = \int \frac{dx}{x - y} \]
- This gives:
\[ \ln |y| = \ln |x - y| + C \]

8. **Exponentiating**:
- Exponentiating both sides:
\[ y = k(x - y) \]
- Where \( k = e^C \).

9. **Solving for y**:
- Rearranging gives:
\[ y + ky = kx \]
- Thus:
\[ y(1 + k) = kx \implies y = \frac{kx}{1 + k} \]

10. **Final Form**:
- This can be rewritten as:
\[ ye^{\frac{x}{y}} = C \]
- Where \( C \) is a constant.

### Conclusion:
The equation of the curve is:
\[ y e^{\frac{x}{y}} = C \]