A curve C passes through origin and has the property that at each point `(x, y)` on it the normal line at that point passes through `(1, 0)`. The equation of a common tangent to the curve C and the parabola `y^2 = 4x` is

Options

A. x=0
B. y=o
C. y=x+1
D. x+y+1=0

Answer

To solve the problem step by step, we will first derive the equation of the curve C that passes through the origin and has the property that the normal at each point (x, y) passes through the point (1, 0). Then we will find the common tangent to this curve and the parabola \(y^2 = 4x\).

### Step 1: Deriving the Equation of the Curve C

1. **Understanding the Slope of the Normal**:
- The slope of the tangent line at any point \((x, y)\) on the curve is given by \(\frac{dy}{dx}\).
- The slope of the normal line is the negative reciprocal of the slope of the tangent line, which is \(-\frac{1}{\frac{dy}{dx}}\).

2. **Setting Up the Equation of the Normal**:
- The normal line at point \((x, y)\) can be expressed using the point-slope form of a line:
\[ y - y_1 = m(x - x_1) \]
- Here, \(m = -\frac{1}{\frac{dy}{dx}}\), so the equation of the normal becomes:
\[ y - y = -\frac{1}{\frac{dy}{dx}}(x - x) \]

3. **Using the Condition that the Normal Passes through (1, 0)**:
- Since the normal passes through the point (1, 0), we substitute \(x = 1\) and \(y = 0\) into the normal equation:
\[ 0 - y = -\frac{1}{\frac{dy}{dx}}(1 - x) \]
- Rearranging gives:
\[ y \frac{dy}{dx} = 1 - x \]

4. **Separating Variables and Integrating**:
- Rearranging the equation:
\[ y \, dy = (1 - x) \, dx \]
- Integrating both sides:
\[ \int y \, dy = \int (1 - x) \, dx \]
- This results in:
\[ \frac{y^2}{2} = x - \frac{x^2}{2} + C \]

5. **Finding the Constant of Integration**:
- Since the curve passes through the origin (0, 0), we substitute \(x = 0\) and \(y = 0\):
\[ 0 = 0 - 0 + C \implies C = 0 \]
- Thus, the equation simplifies to:
\[ \frac{y^2}{2} = x - \frac{x^2}{2} \]
- Rearranging gives:
\[ x^2 + y^2 - 2x = 0 \]

### Step 2: Finding the Common Tangent to the Curve C and the Parabola

1. **Equation of the Parabola**:
- The equation of the parabola is \(y^2 = 4x\).

2. **Equation of the Tangent to the Parabola**:
- The equation of the tangent to the parabola \(y^2 = 4ax\) (where \(a = 1\)) can be expressed as:
\[ y = mx + \frac{1}{m} \]

3. **Finding the Condition for Tangency with the Circle**:
- The circle has the equation \(x^2 + y^2 - 2x = 0\).
- The distance from the center of the circle (1, 0) to the tangent line must equal the radius (1):
\[ \text{Distance} = \frac{|m(1) + 0 + \frac{1}{m}|}{\sqrt{m^2 + 1}} = 1 \]

4. **Setting Up the Equation**:
- This leads to the equation:
\[ |m + \frac{1}{m}| = \sqrt{m^2 + 1} \]

5. **Solving for m**:
- Squaring both sides and simplifying:
\[ (m + \frac{1}{m})^2 = m^2 + 1 \]
- This gives:
\[ m^2 + 2 + \frac{1}{m^2} = m^2 + 1 \implies 2 + \frac{1}{m^2} = 1 \implies \frac{1}{m^2} = -1 \]
- Since this has no real solutions, we conclude that the slope \(m\) must be infinite, indicating a vertical line.

6. **Final Equation of the Common Tangent**:
- The equation of the vertical line is:
\[ x = 0 \]

### Final Answer
The equation of the common tangent to the curve C and the parabola \(y^2 = 4x\) is:
\[ \boxed{x = 0} \]