The Curve possessing the property that the intercept made by the tangent at any point of the curve on they-axis is equal to square of the abscissa of the point of tangency, is given by

Options

A. `y^(2)=x+c`
B. `y=2x^(2)+cx`
C. `Y=-x^(2)+cx`
D. None of these

Answer

To solve the problem, we need to find the equation of a curve such that the y-intercept of the tangent at any point on the curve equals the square of the abscissa of that point.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- Let the curve be represented by \( y = f(x) \).
- At a point \( P(x, y) \) on the curve, the tangent intersects the y-axis at point \( Q(0, x^2) \).
- This means that the y-intercept of the tangent line at point \( P \) is \( x^2 \).

2. **Finding the Slope of the Tangent**:
- The slope of the tangent line at point \( P \) is given by \( \frac{dy}{dx} \).
- The slope of the line segment \( PQ \) can be expressed as:
\[ \text{slope of } PQ = \frac{y - x^2}{x - 0} = \frac{f(x) - x^2}{x} \]

3. **Setting Up the Equation**:
- Since the slope of the tangent equals the slope of \( PQ \), we have:
\[ \frac{dy}{dx} = \frac{f(x) - x^2}{x} \]
- Rearranging gives us:
\[ \frac{dy}{dx} = \frac{f(x)}{x} - x \]

4. **Rearranging to Form a Differential Equation**:
- This can be rewritten as:
\[ \frac{dy}{dx} + \frac{1}{x} f(x) = -x \]
- This is a linear first-order differential equation in the standard form \( \frac{dy}{dx} + P(x)y = Q(x) \) where \( P(x) = \frac{1}{x} \) and \( Q(x) = -x \).

5. **Finding the Integrating Factor**:
- The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{\int P(x) \, dx} = e^{\int \frac{1}{x} \, dx} = e^{\ln |x|} = |x| \]
- We can take \( \mu(x) = \frac{1}{x} \) since we are considering positive values of \( x \).

6. **Multiplying Through by the Integrating Factor**:
- Multiply the entire differential equation by \( \frac{1}{x} \):
\[ \frac{1}{x} \frac{dy}{dx} + \frac{1}{x^2} f(x) = -1 \]

7. **Integrating Both Sides**:
- The left-hand side can be expressed as the derivative of a product:
\[ \frac{d}{dx} \left( \frac{f(x)}{x} \right) = -1 \]
- Integrating both sides gives:
\[ \frac{f(x)}{x} = -x + C \]
- Thus, we have:
\[ f(x) = -x^2 + Cx \]

8. **Final Equation of the Curve**:
- Therefore, the equation of the curve is:
\[ y = -x^2 + Cx \]

### Conclusion:
The required equation of the curve is \( y = -x^2 + Cx \).