Three normals to the parabola `y^2= x` are drawn through a point `(C, O)` then C=

Options

A. `c=1//4`
B. `c=1//2`
C. `cgt1//2`
D. none of these

Answer

To solve the problem of finding the value of \( C \) for which three normals to the parabola \( y^2 = x \) can be drawn through the point \( (C, 0) \), we can follow these steps:

### Step 1: Understand the equation of the normal
The equation of the normal to the parabola \( y^2 = 4ax \) at a point with slope \( m \) is given by:
\[ y = mx - 2am - am^2 \]
For the parabola \( y^2 = x \), we have \( 4a = 1 \), so \( a = \frac{1}{4} \).

### Step 2: Substitute \( a \) into the normal equation
Substituting \( a = \frac{1}{4} \) into the normal equation, we get:
\[ y = mx - 2 \cdot \frac{1}{4} m - \frac{1}{4} m^2 \]
This simplifies to:
\[ y = mx - \frac{1}{2} m - \frac{1}{4} m^2 \]

### Step 3: Set the point through which the normal passes
Since the normal passes through the point \( (C, 0) \), we substitute \( y = 0 \) and \( x = C \):
\[ 0 = mC - \frac{1}{2} m - \frac{1}{4} m^2 \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ mC = \frac{1}{2} m + \frac{1}{4} m^2 \]
Factoring out \( m \) (assuming \( m \neq 0 \)):
\[ 0 = m \left(C - \frac{1}{2} - \frac{1}{4} m\right) \]

### Step 5: Solve for \( m \)
This gives us:
\[ C - \frac{1}{2} - \frac{1}{4} m = 0 \]
Rearranging this, we find:
\[ \frac{1}{4} m = C - \frac{1}{2} \implies m = 4(C - \frac{1}{2}) \]

### Step 6: Condition for real values of \( m \)
For \( m \) to be real, we need:
\[ C - \frac{1}{2} \geq 0 \implies C \geq \frac{1}{2} \]

### Step 7: Finding the condition for three normals
To have three normals, we need \( 4C - 2 \) to be greater than or equal to zero:
\[ 4C - 2 > 0 \implies C > \frac{1}{2} \]

### Conclusion
Thus, the value of \( C \) must be greater than \( \frac{1}{2} \) for three normals to exist through the point \( (C, 0) \).

### Final Answer
The value of \( C \) is \( C > \frac{1}{2} \).

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