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 a 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 (y-coordinate) of \( P \).

### Step-by-Step Solution:

1. **Identify the point P**: Let the coordinates of point \( P \) be \( (x, y) \).

2. **Equation of the tangent at point P**: The equation of the tangent line at point \( P \) can be written using the point-slope form:
\[ y - y_1 = m(x - x_1) \]
where \( m = \frac{dy}{dx} \) is the slope of the curve at point \( P \). Therefore, the equation becomes:
\[ y - y = \frac{dy}{dx}(x - x) \]

3. **Find the x-intercept of the tangent**: To find where the tangent intersects the x-axis, set \( y = 0 \):
\[ 0 - y = \frac{dy}{dx}(x - x) \]
Rearranging gives:
\[ -y = \frac{dy}{dx}(x - x) \]
This implies that:
\[ x_{\text{intercept}} = x - \frac{y}{\frac{dy}{dx}} \]

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:
\[ x = y + \frac{y}{\frac{dy}{dx}} \]
Multiplying through by \( \frac{dy}{dx} \):
\[ x \frac{dy}{dx} = y \frac{dy}{dx} + y \]

6. **Rearranging terms**: This can be rearranged to:
\[ x \frac{dy}{dx} - y \frac{dy}{dx} = y \]
Factoring out \( \frac{dy}{dx} \):
\[ (x - y) \frac{dy}{dx} = y \]

7. **Separating variables**: Now, we can separate the variables:
\[ \frac{dy}{y} = \frac{dx}{x - y} \]

8. **Integrating both sides**: Integrate both sides:
\[ \int \frac{dy}{y} = \int \frac{dx}{x - y} \]
This leads to:
\[ \ln |y| = \ln |x - y| + C \]

9. **Exponentiating both sides**: Exponentiating gives:
\[ y = k(x - y) \quad \text{where } k = e^C \]

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

11. **Final form**: This can be rewritten as:
\[ c = y \cdot e^{\frac{x}{y}} \quad \text{where } c \text{ is an arbitrary constant.} \]

Thus, the equation of the curve is:
\[ c = y \cdot e^{\frac{x}{y}} \]