The x-intercept of the tangent to a curve is equal to the ordinate of the point of contact. The equation of the curve through the point (1,1) is

Options

A. `ye^(x//y)=e`
B. `xe^(x//y)=e`
C. `xe^(y//x)=e`
D. `ye^(y//x)=e`

Explanation

To solve the problem, we need to find the equation of the curve given that the x-intercept of the tangent to the curve is equal to the ordinate (y-coordinate) of the point of contact. The curve passes through the point (1, 1).

### Step-by-step Solution:

1. **Understanding the Problem:**
We need to find a curve \( y = f(x) \) such that the x-intercept of the tangent line at any point \( (x_0, y_0) \) on the curve is equal to \( y_0 \).

2. **Equation of the Tangent Line:**
The equation of the tangent to the curve at point \( (x_0, y_0) \) can be expressed as:
\[ y - y_0 = f'(x_0)(x - x_0) \]
Rearranging gives:
\[ y = f'(x_0)(x - x_0) + y_0 \]

3. **Finding the x-intercept:**
To find the x-intercept, set \( y = 0 \):
\[ 0 = f'(x_0)(x - x_0) + y_0 \]
Solving for \( x \):
\[ f'(x_0)(x - x_0) = -y_0 \implies x - x_0 = -\frac{y_0}{f'(x_0)} \implies x = x_0 - \frac{y_0}{f'(x_0)} \]

4. **Setting the x-intercept equal to the ordinate:**
According to the problem, the x-intercept is equal to the ordinate \( y_0 \):
\[ x_0 - \frac{y_0}{f'(x_0)} = y_0 \]
Rearranging gives:
\[ x_0 = y_0 + \frac{y_0}{f'(x_0)} \implies x_0 = y_0 \left(1 + \frac{1}{f'(x_0)}\right) \]

5. **Differentiating and Rearranging:**
Rearranging leads to:
\[ f'(x_0) = \frac{y_0}{x_0 - y_0} \]
This implies:
\[ \frac{dy}{dx} = \frac{y}{x - y} \]

6. **Separating Variables:**
We can separate the variables:
\[ \frac{dy}{y} = \frac{dx}{x - y} \]

7. **Integrating Both Sides:**
Integrating both sides:
\[ \int \frac{dy}{y} = \int \frac{dx}{x - y} \]
The left side integrates to \( \ln |y| \) and the right side integrates to \( \ln |x - y| + C \):
\[ \ln |y| = \ln |x - y| + C \]

8. **Exponentiating:**
Exponentiating both sides gives:
\[ |y| = e^C |x - y| \]
Let \( e^C = k \), then:
\[ y = k(x - y) \]
Rearranging gives:
\[ y + ky = kx \implies y(1 + k) = kx \implies y = \frac{kx}{1 + k} \]

9. **Finding the Constant k:**
Since the curve passes through the point \( (1, 1) \):
\[ 1 = \frac{k \cdot 1}{1 + k} \implies 1 + k = k \implies k = 1 \]

10. **Final Equation of the Curve:**
Substituting \( k = 1 \) back into the equation:
\[ y = \frac{x}{2} \]

### Conclusion:
The equation of the curve is:
\[ y = \frac{x}{2} \]